Chapter 12 Limits And Derivatives

Solution:

For polynomial functions, the limit as $x$ approaches a specific value can be found by direct substitution of the value into the function, because polynomials are continuous everywhere.

(i) Find the limit: $\lim\limits_{x \to 1} [x^3 - x^2 + 1]$

The function is $f(x) = x^3 - x^2 + 1$, which is a polynomial.

To find the limit as $x$ approaches 1, substitute $x=1$ into the expression:

$\lim\limits_{x \to 1} [x^3 - x^2 + 1] = (1)^3 - (1)^2 + 1$

$= 1 - 1 + 1$

$= 1$

The limit is $\mathbf{1}$.

(ii) Find the limit: $\lim\limits_{x \to 3} [x(x + 1)]$

The function is $f(x) = x(x + 1) = x^2 + x$, which is a polynomial.

To find the limit as $x$ approaches 3, substitute $x=3$ into the expression:

$\lim\limits_{x \to 3} [x(x + 1)] = (3)(3 + 1)$

$= 3 \times 4$

$= 12$

The limit is $\mathbf{12}$.

(iii) Find the limit: $\lim\limits_{x \to -1}$ [1 + x + x² + … + x¹⁰]

The function is $f(x) = 1 + x + x^2 + \dots + x^{10}$, which is a polynomial (a finite sum of powers of $x$).

To find the limit as $x$ approaches -1, substitute $x=-1$ into the expression:

$\lim\limits_{x \to -1} [1 + x + x^2 + … + x^{10}] = 1 + (-1) + (-1)^2 + (-1)^3 + \dots + (-1)^{10}$

Evaluate each term:

$(-1)^0 = 1$

$(-1)^1 = -1$

$(-1)^2 = 1$

$(-1)^3 = -1$

...and so on. The value of $(-1)^n$ is 1 if $n$ is even and -1 if $n$ is odd.

The expression is a sum of 11 terms (from $x^0$ to $x^{10}$). The powers are 0, 1, 2, ..., 10.

There are 6 even powers (0, 2, 4, 6, 8, 10) and 5 odd powers (1, 3, 5, 7, 9).

The sum is: $(1) + (-1) + (1) + (-1) + (1) + (-1) + (1) + (-1) + (1) + (-1) + (1)$

We can group the terms:

$(1 + (-1)) + (1 + (-1)) + (1 + (-1)) + (1 + (-1)) + (1 + (-1)) + 1$

$= 0 + 0 + 0 + 0 + 0 + 1$

$= 1$

Alternatively, sum the terms with even powers and terms with odd powers:

Sum = $\underbrace{1 + 1 + 1 + 1 + 1 + 1}_{\text{6 terms}}$ + $\underbrace{(-1) + (-1) + (-1) + (-1) + (-1)}_{\text{5 terms}}$

Sum = $(6 \times 1) + (5 \times -1) = 6 - 5 = 1$

The limit is $\mathbf{1}$.

Solution:

We will evaluate each limit. For rational functions $\frac{f(x)}{g(x)}$, if $g(a) \neq 0$, then $\lim\limits_{x \to a} \frac{f(x)}{g(x)} = \frac{f(a)}{g(a)}$. If $g(a) = 0$ and $f(a) = 0$, we have an indeterminate form ($0/0$) and need to simplify the expression by factoring.

(i) Find the limit: $\lim\limits_{x \to 1} \left[ \frac{x^2 \;+ \; 1}{x \;+\; 100} \right]$

The function is $f(x) = \frac{x^2 + 1}{x + 100}$. We want to find the limit as $x \to 1$.

The denominator at $x=1$ is $1 + 100 = 101$, which is not zero.

We can find the limit by direct substitution:

$\lim\limits_{x \to 1} \left[ \frac{x^2 \;+ \; 1}{x \;+\; 100} \right] = \frac{(1)^2 + 1}{1 + 100}$

$= \frac{1 + 1}{101}$

$= \frac{2}{101}$

The limit is $\mathbf{\frac{2}{101}}$.

(ii) Find the limit: $\lim\limits_{x \to 2} \left[ \frac{x^{3}\;-\;4x^{2}\;+\;4x}{x^{2}\;-\;4} \right]$

The function is $f(x) = \frac{x^3 - 4x^2 + 4x}{x^2 - 4}$. We want to find the limit as $x \to 2$.

Check the denominator at $x=2$: $2^2 - 4 = 4 - 4 = 0$.

Check the numerator at $x=2$: $2^3 - 4(2)^2 + 4(2) = 8 - 4(4) + 8 = 8 - 16 + 8 = 0$.

Since we have the indeterminate form $\frac{0}{0}$, we need to factor the numerator and denominator.

Numerator: $x^3 - 4x^2 + 4x = x(x^2 - 4x + 4) = x(x - 2)^2 = x(x - 2)(x - 2)$

Denominator: $x^2 - 4 = (x - 2)(x + 2)$ (difference of squares)

Now rewrite the limit expression:

$\lim\limits_{x \to 2} \left[ \frac{x(x - 2)(x - 2)}{(x - 2)(x + 2)} \right]$

For $x \neq 2$, we can cancel the common factor $(x - 2)$: $\lim\limits_{x \to 2} \left[ \frac{x(x - 2)}{x + 2} \right]$

Now substitute $x=2$ into the simplified expression:

$\frac{2(2 - 2)}{2 + 2} = \frac{2(0)}{4} = \frac{0}{4} = 0$

The limit is $\mathbf{0}$.

(iii) Find the limit: $\lim\limits_{x \to 2} \left[ \frac{x^{2}\;-\;4}{x^{3}\;-\;4x^{2}\;+\;4x} \right]$

This is the reciprocal of the function in part (ii). The function is $g(x) = \frac{x^2 - 4}{x^3 - 4x^2 + 4x}$. We want to find the limit as $x \to 2$.

Using the factorizations from part (ii):

Numerator: $x^2 - 4 = (x - 2)(x + 2)$

Denominator: $x^3 - 4x^2 + 4x = x(x - 2)(x - 2)$

Rewrite the limit expression:

$\lim\limits_{x \to 2} \left[ \frac{(x - 2)(x + 2)}{x(x - 2)(x - 2)} \right]$

For $x \neq 2$, we can cancel the common factor $(x - 2)$: $\lim\limits_{x \to 2} \left[ \frac{x + 2}{x(x - 2)} \right]$

Now substitute $x=2$ into the simplified expression:

The numerator approaches $2 + 2 = 4$.

The denominator approaches $2(2 - 2) = 2(0) = 0$.

Since the numerator approaches a non-zero number (4) and the denominator approaches 0, the limit does not exist. We can examine the one-sided limits to determine the behavior.

Consider the denominator $x(x-2)$. As $x \to 2$, $x$ is positive (near 2). The sign of the denominator is determined by $(x-2)$.

As $x \to 2^+$ (x approaches 2 from the right), $x > 2$, so $x-2 > 0$. Denominator $x(x-2) \to 0^+$.

$\lim\limits_{x \to 2^+} \left[ \frac{x + 2}{x(x - 2)} \right] = \frac{4}{0^+} = +\infty$

As $x \to 2^-$ (x approaches 2 from the left), $x < 2$, so $x-2 < 0$. Denominator $x(x-2) \to 0^-$.

$\lim\limits_{x \to 2^-} \left[ \frac{x + 2}{x(x - 2)} \right] = \frac{4}{0^-} = -\infty$

Since the left-hand limit and the right-hand limit are not equal, the overall limit does not exist.

The limit is Does Not Exist ($\infty$ or $-\infty$).

(More specifically, the limit tends towards $\infty$ or $-\infty$ depending on the direction of approach).

(iv) Find the limit: $\lim\limits_{x \to 2} \left[ \frac{x^{3}\;-\;2x^{2}}{x^{2}\;-\;5x\;+\;6} \right]$

The function is $h(x) = \frac{x^3 - 2x^2}{x^2 - 5x + 6}$. We want to find the limit as $x \to 2$.

Check the denominator at $x=2$: $2^2 - 5(2) + 6 = 4 - 10 + 6 = 0$.

Check the numerator at $x=2$: $2^3 - 2(2)^2 = 8 - 2(4) = 8 - 8 = 0$.

Since we have the indeterminate form $\frac{0}{0}$, we need to factor the numerator and denominator.

Numerator: $x^3 - 2x^2 = x^2(x - 2)$

Denominator: $x^2 - 5x + 6$. We look for two numbers that multiply to 6 and add up to -5. These numbers are -2 and -3.

$x^2 - 5x + 6 = (x - 2)(x - 3)$

Now rewrite the limit expression:

$\lim\limits_{x \to 2} \left[ \frac{x^2(x - 2)}{(x - 2)(x - 3)} \right]$

For $x \neq 2$, we can cancel the common factor $(x - 2)$: $\lim\limits_{x \to 2} \left[ \frac{x^2}{x - 3} \right]$

Now substitute $x=2$ into the simplified expression:

$\frac{(2)^2}{2 - 3} = \frac{4}{-1} = -4$

The limit is $\mathbf{-4}$.

(v) Find the limit: $\lim\limits_{x \to 1} \left[ \frac{x\;-\;2}{x^{2}\;-\;x}-\frac{1}{x^3\;-\;3x^2\;+\;2x} \right]$

We have the limit of a difference of two rational functions as $x \to 1$.

Let's consider the denominators of each term at $x=1$:

First denominator: $x^2 - x = 1^2 - 1 = 1 - 1 = 0$.

Second denominator: $x^3 - 3x^2 + 2x = 1^3 - 3(1)^2 + 2(1) = 1 - 3 + 2 = 0$.

Since both denominators are 0 at $x=1$, we need to combine the fractions and simplify the expression before evaluating the limit.

Factor the denominators:

$x^2 - x = x(x - 1)$

$x^3 - 3x^2 + 2x = x(x^2 - 3x + 2)$

Factor the quadratic term in the second denominator. We look for two numbers that multiply to 2 and add up to -3. These are -1 and -2.

$x^2 - 3x + 2 = (x - 1)(x - 2)$

So, the second denominator is $x(x - 1)(x - 2)$.

The expression is:

$\frac{x - 2}{x(x - 1)} - \frac{1}{x(x - 1)(x - 2)}$

The least common multiple of the denominators is $x(x - 1)(x - 2)$.

Combine the fractions:

$\frac{(x - 2)(x - 2)}{x(x - 1)(x - 2)} - \frac{1}{x(x - 1)(x - 2)}$

$= \frac{(x - 2)^2 - 1}{x(x - 1)(x - 2)}$

Expand the numerator $(x - 2)^2 - 1 = (x^2 - 4x + 4) - 1 = x^2 - 4x + 3$.

Factor the numerator $x^2 - 4x + 3$. We look for two numbers that multiply to 3 and add up to -4. These are -1 and -3.

$x^2 - 4x + 3 = (x - 1)(x - 3)$

So, the expression becomes:

$\frac{(x - 1)(x - 3)}{x(x - 1)(x - 2)}$

For $x \neq 1$ (which is true as we take the limit as $x \to 1$), we can cancel the common factor $(x - 1)$:

$\frac{x - 3}{x(x - 2)}$

Now, evaluate the limit of this simplified expression as $x \to 1$ by direct substitution:

$\lim\limits_{x \to 1} \left[ \frac{x - 3}{x(x - 2)} \right] = \frac{1 - 3}{1(1 - 2)}$

$= \frac{-2}{1(-1)}$

$= \frac{-2}{-1}$

$= 2$

The limit is $\mathbf{2}$.

Solution:

(i) Evaluate: $\lim\limits_{x \to 1} \frac{x^{15}\;-\;1}{x^{10}\;-\;1}$

When we substitute $x=1$ into the expression, we get $\frac{1^{15} - 1}{1^{10} - 1} = \frac{1 - 1}{1 - 1} = \frac{0}{0}$, which is an indeterminate form.

We can use the standard limit formula: $\lim\limits_{x \to a} \frac{x^n - a^n}{x - a} = na^{n-1}$.

Divide both the numerator and the denominator by $(x - 1)$:

$\lim\limits_{x \to 1} \frac{\frac{x^{15}\;-\;1}{x-1}}{\frac{x^{10}\;-\;1}{x-1}}$

Now, apply the formula to the numerator and the denominator separately, with $a=1$.

For the numerator: $\lim\limits_{x \to 1} \frac{x^{15} - 1^{15}}{x - 1}$. Here $n=15$ and $a=1$. The limit is $15 \times 1^{15-1} = 15 \times 1^{14} = 15$.

For the denominator: $\lim\limits_{x \to 1} \frac{x^{10} - 1^{10}}{x - 1}$. Here $n=10$ and $a=1$. The limit is $10 \times 1^{10-1} = 10 \times 1^9 = 10$.

So, the limit of the ratio is the ratio of the limits (since both limits exist and the denominator limit is non-zero):

$\lim\limits_{x \to 1} \frac{x^{15}\;-\;1}{x^{10}\;-\;1} = \frac{\lim\limits_{x \to 1} \frac{x^{15}\;-\;1}{x-1}}{\lim\limits_{x \to 1} \frac{x^{10}\;-\;1}{x-1}} = \frac{15}{10}$

Simplify the fraction:

$\frac{\cancel{15}^{3}}{\cancel{10}_{2}} = \frac{3}{2}$

The limit is $\mathbf{\frac{3}{2}}$.

(ii) Evaluate: $\lim\limits_{x \to 0} \frac{\sqrt{1\;+\;x}\;-\;1}{x}$

When we substitute $x=0$ into the expression, we get $\frac{\sqrt{1 + 0} - 1}{0} = \frac{\sqrt{1} - 1}{0} = \frac{1 - 1}{0} = \frac{0}{0}$, which is an indeterminate form.

To evaluate this limit, we can rationalize the numerator by multiplying the numerator and denominator by the conjugate of the numerator, which is $\sqrt{1+x} + 1$.

$\lim\limits_{x \to 0} \frac{\sqrt{1\;+\;x}\;-\;1}{x} = \lim\limits_{x \to 0} \frac{(\sqrt{1\;+\;x}\;-\;1)(\sqrt{1\;+\;x}\;+\;1)}{x(\sqrt{1\;+\;x}\;+\;1)}$

Use the difference of squares formula $(a-b)(a+b) = a^2 - b^2$ in the numerator:

$= \lim\limits_{x \to 0} \frac{(\sqrt{1\;+\;x})^2 - 1^2}{x(\sqrt{1\;+\;x}\;+\;1)}$

$= \lim\limits_{x \to 0} \frac{(1\;+\;x) - 1}{x(\sqrt{1\;+\;x}\;+\;1)}$

$= \lim\limits_{x \to 0} \frac{x}{x(\sqrt{1\;+\;x}\;+\;1)}$

For $x \neq 0$ (as $x$ approaches 0 but is not equal to 0), we can cancel the common factor $x$ from the numerator and denominator:

$= \lim\limits_{x \to 0} \frac{1}{\sqrt{1\;+\;x}\;+\;1}$

Now, substitute $x=0$ into the simplified expression (the denominator is not zero at $x=0$):

$= \frac{1}{\sqrt{1\;+\;0}\;+\;1}$

$= \frac{1}{\sqrt{1}\;+\;1}$

$= \frac{1}{1\;+\;1} = \frac{1}{2}$

The limit is $\mathbf{\frac{1}{2}}$.

Solution:

We will evaluate the given limits using standard limit formulas for trigonometric functions.

The standard limit is $\lim\limits_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$ and $\lim\limits_{\theta \to 0} \frac{\tan \theta}{\theta} = 1$, provided $\theta$ is in radians.

(i) Evaluate: $\lim\limits_{x \to 0} \frac{\sin4x}{\sin2x}$

When $x=0$, the expression is $\frac{\sin(4 \times 0)}{\sin(2 \times 0)} = \frac{\sin 0}{\sin 0} = \frac{0}{0}$, which is an indeterminate form.

To evaluate this limit, we can rewrite the expression using the standard limit $\lim\limits_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$. We need to have the argument of the sine function in the denominator as well.

Multiply the numerator by $4x$ and divide by $4x$. Multiply the denominator by $2x$ and divide by $2x$.

$\lim\limits_{x \to 0} \frac{\sin4x}{\sin2x} = \lim\limits_{x \to 0} \frac{\frac{\sin4x}{4x} \times 4x}{\frac{\sin2x}{2x} \times 2x}$

$= \lim\limits_{x \to 0} \frac{\frac{\sin4x}{4x}}{\frac{\sin2x}{2x}} \times \frac{4x}{2x}$

For $x \neq 0$, we can cancel $x$ in the fraction $\frac{4x}{2x} = 2$.

$= \lim\limits_{x \to 0} \frac{\frac{\sin4x}{4x}}{\frac{\sin2x}{2x}} \times 2$

As $x \to 0$, we have $4x \to 0$ and $2x \to 0$. Let $\theta_1 = 4x$ and $\theta_2 = 2x$. Both $\theta_1$ and $\theta_2$ tend to 0 as $x$ tends to 0.

So, $\lim\limits_{x \to 0} \frac{\sin4x}{4x} = \lim\limits_{\theta_1 \to 0} \frac{\sin\theta_1}{\theta_1} = 1$.

And, $\lim\limits_{x \to 0} \frac{\sin2x}{2x} = \lim\limits_{\theta_2 \to 0} \frac{\sin\theta_2}{\theta_2} = 1$.

Therefore, the limit becomes:

$= \frac{\lim\limits_{x \to 0} \frac{\sin4x}{4x}}{\lim\limits_{x \to 0} \frac{\sin2x}{2x}} \times \lim\limits_{x \to 0} 2$

$= \frac{1}{1} \times 2$

$= 1 \times 2 = 2$

The limit is $\mathbf{2}$.

(ii) Evaluate: $\lim\limits_{x \to 0} \frac{\tan x}{x}$

When $x=0$, the expression is $\frac{\tan 0}{0} = \frac{0}{0}$, which is an indeterminate form.

This is a standard limit formula.

We know that $\lim\limits_{x \to 0} \frac{\tan x}{x} = 1$.

Alternatively, we can write $\tan x = \frac{\sin x}{\cos x}$.

$\lim\limits_{x \to 0} \frac{\tan x}{x} = \lim\limits_{x \to 0} \frac{\frac{\sin x}{\cos x}}{x}$

$= \lim\limits_{x \to 0} \frac{\sin x}{x \cos x}$

$= \lim\limits_{x \to 0} \left( \frac{\sin x}{x} \times \frac{1}{\cos x} \right)$

Using the properties of limits, the limit of a product is the product of the limits (if they exist):

$= \left( \lim\limits_{x \to 0} \frac{\sin x}{x} \right) \times \left( \lim\limits_{x \to 0} \frac{1}{\cos x} \right)$

We know $\lim\limits_{x \to 0} \frac{\sin x}{x} = 1$.

For the second part, substitute $x=0$: $\lim\limits_{x \to 0} \frac{1}{\cos x} = \frac{1}{\cos 0} = \frac{1}{1} = 1$.

So, the limit is:

$= 1 \times 1 = 1$

The limit is $\mathbf{1}$.

Given:

The limit expression $\lim\limits_{x \to 3} (x + 3)$.

To Evaluate:

The given limit.

Solution:

The function is $f(x) = x + 3$. This is a polynomial function.

For a polynomial function, the limit as $x$ approaches a real number $a$ is equal to the value of the function at $a$. That is, $\lim\limits_{x \to a} P(x) = P(a)$ for any polynomial $P(x)$.

In this case, the function is $P(x) = x + 3$ and $a = 3$.

We can find the limit by direct substitution:

The value of the limit is $\mathbf{6}$.

Given:

The limit expression $\lim\limits_{x \to \pi}\left( x - \frac{22}{7} \right)$.

To Evaluate:

The given limit.

Solution:

The function is $f(x) = x - \frac{22}{7}$. This is a polynomial function (specifically, a linear function).

For any polynomial function, the limit as the variable approaches a real number is equal to the value of the function at that number. This is due to the continuity of polynomial functions.

In this case, the function is $f(x) = x - \frac{22}{7}$ and we are evaluating the limit as $x$ approaches $\pi$.

We can find the limit by direct substitution:

The value of the limit is $\mathbf{\pi - \frac{22}{7}}$.

Given:

The limit expression $\lim\limits_{r \to 1} \pi r^2$.

To Evaluate:

The given limit.

Solution:

The function is $f(r) = \pi r^2$. This is a polynomial function of the variable $r$ (where $\pi$ is a constant coefficient).

For any polynomial function $P(x)$, the limit as the variable $x$ approaches a real number $a$ is equal to the value of the function at $a$. That is, $\lim\limits_{x \to a} P(x) = P(a)$.

In this case, the function is $f(r) = \pi r^2$ and we are evaluating the limit as $r$ approaches 1.

We can find the limit by direct substitution of $r=1$ into the expression:

The value of the limit is $\mathbf{\pi}$.

Given:

The limit expression $\lim\limits_{x \to 4}\frac{4x \;+\; 3}{x \;-\; 2}$.

To Evaluate:

The given limit.

Solution:

The function is a rational function $f(x) = \frac{4x + 3}{x - 2}$. We need to find the limit as $x$ approaches 4.

First, check the value of the denominator when $x = 4$:

Denominator at $x=4$ is $4 - 2 = 2$.

Since the denominator is non-zero at $x = 4$, we can evaluate the limit by direct substitution of $x = 4$ into the function.

The value of the limit is $\mathbf{\frac{19}{2}}$.

Given:

The limit expression $\lim\limits_{x \to -1}\frac{x^{10}\;+\;x^{5}\;+\;1}{x\;-\;1}$.

To Evaluate:

The given limit.

Solution:

The function is a rational function $f(x) = \frac{x^{10} + x^5 + 1}{x - 1}$. We need to find the limit as $x$ approaches -1.

First, check the value of the denominator when $x = -1$:

Denominator at $x=-1$ is $-1 - 1 = -2$.

Since the denominator is non-zero at $x = -1$, we can evaluate the limit by direct substitution of $x = -1$ into the function.

Evaluate the terms in the numerator:

$(-1)^{10} = 1$ (since 10 is an even power)

$(-1)^{5} = -1$ (since 5 is an odd power)

So, the numerator at $x=-1$ is $1 + (-1) + 1 = 1 - 1 + 1 = 1$.

The denominator at $x=-1$ is $-2$.

Substituting these values:

The value of the limit is $\mathbf{-\frac{1}{2}}$.

Given:

The limit expression $\lim\limits_{x \to 0}\frac{(x \;+\; 1)^{5} \;-\; 1}{x}$.

To Evaluate:

The given limit.

Solution:

When we substitute $x = 0$ into the expression, we get $\frac{(0 + 1)^5 - 1}{0} = \frac{1^5 - 1}{0} = \frac{1 - 1}{0} = \frac{0}{0}$, which is an indeterminate form.

We can evaluate this limit using the standard limit formula: $\lim\limits_{y \to a} \frac{y^n - a^n}{y - a} = na^{n-1}$.

Let $y = x + 1$. As $x \to 0$, $y \to 0 + 1$, so $y \to 1$.

The expression becomes $\frac{y^5 - 1}{x}$. We need the denominator in terms of $y$. Since $y = x + 1$, we have $x = y - 1$.

Substituting these into the limit expression:

$\lim\limits_{y \to 1}\frac{y^{5} \;-\; 1}{y \;-\; 1}$

This matches the standard limit formula with $y$ instead of $x$, $a = 1$, and $n = 5$.

Using the formula, the limit is $na^{n-1} = 5 \times 1^{5-1} = 5 \times 1^4 = 5 \times 1 = 5$.

The value of the limit is $\mathbf{5}$.

Alternate Solution: Using Binomial Expansion

We can expand $(x + 1)^5$ using the binomial theorem:

$(x + 1)^5 = \binom{5}{0}x^0 1^5 + \binom{5}{1}x^1 1^4 + \binom{5}{2}x^2 1^3 + \binom{5}{3}x^3 1^2 + \binom{5}{4}x^4 1^1 + \binom{5}{5}x^5 1^0$

$= 1 \times 1 \times 1 + 5 \times x \times 1 + 10 \times x^2 \times 1 + 10 \times x^3 \times 1 + 5 \times x^4 \times 1 + 1 \times x^5 \times 1$

$= 1 + 5x + 10x^2 + 10x^3 + 5x^4 + x^5$

Now substitute this into the numerator of the limit expression:

$(x + 1)^5 - 1 = (1 + 5x + 10x^2 + 10x^3 + 5x^4 + x^5) - 1$

$= 5x + 10x^2 + 10x^3 + 5x^4 + x^5$

The limit expression becomes:

$\lim\limits_{x \to 0}\frac{5x + 10x^2 + 10x^3 + 5x^4 + x^5}{x}$

For $x \neq 0$, we can divide each term in the numerator by $x$:

$= \lim\limits_{x \to 0} \left( \frac{5x}{x} + \frac{10x^2}{x} + \frac{10x^3}{x} + \frac{5x^4}{x} + \frac{x^5}{x} \right)$

$= \lim\limits_{x \to 0} (5 + 10x + 10x^2 + 5x^3 + x^4)$

Now, this is a polynomial in $x$. We can evaluate the limit by direct substitution of $x=0$:

$= 5 + 10(0) + 10(0)^2 + 5(0)^3 + (0)^4$

$= 5 + 0 + 0 + 0 + 0$

$= 5$

The value of the limit is $\mathbf{5}$.

Given:

The limit expression $\lim\limits_{x \to 2}\frac{3x^2 \;-\; x \;-\; 10}{x^2 \;-\; 4}$.

To Evaluate:

The given limit.

Solution:

The function is $f(x) = \frac{3x^2 - x - 10}{x^2 - 4}$. We want to find the limit as $x$ approaches 2.

First, check the value of the numerator and denominator when $x = 2$:

Numerator at $x=2$: $3(2)^2 - (2) - 10 = 3(4) - 2 - 10 = 12 - 2 - 10 = 0$.

Denominator at $x=2$: $(2)^2 - 4 = 4 - 4 = 0$.

Since we have the indeterminate form $\frac{0}{0}$, we need to factor the numerator and the denominator to cancel the common factor $(x - 2)$.

Factor the denominator using the difference of squares formula $a^2 - b^2 = (a - b)(a + b)$:

$x^2 - 4 = x^2 - 2^2 = (x - 2)(x + 2)$

Factor the numerator $3x^2 - x - 10$. Since the value is 0 at $x=2$, $(x-2)$ is a factor. We can use polynomial division or recognize that we need terms that multiply to $3x^2$ and $-10$ and combine to $-x$. By inspection or trial and error, or by noting that $3x^2 - x - 10 = (x-2)(Ax+B)$, we find the other factor.

Let's perform polynomial division or factoring by grouping. We look for two numbers that multiply to $3 \times (-10) = -30$ and add up to -1. These numbers are -6 and 5.

$3x^2 - x - 10 = 3x^2 - 6x + 5x - 10$

$= 3x(x - 2) + 5(x - 2)$

$= (x - 2)(3x + 5)$

Now substitute the factored forms into the limit expression:

For $x \neq 2$, the factor $(x - 2)$ is non-zero, so we can cancel it from the numerator and the denominator:

Now, the denominator $x+2$ is non-zero when $x=2$ (since $2+2=4$). We can evaluate the limit by direct substitution of $x=2$ into the simplified expression:

The value of the limit is $\mathbf{\frac{11}{4}}$.

Given:

The limit expression $\lim\limits_{x \to 3}\frac{x^4 \;-\; 81}{2x^2 \;-\; 5x \;-\; 3}$.

To Evaluate:

The given limit.

Solution:

The function is $f(x) = \frac{x^4 - 81}{2x^2 - 5x - 3}$. We want to find the limit as $x$ approaches 3.

First, check the value of the numerator and denominator when $x = 3$:

Numerator at $x=3$: $3^4 - 81 = 81 - 81 = 0$.

Denominator at $x=3$: $2(3)^2 - 5(3) - 3 = 2(9) - 15 - 3 = 18 - 15 - 3 = 0$.

Since we have the indeterminate form $\frac{0}{0}$, we need to factor the numerator and the denominator to cancel the common factor $(x - 3)$.

Factor the numerator using the difference of squares formula $a^2 - b^2 = (a - b)(a + b)$ twice:

$x^4 - 81 = (x^2)^2 - 9^2 = (x^2 - 9)(x^2 + 9)$

$= (x^2 - 3^2)(x^2 + 9) = (x - 3)(x + 3)(x^2 + 9)$

Factor the denominator $2x^2 - 5x - 3$. Since the value is 0 at $x=3$, $(x-3)$ is a factor. We look for two numbers that multiply to $2 \times (-3) = -6$ and add up to -5. These numbers are -6 and 1.

$2x^2 - 5x - 3 = 2x^2 - 6x + x - 3$

$= 2x(x - 3) + 1(x - 3)$

$= (x - 3)(2x + 1)$

Now substitute the factored forms into the limit expression:

For $x \neq 3$, the factor $(x - 3)$ is non-zero, so we can cancel it from the numerator and the denominator:

Now, the denominator $2x+1$ is non-zero when $x=3$ (since $2(3)+1=7$). We can evaluate the limit by direct substitution of $x=3$ into the simplified expression:

The value of the limit is $\mathbf{\frac{108}{7}}$.

Given:

The limit expression $\lim\limits_{x \to 0} \frac{ax \;+\; b}{cx \;+\; 1}$.

To Evaluate:

The given limit.

Solution:

The function is a rational function $f(x) = \frac{ax + b}{cx + 1}$. We want to find the limit as $x$ approaches 0.

First, check the value of the denominator when $x = 0$:

Denominator at $x=0$ is $c(0) + 1 = 0 + 1 = 1$.

Since the denominator is non-zero at $x = 0$, we can evaluate the limit by direct substitution of $x = 0$ into the function.

The value of the limit is $\mathbf{b}$, provided that $cx+1 \neq 0$ in the neighbourhood of $x=0$. At $x=0$, the denominator is 1, so this condition is met.

Given:

The limit expression $\lim\limits_{z \to 1} \frac{z^{\frac{1}{3}} \;-\; 1}{z^{\frac{1}{6}} \;-\; 1}$.

To Evaluate:

The given limit.

Solution:

When we substitute $z = 1$ into the expression, we get $\frac{1^{1/3} - 1}{1^{1/6} - 1} = \frac{1 - 1}{1 - 1} = \frac{0}{0}$, which is an indeterminate form.

We can evaluate this limit using the standard limit formula: $\lim\limits_{x \to a} \frac{x^n - a^n}{x - a} = na^{n-1}$.

We can rewrite the expression by dividing both the numerator and the denominator by $(z - 1)$: $\lim\limits_{z \to 1} \frac{\frac{z^{\frac{1}{3}} \;-\; 1}{z-1}}{\frac{z^{\frac{1}{6}} \;-\; 1}{z-1}}$

Apply the standard limit formula $\lim\limits_{z \to 1} \frac{z^n - 1^n}{z - 1} = n(1)^{n-1} = n$ to the numerator and denominator.

For the numerator: $\lim\limits_{z \to 1} \frac{z^{\frac{1}{3}} - 1^{\frac{1}{3}}}{z - 1}$. Here $n = \frac{1}{3}$ and $a = 1$. The limit is $\frac{1}{3} \times 1^{\frac{1}{3}-1} = \frac{1}{3} \times 1^{-\frac{2}{3}} = \frac{1}{3} \times 1 = \frac{1}{3}$.

For the denominator: $\lim\limits_{z \to 1} \frac{z^{\frac{1}{6}} - 1^{\frac{1}{6}}}{z - 1}$. Here $n = \frac{1}{6}$ and $a = 1$. The limit is $\frac{1}{6} \times 1^{\frac{1}{6}-1} = \frac{1}{6} \times 1^{-\frac{5}{6}} = \frac{1}{6} \times 1 = \frac{1}{6}$.

The limit of the ratio is the ratio of the limits:

The value of the limit is $\mathbf{2}$.

Alternate Solution: Using Substitution

Let $y = z^{1/6}$. As $z \to 1$, $y \to 1^{1/6} = 1$.

Also, $z^{1/3} = (z^{1/6})^2 = y^2$.

Substitute these into the limit expression:

$\lim\limits_{y \to 1} \frac{y^2 - 1}{y - 1}$

The numerator is a difference of squares: $y^2 - 1 = (y - 1)(y + 1)$.

So the limit becomes:

$\lim\limits_{y \to 1} \frac{(y - 1)(y + 1)}{y - 1}$

For $y \neq 1$ (as $y$ approaches 1 but is not equal to 1), we can cancel the common factor $(y - 1)$:

$= \lim\limits_{y \to 1} (y + 1)$

Now, evaluate the limit by direct substitution of $y = 1$ into the simplified expression:

$= 1 + 1 = 2$

The value of the limit is $\mathbf{2}$.

Given:

The limit expression $\lim\limits_{x \to 1} \frac{ax^2 \;+\; bx \;+\; c}{cx^2 \;+\; bx \;+\; a}$, with the condition $a + b + c \neq 0$.

To Evaluate:

The given limit.

Solution:

The function is a rational function $f(x) = \frac{ax^2 + bx + c}{cx^2 + bx + a}$. We want to find the limit as $x$ approaches 1.

First, check the value of the numerator and denominator when $x = 1$:

Numerator at $x=1$: $a(1)^2 + b(1) + c = a + b + c$.

Denominator at $x=1$: $c(1)^2 + b(1) + a = c + b + a = a + b + c$.

We are given that $a + b + c \neq 0$.

Since the denominator is non-zero at $x = 1$, we can evaluate the limit by direct substitution of $x = 1$ into the function.

Since $a + b + c \neq 0$, the fraction simplifies to 1.

The value of the limit is $\mathbf{1}$.

Given:

The limit expression $\lim\limits_{x \to -2} \frac{\frac{1}{x}+\frac{1}{2}}{x \;+\; 2}$.

To Evaluate:

The given limit.

Solution:

When we substitute $x = -2$ into the expression, the numerator becomes $\frac{1}{-2} + \frac{1}{2} = -\frac{1}{2} + \frac{1}{2} = 0$, and the denominator becomes $-2 + 2 = 0$.

Since we have the indeterminate form $\frac{0}{0}$, we need to simplify the expression.

First, simplify the numerator by finding a common denominator:

$\frac{1}{x} + \frac{1}{2} = \frac{1 \times 2}{x \times 2} + \frac{1 \times x}{2 \times x} = \frac{2}{2x} + \frac{x}{2x} = \frac{2 + x}{2x}$

Now rewrite the limit expression with the simplified numerator:

Rewrite the complex fraction as a division:

For $x \neq -2$, the factor $(x + 2)$ is non-zero, so we can cancel it from the numerator and the denominator:

Now, evaluate the limit by direct substitution of $x=-2$ into the simplified expression:

The value of the limit is $\mathbf{-\frac{1}{4}}$.

Given:

To Evaluate:

Evaluate the given limit.

Solution:

Let the given limit be $L$.

$L = \lim\limits_{x \to 0} \frac{\sin ax}{bx}$

When we substitute $x = 0$ into the expression, the numerator becomes $\sin(a \times 0) = \sin 0 = 0$ and the denominator becomes $b \times 0 = 0$. This gives the indeterminate form $\frac{0}{0}$.

We can evaluate this limit using the standard limit formula: $\lim\limits_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$.

To apply this formula to our expression, we need the term $ax$ in the denominator corresponding to $\sin ax$.

Assume $b \neq 0$. We can rewrite the expression by multiplying the numerator and the denominator by $a$ (assuming $a \neq 0$ initially):

$L = \lim\limits_{x \to 0} \frac{\sin ax}{bx} = \lim\limits_{x \to 0} \frac{\sin ax}{bx} \times \frac{a}{a}$

Rearrange the terms to group $\frac{\sin ax}{ax}$:

$L = \lim\limits_{x \to 0} \left( \frac{\sin ax}{ax} \times \frac{a}{b} \right)$

Using the property that the limit of a product is the product of the limits (if they exist), we can write:

$L = \left( \lim\limits_{x \to 0} \frac{\sin ax}{ax} \right) \times \left( \lim\limits_{x \to 0} \frac{a}{b} \right)$

For the first limit, let $\theta = ax$. As $x \to 0$, $ax \to a \times 0 = 0$, so $\theta \to 0$.

$\lim\limits_{x \to 0} \frac{\sin ax}{ax} = \lim\limits_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$ (This is valid if $ax \neq 0$ as $x \to 0$, which is true if $a \neq 0$).

For the second limit, $\frac{a}{b}$ is a constant with respect to $x$ (assuming $b \neq 0$).

$\lim\limits_{x \to 0} \frac{a}{b} = \frac{a}{b}$

So, if $a \neq 0$ and $b \neq 0$, the limit is:

$L = 1 \times \frac{a}{b} = \frac{a}{b}$

Now, let's consider the special cases:

Case 1: $a = 0$.

The expression becomes $\lim\limits_{x \to 0} \frac{\sin(0x)}{bx} = \lim\limits_{x \to 0} \frac{\sin 0}{bx} = \lim\limits_{x \to 0} \frac{0}{bx}$.

If $b \neq 0$, then for any $x \neq 0$, $\frac{0}{bx} = 0$. So, $\lim\limits_{x \to 0} 0 = 0$. In this case, the result $\frac{a}{b} = \frac{0}{b} = 0$ is still correct.

If $a=0$ and $b=0$, the expression is $\lim\limits_{x \to 0} \frac{\sin 0x}{0x} = \lim\limits_{x \to 0} \frac{0}{0}$, which is an indeterminate form. This typically requires further analysis (like L'Hopital's rule or series expansion) but given the basic nature of the expression, it's likely assumed that $b \neq 0$. If $a=0$ and $b=0$, the expression is usually considered undefined unless specified otherwise.

Case 2: $b = 0$ and $a \neq 0$.

The expression is $\lim\limits_{x \to 0} \frac{\sin ax}{0x}$. The denominator is $0$ for all $x \neq 0$. The numerator $\sin ax$ approaches $\sin 0 = 0$ as $x \to 0$. However, for $x$ close to 0 but not equal to 0, $\sin ax$ is non-zero (if $a \neq 0$). The expression is of the form $\frac{\text{non-zero}}{\text{zero}}$ (as $x \to 0$, $ax \approx 0$, so $\sin ax \approx ax \neq 0$), which implies the limit does not exist (it would be $\infty$ or $-\infty$). The result $\frac{a}{b}$ is $\frac{a}{0}$ which is undefined, consistent with the limit not existing.

Considering the usual context of such limit problems at this level, it is typically implied that the denominator coefficient $b$ is non-zero so that the limit exists and can be evaluated using the standard formula.

Assuming $b \neq 0$, the limit is $\frac{a}{b}$. This also covers the case when $a=0$ and $b \neq 0$.

Answer:

The value of the limit is $\mathbf{\frac{a}{b}}$ (provided $b \neq 0$).

Given:

The limit expression $\lim\limits_{x \to 0} \frac{\sin ax}{\sin bx}$, with conditions $a \neq 0$ and $b \neq 0$.

To Evaluate:

The given limit.

Solution:

When we substitute $x = 0$ into the expression, we get $\frac{\sin(a \times 0)}{\sin(b \times 0)} = \frac{\sin 0}{\sin 0} = \frac{0}{0}$, which is an indeterminate form.

We will use the standard limit formula: $\lim\limits_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$.

To apply this formula, we can divide the numerator and the denominator by $x$.

Now, manipulate the numerator and the denominator to match the standard form. Multiply and divide the numerator by $a$ and the denominator by $b$ (since $a \neq 0$ and $b \neq 0$):

(Assuming $x \neq 0$, which is true as $x$ approaches 0)

Using the properties of limits, the limit of a quotient is the quotient of the limits, and the limit of a product is the product of the limits, provided the individual limits exist and the denominator limit is non-zero.

As $x \to 0$, $ax \to 0$ and $bx \to 0$ (since $a, b \neq 0$).

Using the standard limit $\lim\limits_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$:

$\lim\limits_{x \to 0} \frac{\sin ax}{ax} = 1$

$\lim\limits_{x \to 0} \frac{\sin bx}{bx} = 1$

The limits of the constants are:

$\lim\limits_{x \to 0} a = a$

$\lim\limits_{x \to 0} b = b$

Substitute these values back into the limit expression:

The value of the limit is $\mathbf{\frac{a}{b}}$. This is valid since $b \neq 0$ was given.

Given:

The limit expression $\lim\limits_{x \to \pi} \frac{\sin \;(\pi \;-\; x)}{\pi \;(\pi \;-\; x)}$.

To Evaluate:

The given limit.

Solution:

When we substitute $x = \pi$ into the expression, we get $\frac{\sin (\pi - \pi)}{\pi (\pi - \pi)} = \frac{\sin 0}{\pi \times 0} = \frac{0}{0}$, which is an indeterminate form.

We can use the standard limit formula: $\lim\limits_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$.

Let $\theta = \pi - x$. As $x \to \pi$, $\theta \to \pi - \pi = 0$.

The expression involves $\sin(\pi - x)$ in the numerator and $\pi (\pi - x)$ in the denominator. The argument of the sine function is $(\pi - x)$. The term $(\pi - x)$ is present in the denominator, along with a constant factor $\pi$.

Rewrite the limit in terms of $\theta$:

$\lim\limits_{x \to \pi} \frac{\sin \;(\pi \;-\; x)}{\pi \;(\pi \;-\; x)} = \lim\limits_{\theta \to 0} \frac{\sin \theta}{\pi \theta}$

We can take the constant factor $\frac{1}{\pi}$ out of the limit:

Using the standard limit $\lim\limits_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$:

The value of the limit is $\mathbf{\frac{1}{\pi}}$.

Given:

The limit expression $\lim\limits_{x \to 0} \frac{\cos x}{\pi \;-\; x}$.

To Evaluate:

The given limit.

Solution:

The function is $f(x) = \frac{\cos x}{\pi - x}$. We want to find the limit as $x$ approaches 0.

First, check the value of the denominator when $x = 0$:

Denominator at $x=0$ is $\pi - 0 = \pi$.

Since the denominator is non-zero at $x = 0$, we can evaluate the limit by direct substitution of $x = 0$ into the function.

We know that $\cos 0 = 1$.

The value of the limit is $\mathbf{\frac{1}{\pi}}$.

Given:

The limit expression $\lim\limits_{x \to 0} \frac{\cos 2x \;-\; 1}{\cos x \;-\; 1}$.

To Evaluate:

The given limit.

Solution:

When we substitute $x = 0$ into the expression, we get $\frac{\cos(2 \times 0) - 1}{\cos 0 - 1} = \frac{\cos 0 - 1}{\cos 0 - 1} = \frac{1 - 1}{1 - 1} = \frac{0}{0}$, which is an indeterminate form.

We can use trigonometric identities to simplify the expression.

Recall the identity $\cos 2\theta = 1 - 2\sin^2 \theta$. Rearranging, we get $\cos 2\theta - 1 = -2\sin^2 \theta$.

Apply this identity to the numerator: $\cos 2x - 1 = -2\sin^2 x$.

Apply this identity to the denominator: $\cos x - 1$. To use the identity form, we can think of $x$ as $2(x/2)$. So, $\cos x = 1 - 2\sin^2(x/2)$. Rearranging, $\cos x - 1 = -2\sin^2(x/2)$.

Substitute these into the limit expression:

Cancel the -2 terms:

Now, we can evaluate the limit of the inner expression $\frac{\sin x}{\sin (x/2)}$. We use the standard limit $\lim\limits_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$.

Divide the numerator and denominator of the inner expression by $x$:

Manipulate the denominator to match the standard form. Multiply the denominator by $2/2$:

As $x \to 0$, $x \to 0$ and $x/2 \to 0$.

Using the standard limit $\lim\limits_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$:

$\lim\limits_{x \to 0} \frac{\sin x}{x} = 1$

$\lim\limits_{x \to 0} \frac{\sin (x/2)}{x/2} = 1$

So the inner limit is:

Wait, this is not correct. Let's re-evaluate the inner limit carefully.

Let's go back to $\lim\limits_{x \to 0} \frac{\sin x}{\sin (x/2)}$. We can divide numerator by $x$ and denominator by $x$.

In the denominator, we need $x/2$ instead of $x$. So, multiply and divide the denominator by $1/2$:

Taking limits of numerator and denominator:

Ah, I see the error in my previous step. I was calculating the limit of $\frac{\sin x}{\sin (x/2)}$ as $\frac{1}{2}$. Let's re-evaluate the squared limit.

The original limit was $\lim\limits_{x \to 0} \left( \frac{\sin x}{\sin (x/2)} \right)^2$.

Let's evaluate the limit of $\frac{\sin x}{\sin (x/2)}$ more carefully. We need $\frac{\sin \theta}{\theta}$ terms.

For $x \neq 0$, cancel $x$:

Taking limits:

So, $\lim\limits_{x \to 0} \frac{\sin x}{\sin (x/2)} = 2$.

Now, square this result for the original limit:

The value of the limit is $\mathbf{4}$.

Alternate Solution: Using Double Angle Identity

We can use the identity $\sin 2\theta = 2\sin\theta\cos\theta$. Let $\theta = x/2$, so $2\theta = x$.

$\sin x = \sin(2(x/2)) = 2\sin(x/2)\cos(x/2)$.

Substitute this into the expression $\frac{\sin^2 x}{\sin^2 (x/2)}$:

For $x \neq 0$, $\sin(x/2) \neq 0$ for small $x$. We can cancel $\sin^2(x/2)$:

Now evaluate the limit by direct substitution of $x=0$:

We know $\cos 0 = 1$.

The value of the limit is $\mathbf{4}$. This confirms the previous result.

Given:

The limit expression $\lim\limits_{x \to 0} \frac{ax \;+\; x\cos x}{b\sin x}$.

To Evaluate:

The given limit.

Solution:

When we substitute $x = 0$ into the expression, we get $\frac{a(0) + 0\cos 0}{b\sin 0} = \frac{0 + 0 \times 1}{b \times 0} = \frac{0}{0}$, which is an indeterminate form.

We need to simplify the expression. We can factor out $x$ from the numerator and use the standard limit $\lim\limits_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$.

Rewrite the numerator: $ax + x\cos x = x(a + \cos x)$.

Rewrite the limit expression:

Rearrange the terms to get the standard sine limit form:

Using the properties of limits, the limit of a product is the product of the limits (if they exist), and the limit of a quotient is the quotient of the limits (if they exist and the denominator limit is non-zero):

Evaluate the first limit by direct substitution of $x=0$, assuming $b \neq 0$:

$\lim\limits_{x \to 0} \frac{a \;+\; \cos x}{b} = \frac{a \;+\; \cos 0}{b} = \frac{a \;+\; 1}{b}$

Evaluate the second limit using the standard limit $\lim\limits_{x \to 0} \frac{\sin x}{x} = 1$:

$\lim\limits_{x \to 0} \frac{1}{\frac{\sin x}{x}} = \frac{1}{\lim\limits_{x \to 0} \frac{\sin x}{x}} = \frac{1}{1} = 1$

Now, multiply the results of the two limits:

This result is valid provided $b \neq 0$. If $b=0$, the original expression's denominator is 0, and the limit does not exist unless the numerator is also 0 in a way that allows cancellation. However, if $b=0$, the limit is of the form $\lim\limits_{x \to 0} \frac{x(a+\cos x)}{0}$. If $a+1 \neq 0$, the numerator is non-zero near 0, so the limit does not exist.

Assuming $b \neq 0$, the value of the limit is $\mathbf{\frac{a \;+\; 1}{b}}$.

Given:

The limit expression $\lim\limits_{x \to 0} x\sec x$.

To Evaluate:

The given limit.

Solution:

The function is $f(x) = x\sec x$. We know that $\sec x = \frac{1}{\cos x}$.

So, the expression can be written as:

When we substitute $x = 0$ into this expression, we get $\frac{0}{\cos 0}$.

We know that $\cos 0 = 1$.

So, the value is $\frac{0}{1} = 0$.

Since the denominator $\cos x$ is non-zero at $x=0$ (it is 1), and the numerator approaches 0, the limit can be found by direct substitution.

The limit of $x$ as $x \to 0$ is $0$.

The limit of $\cos x$ as $x \to 0$ is $\cos 0 = 1$ (since $\cos x$ is a continuous function).

The value of the limit is $\mathbf{0}$.

Given:

The limit expression $\lim\limits_{x \to 0} \frac{\sin ax\;+\;bx}{ax\;+\;\sin bx}$, with conditions $a \neq 0$, $b \neq 0$, and $a+b \neq 0$.

To Evaluate:

The given limit.

Solution:

When we substitute $x = 0$ into the expression, the numerator becomes $\sin(a \times 0) + b(0) = \sin 0 + 0 = 0$. The denominator becomes $a(0) + \sin(b \times 0) = 0 + \sin 0 = 0$.

Since we have the indeterminate form $\frac{0}{0}$, we need to simplify the expression.

We can divide both the numerator and the denominator by $x$ (since $x \to 0$, $x \neq 0$).

For $x \neq 0$, $\frac{bx}{x} = b$ and $\frac{ax}{x} = a$. Also, we can manipulate the sine terms to use the standard limit $\lim\limits_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$.

$\frac{\sin ax}{x} = \frac{\sin ax}{ax} \times a$ (since $a \neq 0$)

$\frac{\sin bx}{x} = \frac{\sin bx}{bx} \times b$ (since $b \neq 0$)

Substitute these back into the expression:

Using the properties of limits, the limit of a quotient is the quotient of the limits, and the limit of a sum/product is the sum/product of the limits, provided the individual limits exist and the denominator limit is non-zero.

As $x \to 0$, $ax \to 0$ and $bx \to 0$. Thus, $\lim\limits_{x \to 0} \frac{\sin ax}{ax} = 1$ and $\lim\limits_{x \to 0} \frac{\sin bx}{bx} = 1$.

Evaluate the limit:

Since we are given that $a + b \neq 0$, we can cancel the term $(a+b)$.

The value of the limit is $\mathbf{1}$.

Given:

The limit expression $\lim\limits_{x \to 0} (\text{cosec}\;x-\cot x)$.

To Evaluate:

The given limit.

Solution:

The given expression involves trigonometric functions $\text{cosec}\;x$ and $\cot x$. We can rewrite these in terms of $\sin x$ and $\cos x$:

$\text{cosec}\;x = \frac{1}{\sin x}$

$\cot x = \frac{\cos x}{\sin x}$

Substitute these into the limit expression:

Combine the terms into a single fraction since they have a common denominator:

When we substitute $x = 0$ into this expression, we get $\frac{1 - \cos 0}{\sin 0} = \frac{1 - 1}{0} = \frac{0}{0}$, which is an indeterminate form. This indicates that we need to simplify the expression further.

We can use the half-angle identity for the numerator: $1 - \cos x = 2\sin^2\left(\frac{x}{2}\right)$.

We can use the double-angle identity for the denominator: $\sin x = 2\sin\left(\frac{x}{2}\right)\cos\left(\frac{x}{2}\right)$.

Substitute these identities into the fraction:

For $x$ approaching 0 but $x \neq 0$, $\frac{x}{2}$ approaches 0 but is not 0, so $\sin\left(\frac{x}{2}\right) \neq 0$. We can cancel one factor of $2\sin\left(\frac{x}{2}\right)$ from the numerator and the denominator:

Now, evaluate the limit of this simplified expression by direct substitution of $x=0$ (since the denominator $\cos(x/2)$ is non-zero at $x=0$):

We know that $\sin 0 = 0$ and $\cos 0 = 1$.

The value of the limit is $\mathbf{0}$.

Given:

The limit expression $\lim\limits_{x \to \frac{\pi}{2}} \frac{\tan2x}{x\;-\;\frac{\pi}{2}}$.

To Evaluate:

The given limit.

Solution:

When we substitute $x = \frac{\pi}{2}$ into the expression, the numerator is $\tan\left(2 \times \frac{\pi}{2}\right) = \tan(\pi) = 0$, and the denominator is $\frac{\pi}{2} - \frac{\pi}{2} = 0$.

Since we have the indeterminate form $\frac{0}{0}$, we need to use a substitution or algebraic manipulation.

Let $y = x - \frac{\pi}{2}$.

As $x \to \frac{\pi}{2}$, $y \to \frac{\pi}{2} - \frac{\pi}{2}$, which means $y \to 0$.

From the substitution, we have $x = y + \frac{\pi}{2}$.

Substitute $x$ in the numerator:

$\tan(2x) = \tan\left(2\left(y + \frac{\pi}{2}\right)\right)$

$\tan(2x) = \tan(2y + \pi)$

Using the trigonometric identity $\tan(\theta + \pi) = \tan \theta$, where $\theta = 2y$:

$\tan(2y + \pi) = \tan(2y)$

Now substitute $y = x - \frac{\pi}{2}$ and $\tan(2x) = \tan(2y)$ into the limit expression. The limit variable changes from $x$ to $y$ as $x \to \frac{\pi}{2}$ implies $y \to 0$.

We can use the standard limit formula $\lim\limits_{\theta \to 0} \frac{\tan \theta}{\theta} = 1$. To match this form with $\tan(2y)$, the denominator should be $2y$. We can achieve this by multiplying the numerator and the denominator by 2.

Using the property that the limit of a constant times a function is the constant times the limit of the function:

Let $\theta = 2y$. As $y \to 0$, $\theta = 2y \to 0$. The limit becomes:

Using the standard limit $\lim\limits_{\theta \to 0} \frac{\tan \theta}{\theta} = 1$:

The value of the limit is $\mathbf{2}$.

Given:

The piecewise function $f(x) = \begin{cases} 2x + 3 , & x \leq0 \\ 3(x + 1) , & x > 0 \end{cases}$

To Find:

$\lim\limits_{x \to 0} f(x)$ and $\lim\limits_{x \to 1} f(x)$.

Solution:

Finding $\lim\limits_{x \to 0} f(x)$:

To find the limit as $x \to 0$, we need to consider the left-hand limit (as $x$ approaches 0 from the left, $x < 0$) and the right-hand limit (as $x$ approaches 0 from the right, $x > 0$). The limit exists if and only if both one-sided limits exist and are equal.

For the left-hand limit, as $x \to 0^-$, $x \leq 0$. So we use the first part of the function definition: $f(x) = 2x + 3$.

Since $2x + 3$ is a polynomial, we can find the limit by direct substitution:

For the right-hand limit, as $x \to 0^+$, $x > 0$. So we use the second part of the function definition: $f(x) = 3(x + 1)$.

Since $3(x + 1) = 3x + 3$ is a polynomial, we can find the limit by direct substitution:

Since the left-hand limit and the right-hand limit are equal ($\lim\limits_{x \to 0^-} f(x) = 3$ and $\lim\limits_{x \to 0^+} f(x) = 3$), the limit as $x \to 0$ exists and is equal to 3.

$\lim\limits_{x \to 0} f(x) = \mathbf{3}$.

Finding $\lim\limits_{x \to 1} f(x)$:

To find the limit as $x \to 1$, we need to consider the function definition for values of $x$ near 1.

Since $1 > 0$, for values of $x$ approaching 1 from both the left ($x < 1$ but close to 1) and the right ($x > 1$ but close to 1), we are in the domain where $x > 0$.

So, for $x$ values around 1, we use the second part of the function definition: $f(x) = 3(x + 1)$.

Since $3(x + 1)$ is a polynomial, we can find the limit by direct substitution:

The limit as $x \to 1$ is $\mathbf{6}$.

Given:

The piecewise function $f(x) = \begin{cases} x^2 - 1 & , & x \leq1 \\ -x^2 - 1 & , & x > 1 \end{cases}$

To Find:

$\lim\limits_{x \to 1} f(x)$.

Solution:

To find the limit of $f(x)$ as $x$ approaches a point where the function definition changes (in this case, at $x=1$), we need to evaluate the left-hand limit and the right-hand limit at that point.

The limit $\lim\limits_{x \to 1} f(x)$ exists if and only if the left-hand limit $\lim\limits_{x \to 1^-} f(x)$ and the right-hand limit $\lim\limits_{x \to 1^+} f(x)$ both exist and are equal.

Left-hand limit: $\lim\limits_{x \to 1^-} f(x)$

As $x$ approaches 1 from the left side, $x < 1$. According to the definition of $f(x)$, for $x \leq 1$, $f(x) = x^2 - 1$.

Since $x^2 - 1$ is a polynomial function, its limit as $x \to 1$ can be found by direct substitution:

So, the left-hand limit is $0$.

Right-hand limit: $\lim\limits_{x \to 1^+} f(x)$

As $x$ approaches 1 from the right side, $x > 1$. According to the definition of $f(x)$, for $x > 1$, $f(x) = -x^2 - 1$.

Since $-x^2 - 1$ is a polynomial function, its limit as $x \to 1$ can be found by direct substitution:

So, the right-hand limit is $-2$.

Comparing the one-sided limits:

Left-hand limit: $\lim\limits_{x \to 1^-} f(x) = 0$

Right-hand limit: $\lim\limits_{x \to 1^+} f(x) = -2$

Since the left-hand limit is not equal to the right-hand limit ($0 \neq -2$), the limit of $f(x)$ as $x$ approaches 1 does not exist.

$\lim\limits_{x \to 1} f(x) = \mathbf{Does\;Not\;Exist}$.

Given:

The piecewise function $f(x) = \begin{cases} \frac{|x|}{x} & , & x \neq 0 \\ 0 & , & x = 0 \end{cases}$

To Evaluate:

$\lim\limits_{x \to 0} f(x)$.

Solution:

To find the limit as $x \to 0$, we need to consider the left-hand limit (as $x$ approaches 0 from the left, $x < 0$) and the right-hand limit (as $x$ approaches 0 from the right, $x > 0$). The limit exists if and only if both one-sided limits exist and are equal.

The function definition for $x \neq 0$ is $f(x) = \frac{|x|}{x}$. We need to understand the definition of $|x|$.

$|x| = \begin{cases} x , & x \geq 0 \\ -x , & x < 0 \end{cases}$

Left-hand limit: $\lim\limits_{x \to 0^-} f(x)$

As $x$ approaches 0 from the left side, $x < 0$. In this case, $|x| = -x$.

So, for $x < 0$, $f(x) = \frac{|x|}{x} = \frac{-x}{x}$.

For $x \neq 0$, we can simplify $\frac{-x}{x} = -1$.

The limit of a constant is the constant itself.

So, the left-hand limit is $-1$.

Right-hand limit: $\lim\limits_{x \to 0^+} f(x)$

As $x$ approaches 0 from the right side, $x > 0$. In this case, $|x| = x$.

So, for $x > 0$, $f(x) = \frac{|x|}{x} = \frac{x}{x}$.

For $x \neq 0$, we can simplify $\frac{x}{x} = 1$.

The limit of a constant is the constant itself.

So, the right-hand limit is $1$.

Comparing the one-sided limits:

Left-hand limit: $\lim\limits_{x \to 0^-} f(x) = -1$

Right-hand limit: $\lim\limits_{x \to 0^+} f(x) = 1$

Since the left-hand limit is not equal to the right-hand limit ($-1 \neq 1$), the limit of $f(x)$ as $x$ approaches 0 does not exist.

Note that the value of the function at $x=0$, $f(0) = 0$, does not affect the existence or value of the limit as $x \to 0$. The limit is about the behavior of the function as $x$ gets arbitrarily close to 0, but not equal to 0.

$\lim\limits_{x \to 0} f(x) = \mathbf{Does\;Not\;Exist}$.

Given:

The piecewise function $f(x) = \begin{cases} \frac{x}{|x|} & , & x \neq 0 \\ 0 & , & x = 0 \end{cases}$

To Evaluate:

$\lim\limits_{x \to 0} f(x)$.

Solution:

To find the limit as $x \to 0$, we need to examine the left-hand limit (as $x \to 0^-$) and the right-hand limit (as $x \to 0^+$). The limit exists if and only if these one-sided limits exist and are equal.

The function definition for $x \neq 0$ is $f(x) = \frac{x}{|x|}$. Recall the definition of the absolute value function:

$|x| = \begin{cases} x , & x \geq 0 \\ -x , & x < 0 \end{cases}$

Left-hand limit: $\lim\limits_{x \to 0^-} f(x)$

As $x$ approaches 0 from the left side, $x < 0$. In this case, $|x| = -x$.

So, for $x < 0$, $f(x) = \frac{x}{|x|} = \frac{x}{-x}$.

For $x \neq 0$, we can simplify $\frac{x}{-x} = -1$.

The limit of a constant is the constant itself.

So, the left-hand limit is $-1$.

Right-hand limit: $\lim\limits_{x \to 0^+} f(x)$

As $x$ approaches 0 from the right side, $x > 0$. In this case, $|x| = x$.

So, for $x > 0$, $f(x) = \frac{x}{|x|} = \frac{x}{x}$.

For $x \neq 0$, we can simplify $\frac{x}{x} = 1$.

The limit of a constant is the constant itself.

So, the right-hand limit is $1$.

Comparing the one-sided limits:

Left-hand limit: $\lim\limits_{x \to 0^-} f(x) = -1$

Right-hand limit: $\lim\limits_{x \to 0^+} f(x) = 1$

Since the left-hand limit is not equal to the right-hand limit ($-1 \neq 1$), the limit of $f(x)$ as $x$ approaches 0 does not exist.

The value of the function at $x=0$, $f(0) = 0$, is irrelevant to the existence or value of the limit as $x \to 0$.

$\lim\limits_{x \to 0} f(x) = \mathbf{Does\;Not\;Exist}$.

We are asked to find the limit of the function $f(x) = |x| - 5$ as $x$ approaches $5$.

The function is given by $f(x) = |x| - 5$.

We need to evaluate $\lim\limits_{x \to 5} f(x)$.

Recall the definition of the absolute value function $|x|$:

$|x| = \begin{cases} x & , & x \geq 0 \\ -x & , & x < 0 \end{cases}$

As $x$ approaches $5$, the values of $x$ are close to $5$. Since $5$ is a positive number, for $x$ values sufficiently close to $5$, $x$ will be positive.

Therefore, for $x$ approaching $5$, we have $|x| = x$.

So, for $x$ near $5$, the function $f(x)$ can be written as:

$f(x) = x - 5$

Now we can find the limit of this simplified expression as $x$ approaches $5$:

$\lim\limits_{x \to 5} f(x) = \lim\limits_{x \to 5} (x - 5)$

Since $g(x) = x - 5$ is a polynomial function (which is continuous everywhere), the limit as $x$ approaches a value can be found by direct substitution.

$\lim\limits_{x \to 5} (x - 5) = 5 - 5 = 0$

Therefore, the limit of the given function as $x$ approaches $5$ is $0$.

$\lim\limits_{x \to 5} (|x| - 5) = 0$

Given:

To Find:

The possible values of $a$ and $b$.

Solution:

The given condition $\lim\limits_{x \to 1} f(x) = f(1)$ means that the function $f(x)$ is continuous at $x = 1$.

For the limit $\lim\limits_{x \to 1} f(x)$ to exist, the left-hand limit (LHL) and the right-hand limit (RHL) at $x=1$ must be equal.

Also, according to the given condition, this common limit value must be equal to the function's value at $x=1$, i.e., $f(1)$.

So, we must have: LHL at $x=1$ = RHL at $x=1$ = $f(1)$.

First, find the value of $f(1)$ from the definition of $f(x)$:

Next, calculate the left-hand limit (LHL) at $x=1$. This involves using the part of the function defined for $x < 1$, which is $f(x) = a + bx$.

Substitute $x=1$ into the expression:

Now, calculate the right-hand limit (RHL) at $x=1$. This involves using the part of the function defined for $x > 1$, which is $f(x) = b - ax$.

Substitute $x=1$ into the expression:

According to the given condition, LHL = RHL = $f(1)$. Using equations (i), (ii), and (iii), we get:

$a + b = 4$

and

$b - a = 4$

Now, we solve the system of linear equations formed by (iv) and (v) for $a$ and $b$.

Equation (iv): $a + b = 4$

Equation (v): $-a + b = 4$

Add equation (iv) and equation (v):

Divide by 2:

Substitute the value of $b=4$ into equation (iv):

Subtract 4 from both sides:

So, the unique solution is $a = 0$ and $b = 4$. These are the possible values of $a$ and $b$ for the given condition to hold.

Answer:

The possible values of $a$ and $b$ are $\mathbf{a = 0}$ and $\mathbf{b = 4}$.

Given the function $f(x)$ defined as:

$f(x) = (x - a_1) (x - a_2) \cdots (x - a_n)$

This function is a product of linear terms $(x - a_i)$. Expanding this product would result in a polynomial of degree $n$. For example, if $n=2$, $f(x) = (x-a_1)(x-a_2) = x^2 - (a_1+a_2)x + a_1a_2$, which is a polynomial.

Polynomial functions are continuous everywhere in their domain, which is all real numbers.

Part 1: Find $\lim\limits_{x \to a_1} f(x)$

Since $f(x)$ is a polynomial function, it is continuous at $x = a_1$.

For a continuous function, the limit as $x$ approaches a point is equal to the function's value at that point.

So, $\lim\limits_{x \to a_1} f(x) = f(a_1)$.

Let's substitute $x = a_1$ into the expression for $f(x)$:

$f(a_1) = (a_1 - a_1) (a_1 - a_2) (a_1 - a_3) \cdots (a_1 - a_n)$

The first term in the product is $(a_1 - a_1)$, which is equal to $0$.

$f(a_1) = (0) (a_1 - a_2) (a_1 - a_3) \cdots (a_1 - a_n)$

Any product that includes a factor of $0$ is equal to $0$.

$f(a_1) = 0$

Therefore, $\lim\limits_{x \to a_1} f(x) = 0$.

Part 2: Compute $\lim\limits_{x \to a} f(x)$ for some $a \neq a_1, a_2, \ldots, a_n$

As established earlier, $f(x)$ is a polynomial function and is continuous everywhere.

We need to find the limit as $x$ approaches a value $a$, where $a$ is not equal to any of $a_1, a_2, \ldots, a_n$.

Since $f(x)$ is continuous at $x=a$, the limit is equal to the function's value at $a$.

So, $\lim\limits_{x \to a} f(x) = f(a)$.

Let's substitute $x = a$ into the expression for $f(x)$:

$f(a) = (a - a_1) (a - a_2) (a - a_3) \cdots (a - a_n)$

Since $a \neq a_i$ for any $i=1, 2, \ldots, n$, none of the factors $(a - a_i)$ are equal to $0$.

The limit is the value of the function evaluated at $x=a$.

Therefore, $\lim\limits_{x \to a} f(x) = (a - a_1) (a - a_2) \cdots (a - a_n)$.

Summary of results:

The limit of $f(x)$ as $x$ approaches $a_1$ is $0$.

The limit of $f(x)$ as $x$ approaches $a$ (where $a \neq a_1, \ldots, a_n$) is $(a - a_1) (a - a_2) \cdots (a - a_n)$.

Given the function $f(x)$ defined as:

$f(x) = \begin{cases} |x| + 1 & , & x < 0 \\ 0 & , & x = 0 \\ |x| - 1 & , & x > 0 \end{cases}$

We need to find the values of 'a' for which the limit $\lim\limits_{x \to a} f(x)$ exists.

The limit $\lim\limits_{x \to a} f(x)$ exists if and only if the left-hand limit (LHL) and the right-hand limit (RHL) at $x=a$ are equal, i.e., $\lim\limits_{x \to a^-} f(x) = \lim\limits_{x \to a^+} f(x)$.

We analyze the existence of the limit by considering different cases for the value of 'a'.

Case 1: $a < 0$

For any value of $a$ such that $a < 0$, we consider the limit as $x$ approaches $a$. For values of $x$ close to $a$ (both from the left and the right), $x$ is less than $0$.

In this range ($x < 0$), the function is defined as $f(x) = |x| + 1$.

Since $x < 0$, the absolute value $|x|$ is equal to $-x$.

So, for $x < 0$, $f(x) = -x + 1$.

The function $-x+1$ is a polynomial, which is continuous for all real numbers.

Therefore, for any $a < 0$, the limit exists and can be found by direct substitution:

$\lim\limits_{x \to a} f(x) = \lim\limits_{x \to a} (-x + 1) = -a + 1$.

The limit exists for all $a < 0$.

Case 2: $a > 0$

For any value of $a$ such that $a > 0$, we consider the limit as $x$ approaches $a$. For values of $x$ close to $a$ (both from the left and the right), $x$ is greater than $0$.

In this range ($x > 0$), the function is defined as $f(x) = |x| - 1$.

Since $x > 0$, the absolute value $|x|$ is equal to $x$.

So, for $x > 0$, $f(x) = x - 1$.

The function $x-1$ is a polynomial, which is continuous for all real numbers.

Therefore, for any $a > 0$, the limit exists and can be found by direct substitution:

$\lim\limits_{x \to a} f(x) = \lim\limits_{x \to a} (x - 1) = a - 1$.

The limit exists for all $a > 0$.

Case 3: $a = 0$

We need to check if the left-hand limit and the right-hand limit at $x=0$ are equal.

Let's compute the LHL at $x=0$:

$\lim\limits_{x \to 0^-} f(x)$

For $x$ approaching $0$ from the left, $x < 0$. The function is $f(x) = |x| + 1 = -x + 1$.

$\lim\limits_{x \to 0^-} f(x) = \lim\limits_{x \to 0^-} (-x + 1)$

Substituting $x=0$:

$\lim\limits_{x \to 0^-} f(x) = -(0) + 1 = 1$

Let's compute the RHL at $x=0$:

$\lim\limits_{x \to 0^+} f(x)$

For $x$ approaching $0$ from the right, $x > 0$. The function is $f(x) = |x| - 1 = x - 1$.

$\lim\limits_{x \to 0^+} f(x) = \lim\limits_{x \to 0^+} (x - 1)$

Substituting $x=0$:

$\lim\limits_{x \to 0^+} f(x) = (0) - 1 = -1$

Now, we compare the LHL and RHL at $x=0$.

LHL = 1

RHL = -1

Since LHL $\neq$ RHL ($1 \neq -1$), the limit $\lim\limits_{x \to 0} f(x)$ does not exist.

Combining the results from all cases:

The limit $\lim\limits_{x \to a} f(x)$ exists for all $a < 0$ and for all $a > 0$.

The limit $\lim\limits_{x \to a} f(x)$ does not exist for $a = 0$.

Therefore, the limit exists for all real numbers 'a' except $a=0$.

The set of values of 'a' for which the limit exists is $(-\infty, 0) \cup (0, \infty)$.

The limit exists for all values of a $\in \mathbb{R} \setminus \{0\}$.

We are given the limit:

$\lim\limits_{x \to 1} \frac{f(x) \;-\; 2}{x^2 \;-\; 1} = \pi$

Let $L = \lim\limits_{x \to 1} \frac{f(x) \;-\; 2}{x^2 \;-\; 1}$. We are given that $L = \pi$.

Consider the denominator of the expression inside the limit: $x^2 - 1$.

As $x \to 1$, the denominator approaches:

$\lim\limits_{x \to 1} (x^2 - 1) = 1^2 - 1 = 1 - 1 = 0$.

We have a situation where the limit of a fraction is a finite non-zero value ($\pi$), and the denominator approaches $0$.

For the limit of the fraction to be finite when the denominator approaches $0$, the numerator must also approach $0$. This leads to an indeterminate form of type $\frac{0}{0}$, which can have a finite limit.

Therefore, the limit of the numerator as $x \to 1$ must be $0$.

$\lim\limits_{x \to 1} (f(x) - 2) = 0$

Using the properties of limits, the limit of a difference is the difference of the limits:

$\lim\limits_{x \to 1} f(x) - \lim\limits_{x \to 1} 2 = 0$

The limit of a constant function is the constant itself:

$\lim\limits_{x \to 1} 2 = 2$

Substituting this back into the equation:

$\lim\limits_{x \to 1} f(x) - 2 = 0$

Now, we can solve for $\lim\limits_{x \to 1} f(x)$:

$\lim\limits_{x \to 1} f(x) = 2$

Thus, the value of $\lim\limits_{x \to 1} f(x)$ is $2$.

$\lim\limits_{x \to 1} f(x) = 2$.

Given the function $f(x)$ defined as:

$f(x) = \begin{cases} mx^2 + n , & x < 0 \\ nx + m , & 0 \leq x \leq 1 \\ nx^3 + m , & x > 1 \end{cases}$

We need to find integer values of $m$ and $n$ such that $\lim\limits_{x \to 0} f(x)$ and $\lim\limits_{x \to 1} f(x)$ both exist.

For a limit $\lim\limits_{x \to a} f(x)$ to exist, the left-hand limit (LHL) and the right-hand limit (RHL) at $x=a$ must be equal, i.e., $\lim\limits_{x \to a^-} f(x) = \lim\limits_{x \to a^+} f(x)$.

Condition for $\lim\limits_{x \to 0} f(x)$ to exist:

We need to evaluate the LHL and RHL at $x=0$.

Left-Hand Limit at $x=0$: As $x \to 0^-$, $x$ is slightly less than $0$. For $x < 0$, $f(x) = mx^2 + n$.

$\lim\limits_{x \to 0^-} f(x) = \lim\limits_{x \to 0^-} (mx^2 + n)$

Substituting $x=0$ into the polynomial expression:

$\lim\limits_{x \to 0^-} f(x) = m(0)^2 + n = 0 + n = n$

Right-Hand Limit at $x=0$: As $x \to 0^+$, $x$ is slightly greater than $0$. For $0 \leq x \leq 1$, $f(x) = nx + m$.

$\lim\limits_{x \to 0^+} f(x) = \lim\limits_{x \to 0^+} (nx + m)$

Substituting $x=0$ into the polynomial expression:

$\lim\limits_{x \to 0^+} f(x) = n(0) + m = 0 + m = m$

For $\lim\limits_{x \to 0} f(x)$ to exist, LHL = RHL at $x=0$.

$n = m$

Condition for $\lim\limits_{x \to 1} f(x)$ to exist:

We need to evaluate the LHL and RHL at $x=1$.

Left-Hand Limit at $x=1$: As $x \to 1^-$, $x$ is slightly less than $1$. For $0 \leq x \leq 1$, $f(x) = nx + m$.

$\lim\limits_{x \to 1^-} f(x) = \lim\limits_{x \to 1^-} (nx + m)$

Substituting $x=1$ into the polynomial expression:

$\lim\limits_{x \to 1^-} f(x) = n(1) + m = n + m$

Right-Hand Limit at $x=1$: As $x \to 1^+$, $x$ is slightly greater than $1$. For $x > 1$, $f(x) = nx^3 + m$.

$\lim\limits_{x \to 1^+} f(x) = \lim\limits_{x \to 1^+} (nx^3 + m)$

Substituting $x=1$ into the polynomial expression:

$\lim\limits_{x \to 1^+} f(x) = n(1)^3 + m = n(1) + m = n + m$

For $\lim\limits_{x \to 1} f(x)$ to exist, LHL = RHL at $x=1$.

$n + m = n + m$

This equation is always true for any values of $m$ and $n$. It does not impose any additional constraint on $m$ and $n$.

For both limits to exist, the conditions from evaluating the limits at $x=0$ and $x=1$ must simultaneously hold.

The condition from $x=0$ is $n=m$.

The condition from $x=1$ is $n+m = n+m$, which is always true.

Therefore, the only requirement for both limits to exist is $n=m$.

We are looking for integer values of $m$ and $n$.

If $m$ is any integer, then $n$ must be equal to that same integer.

So, $m$ and $n$ can be any pair of equal integers.

The integers $m$ and $n$ for which both limits exist are those where $m = n$, where $m$ and $n$ are any integers.

Given:

The function is $f(x) = 3x$.

The point at which the derivative is to be found is $x = 2$.

To Find:

The derivative of $f(x)$ at $x=2$, denoted as $f'(2)$.

Solution:

The derivative of a function $f(x)$ at a point $x=c$ is defined by the limit:

$f'(c) = \lim\limits_{h \to 0} \frac{f(c+h) - f(c)}{h}$

In this problem, $c = 2$ and $f(x) = 3x$.

First, let's find $f(c) = f(2)$:

$f(2) = 3 \times 2 = 6$

Next, let's find $f(c+h) = f(2+h)$:

$f(2+h) = 3 \times (2+h) = 6 + 3h$

Now, substitute these expressions into the limit definition:

$f'(2) = \lim\limits_{h \to 0} \frac{(6 + 3h) - 6}{h}$

Simplify the numerator:

$f'(2) = \lim\limits_{h \to 0} \frac{6 + 3h - 6}{h}$

$f'(2) = \lim\limits_{h \to 0} \frac{3h}{h}$

Since the limit is taken as $h \to 0$, $h$ is a non-zero value approaching $0$. Therefore, we can cancel $h$ from the numerator and the denominator:

$f'(2) = \lim\limits_{h \to 0} 3$

The limit of a constant is the constant itself.

$f'(2) = 3$

Alternatively, we know that the derivative of $f(x) = ax$ is $f'(x) = a$.

For $f(x) = 3x$, the derivative is $f'(x) = 3$.

The derivative at $x=2$ is then $f'(2) = 3$.

The derivative of the function $f(x) = 3x$ at $x=2$ is $3$.

Given:

The function is $f(x) = 2x^2 + 3x - 5$.

We need to find the derivative at $x = -1$.

We also need to prove the relation $f'(0) + 3f'(-1) = 0$.

To Find:

1. The derivative $f'(-1)$.

2. Prove that $f'(0) + 3f'(-1) = 0$.

Solution:

The derivative of a function $f(x)$ at a point $x=c$ is given by the limit definition:

$f'(c) = \lim\limits_{h \to 0} \frac{f(c+h) - f(c)}{h}$

Let's find the derivative at $x = -1$. Here, $c = -1$.

First, calculate $f(-1)$:

$f(-1) = 2(-1)^2 + 3(-1) - 5$

$f(-1) = 2(1) - 3 - 5$

$f(-1) = 2 - 3 - 5 = -6$

Next, calculate $f(-1+h)$:

$f(-1+h) = 2(-1+h)^2 + 3(-1+h) - 5$

$f(-1+h) = 2(1 - 2h + h^2) - 3 + 3h - 5$

$f(-1+h) = 2 - 4h + 2h^2 - 3 + 3h - 5$

Combine like terms:

$f(-1+h) = 2h^2 + (-4h + 3h) + (2 - 3 - 5)$

$f(-1+h) = 2h^2 - h - 6$

Now, form the difference quotient $\frac{f(-1+h) - f(-1)}{h}$:

$f(-1+h) - f(-1) = (2h^2 - h - 6) - (-6)$

$f(-1+h) - f(-1) = 2h^2 - h - 6 + 6 = 2h^2 - h$

$\frac{f(-1+h) - f(-1)}{h} = \frac{2h^2 - h}{h}$

Factor out $h$ from the numerator:

$\frac{2h^2 - h}{h} = \frac{h(2h - 1)}{h}$

For $h \neq 0$, we can cancel $h$:

$\frac{h(2h - 1)}{h} = 2h - 1$

Now, evaluate the limit as $h \to 0$:

$f'(-1) = \lim\limits_{h \to 0} (2h - 1)$

Since $2h-1$ is a polynomial in $h$, we can find the limit by direct substitution:

$f'(-1) = 2(0) - 1 = 0 - 1 = -1$

So, the derivative at $x=-1$ is $-1$.

Next, let's find the derivative at $x = 0$. Here, $c = 0$.

First, calculate $f(0)$:

$f(0) = 2(0)^2 + 3(0) - 5$

$f(0) = 2(0) + 0 - 5 = 0 + 0 - 5 = -5$

Next, calculate $f(0+h) = f(h)$:

$f(h) = 2(h)^2 + 3(h) - 5$

$f(h) = 2h^2 + 3h - 5$

Now, form the difference quotient $\frac{f(0+h) - f(0)}{h}$:

$f(0+h) - f(0) = (2h^2 + 3h - 5) - (-5)$

$f(0+h) - f(0) = 2h^2 + 3h - 5 + 5 = 2h^2 + 3h$

$\frac{f(0+h) - f(0)}{h} = \frac{2h^2 + 3h}{h}$

Factor out $h$ from the numerator:

$\frac{2h^2 + 3h}{h} = \frac{h(2h + 3)}{h}$

For $h \neq 0$, we can cancel $h$:

$\frac{h(2h + 3)}{h} = 2h + 3$

Now, evaluate the limit as $h \to 0$:

$f'(0) = \lim\limits_{h \to 0} (2h + 3)$

Since $2h+3$ is a polynomial in $h$, we can find the limit by direct substitution:

$f'(0) = 2(0) + 3 = 0 + 3 = 3$

So, the derivative at $x=0$ is $3$.

Now, we need to prove that $f'(0) + 3f'(-1) = 0$.

Substitute the values we found for $f'(0)$ and $f'(-1)$:

$f'(0) + 3f'(-1) = (3) + 3(-1)$

$f'(0) + 3f'(-1) = 3 - 3$

$f'(0) + 3f'(-1) = 0$

This proves the required relation.

Alternate Solution (Using Derivative Rules):

We can find the general derivative of $f(x) = 2x^2 + 3x - 5$ using the power rule ($\frac{d}{dx}(x^n) = nx^{n-1}$) and the linearity of differentiation.

$f'(x) = \frac{d}{dx}(2x^2 + 3x - 5)$

$f'(x) = \frac{d}{dx}(2x^2) + \frac{d}{dx}(3x) - \frac{d}{dx}(5)$

$f'(x) = 2 \frac{d}{dx}(x^2) + 3 \frac{d}{dx}(x) - 0$

$f'(x) = 2(2x^{2-1}) + 3(1x^{1-1})$

$f'(x) = 4x^1 + 3x^0$

$f'(x) = 4x + 3$

Now, evaluate the derivative at $x=-1$:

$f'(-1) = 4(-1) + 3 = -4 + 3 = -1$

Evaluate the derivative at $x=0$:

$f'(0) = 4(0) + 3 = 0 + 3 = 3$

Finally, check the relation $f'(0) + 3f'(-1) = 0$:

$f'(0) + 3f'(-1) = 3 + 3(-1) = 3 - 3 = 0$

The relation is proven using the general derivative as well.

The derivative of $f(x) = 2x^2 + 3x - 5$ at $x=-1$ is $-1$.

The relation $f'(0) + 3f'(-1) = 0$ is proven, as $3 + 3(-1) = 0$.

Given:

The function is $f(x) = \sin x$.

The point at which the derivative is to be found is $x = 0$.

To Find:

The derivative of $f(x)$ at $x=0$, denoted as $f'(0)$.

Solution:

The derivative of a function $f(x)$ at a point $x=c$ is defined by the limit:

$f'(c) = \lim\limits_{h \to 0} \frac{f(c+h) - f(c)}{h}$

In this problem, $c = 0$ and $f(x) = \sin x$.

First, let's find $f(c) = f(0)$:

$f(0) = \sin(0) = 0$

Next, let's find $f(c+h) = f(0+h) = f(h)$:

$f(h) = \sin(h)$

Now, substitute these expressions into the limit definition:

$f'(0) = \lim\limits_{h \to 0} \frac{f(h) - f(0)}{h}$

$f'(0) = \lim\limits_{h \to 0} \frac{\sin(h) - 0}{h}$

$f'(0) = \lim\limits_{h \to 0} \frac{\sin h}{h}$

We use the standard trigonometric limit:

$\lim\limits_{x \to 0} \frac{\sin x}{x} = 1$

Replacing $x$ with $h$ in this standard limit, we get:

$\lim\limits_{h \to 0} \frac{\sin h}{h} = 1$

Therefore,

$f'(0) = 1$

The derivative of $\sin x$ at $x=0$ is $1$.

Given:

The function is $f(x) = 3$.

We need to find the derivative at $x=0$ and at $x=3$.

To Find:

The derivative $f'(0)$ and $f'(3)$.

Solution:

The derivative of a function $f(x)$ at a point $x=c$ is defined by the limit:

$f'(c) = \lim\limits_{h \to 0} \frac{f(c+h) - f(c)}{h}$

The function $f(x) = 3$ is a constant function. This means that for any value of $x$, the value of the function is always $3$.

So, $f(c) = 3$ for any value of $c$.

Also, $f(c+h) = 3$ for any value of $c$ and $h$.

Let's find the derivative at $x=0$. Here, $c=0$.

$f'(0) = \lim\limits_{h \to 0} \frac{f(0+h) - f(0)}{h}$

$f'(0) = \lim\limits_{h \to 0} \frac{f(h) - f(0)}{h}$

Substitute the function values:

$f'(0) = \lim\limits_{h \to 0} \frac{3 - 3}{h}$

$f'(0) = \lim\limits_{h \to 0} \frac{0}{h}$

For the limit as $h \to 0$, $h$ is a non-zero value approaching $0$. The expression $\frac{0}{h}$ for $h \neq 0$ is equal to $0$.

$f'(0) = \lim\limits_{h \to 0} 0$

The limit of a constant (which is $0$) is the constant itself.

$f'(0) = 0$

Now, let's find the derivative at $x=3$. Here, $c=3$.

$f'(3) = \lim\limits_{h \to 0} \frac{f(3+h) - f(3)}{h}$

Substitute the function values:

$f'(3) = \lim\limits_{h \to 0} \frac{3 - 3}{h}$

$f'(3) = \lim\limits_{h \to 0} \frac{0}{h}$

Again, for $h \neq 0$, $\frac{0}{h} = 0$.

$f'(3) = \lim\limits_{h \to 0} 0$

$f'(3) = 0$

Alternate Solution (Using Derivative Rules):

The derivative of a constant function $f(x) = c$ is always $f'(x) = 0$.

In this case, $f(x) = 3$, which is a constant function.

So, its derivative is $f'(x) = 0$ for all values of $x$.

Therefore, the derivative at $x=0$ is $f'(0) = 0$.

The derivative at $x=3$ is $f'(3) = 0$.

The derivative of $f(x) = 3$ at $x=0$ is $0$.

The derivative of $f(x) = 3$ at $x=3$ is also $0$.

Given:

The function is $f(x) = 10x$.

To Find:

The derivative of $f(x)$, denoted as $f'(x)$.

Solution:

The derivative of a function $f(x)$ with respect to $x$ is defined by the limit:

$f'(x) = \lim\limits_{h \to 0} \frac{f(x+h) - f(x)}{h}$

First, let's find $f(x+h)$:

$f(x+h) = 10(x+h) = 10x + 10h$

The function value $f(x)$ is given as $10x$.

Now, form the difference $f(x+h) - f(x)$:

$f(x+h) - f(x) = (10x + 10h) - (10x)$

$f(x+h) - f(x) = 10x + 10h - 10x = 10h$

Next, form the difference quotient $\frac{f(x+h) - f(x)}{h}$:

$\frac{f(x+h) - f(x)}{h} = \frac{10h}{h}$

For the limit as $h \to 0$, $h$ is a non-zero value approaching $0$. Thus, we can cancel $h$:

$\frac{\cancel{h}^{10}}{\cancel{h}_{1}} = 10$

Now, evaluate the limit as $h \to 0$:

$f'(x) = \lim\limits_{h \to 0} \frac{f(x+h) - f(x)}{h} = \lim\limits_{h \to 0} 10$

The limit of a constant is the constant itself.

$f'(x) = 10$

Alternate Solution (Using Derivative Rules):

We use the constant multiple rule and the power rule for differentiation.

The derivative of a constant times a function is the constant times the derivative of the function: $\frac{d}{dx}(c \cdot g(x)) = c \cdot g'(x)$.

The derivative of $x$ is $1$: $\frac{d}{dx}(x) = 1$ (using the power rule $\frac{d}{dx}(x^1) = 1 \cdot x^{1-1} = x^0 = 1$).

For $f(x) = 10x$, we have $c=10$ and $g(x) = x$.

$f'(x) = \frac{d}{dx}(10x)$

$f'(x) = 10 \cdot \frac{d}{dx}(x)$

$f'(x) = 10 \cdot 1$

$f'(x) = 10$

The derivative of the function $f(x) = 10x$ is $10$.

Given:

The function is $f(x) = x^2$.

To Find:

The derivative of $f(x)$, denoted as $f'(x)$.

Solution (using the First Principle):

The derivative of a function $f(x)$ with respect to $x$ is defined by the limit:

$f'(x) = \lim\limits_{h \to 0} \frac{f(x+h) - f(x)}{h}$

First, let's find $f(x+h)$:

$f(x+h) = (x+h)^2$

Using the binomial expansion $(a+b)^2 = a^2 + 2ab + b^2$:

$f(x+h) = x^2 + 2xh + h^2$

The function value $f(x)$ is given as $x^2$.

Now, form the difference $f(x+h) - f(x)$:

$f(x+h) - f(x) = (x^2 + 2xh + h^2) - (x^2)$

$f(x+h) - f(x) = x^2 + 2xh + h^2 - x^2 = 2xh + h^2$

Next, form the difference quotient $\frac{f(x+h) - f(x)}{h}$:

$\frac{f(x+h) - f(x)}{h} = \frac{2xh + h^2}{h}$

Factor out $h$ from the numerator:

$\frac{2xh + h^2}{h} = \frac{h(2x + h)}{h}$

For the limit as $h \to 0$, $h$ is a non-zero value approaching $0$. Thus, we can cancel $h$:

$\frac{\cancel{h}^{1}(2x + h)}{\cancel{h}_{1}} = 2x + h$

Now, evaluate the limit as $h \to 0$:

$f'(x) = \lim\limits_{h \to 0} \frac{f(x+h) - f(x)}{h} = \lim\limits_{h \to 0} (2x + h)$

Since $2x+h$ is a polynomial in $h$, we can find the limit by direct substitution of $h=0$:

$f'(x) = 2x + 0$

$f'(x) = 2x$

Alternate Solution (Using Derivative Rules):

We use the power rule for differentiation, which states that for $f(x) = x^n$, the derivative is $f'(x) = nx^{n-1}$.

In this case, $f(x) = x^2$, so $n=2$.

$f'(x) = \frac{d}{dx}(x^2)$

$f'(x) = 2 \cdot x^{2-1}$

$f'(x) = 2x^1$

$f'(x) = 2x$

The derivative of the function $f(x) = x^2$ is $2x$.

Given:

The function is $f(x) = a$, where $a$ is a fixed real number.

To Find:

The derivative of $f(x)$, denoted as $f'(x)$.

Solution (using the First Principle):

The derivative of a function $f(x)$ with respect to $x$ is defined by the limit:

$f'(x) = \lim\limits_{h \to 0} \frac{f(x+h) - f(x)}{h}$

Since $f(x) = a$ is a constant function, the value of the function is always $a$, regardless of the input value of $x$.

So, $f(x) = a$.

And $f(x+h) = a$.

Now, form the difference $f(x+h) - f(x)$:

$f(x+h) - f(x) = a - a = 0$

Next, form the difference quotient $\frac{f(x+h) - f(x)}{h}$:

$\frac{f(x+h) - f(x)}{h} = \frac{0}{h}$

For the limit as $h \to 0$, $h$ is a non-zero value approaching $0$. The expression $\frac{0}{h}$ for $h \neq 0$ is equal to $0$.

Now, evaluate the limit as $h \to 0$:

$f'(x) = \lim\limits_{h \to 0} \frac{f(x+h) - f(x)}{h} = \lim\limits_{h \to 0} 0$

The limit of a constant (which is $0$) is the constant itself.

$f'(x) = 0$

Alternate Solution (Using Derivative Rules):

A fundamental rule of differentiation states that the derivative of any constant function is always zero.

If $f(x) = c$, where $c$ is a constant, then $\frac{d}{dx}(c) = 0$.

In this case, $f(x) = a$, where $a$ is a fixed real number (a constant).

Therefore, the derivative of $f(x)$ is $0$.

$f'(x) = \frac{d}{dx}(a) = 0$

The derivative of the constant function $f(x) = a$ is $0$.

Given:

The function is $f(x) = \frac{1}{x}$.

To Find:

The derivative of $f(x)$, denoted as $f'(x)$.

Solution (using the First Principle):

The derivative of a function $f(x)$ with respect to $x$ is defined by the limit:

$f'(x) = \lim\limits_{h \to 0} \frac{f(x+h) - f(x)}{h}$

First, let's find $f(x+h)$:

$f(x+h) = \frac{1}{x+h}$

The function value $f(x)$ is given as $\frac{1}{x}$.

Now, form the difference $f(x+h) - f(x)$:

$f(x+h) - f(x) = \frac{1}{x+h} - \frac{1}{x}$

To subtract the fractions, find a common denominator, which is $x(x+h)$:

$\frac{1}{x+h} - \frac{1}{x} = \frac{1 \cdot x}{x(x+h)} - \frac{1 \cdot (x+h)}{x(x+h)}$

$= \frac{x - (x+h)}{x(x+h)}$

$= \frac{x - x - h}{x(x+h)}$

$= \frac{-h}{x(x+h)}$

Next, form the difference quotient $\frac{f(x+h) - f(x)}{h}$:

$\frac{f(x+h) - f(x)}{h} = \frac{\frac{-h}{x(x+h)}}{h}$

Dividing by $h$ is equivalent to multiplying by $\frac{1}{h}$:

$\frac{-h}{x(x+h)} \times \frac{1}{h}$

For the limit as $h \to 0$, $h$ is a non-zero value approaching $0$. Thus, we can cancel $h$ (provided $h \neq 0$):

$\frac{-\cancel{h}^{1}}{x(x+h) \cancel{h}_{1}} = \frac{-1}{x(x+h)}$

Now, evaluate the limit as $h \to 0$:

$f'(x) = \lim\limits_{h \to 0} \frac{-1}{x(x+h)}$

Since the expression is a rational function of $h$ and the denominator is non-zero at $h=0$ (assuming $x \neq 0$), we can find the limit by direct substitution of $h=0$:

$f'(x) = \frac{-1}{x(x+0)}$

$f'(x) = \frac{-1}{x(x)}$

$f'(x) = -\frac{1}{x^2}$

Alternate Solution (Using Derivative Rules):

We can rewrite the function using negative exponents: $f(x) = x^{-1}$.

We use the power rule for differentiation, which states that for $f(x) = x^n$, the derivative is $f'(x) = nx^{n-1}$.

In this case, $n=-1$.

$f'(x) = \frac{d}{dx}(x^{-1})$

$f'(x) = (-1) \cdot x^{-1-1}$

$f'(x) = -1 \cdot x^{-2}$

$f'(x) = -x^{-2}$

We can rewrite $x^{-2}$ as $\frac{1}{x^2}$:

$f'(x) = -\frac{1}{x^2}$

The derivative of the function $f(x) = \frac{1}{x}$ is $-\frac{1}{x^2}$.

Given:

The function is $f(x) = 6x^{100} - x^{55} + x$.

To Find:

The derivative of $f(x)$, denoted as $f'(x)$ or $\frac{d}{dx}(f(x))$.

Solution:

We will use the rules of differentiation to find the derivative of the given function. The relevant rules are:

1. The **Sum/Difference Rule**: The derivative of a sum or difference of functions is the sum or difference of their derivatives: $\frac{d}{dx}(u(x) \pm v(x)) = \frac{d}{dx}(u(x)) \pm \frac{d}{dx}(v(x))$.

2. The **Constant Multiple Rule**: The derivative of a constant times a function is the constant times the derivative of the function: $\frac{d}{dx}(c \cdot g(x)) = c \cdot \frac{d}{dx}(g(x))$.

3. The **Power Rule**: The derivative of $x^n$ is $nx^{n-1}$, where $n$ is any real number: $\frac{d}{dx}(x^n) = nx^{n-1}$.

We need to find $\frac{d}{dx}(6x^{100} - x^{55} + x)$.

Using the Sum/Difference Rule, we can differentiate each term separately:

$f'(x) = \frac{d}{dx}(6x^{100}) - \frac{d}{dx}(x^{55}) + \frac{d}{dx}(x)$

Now, we differentiate each term:

For the first term, $\frac{d}{dx}(6x^{100})$, we use the Constant Multiple Rule and the Power Rule ($n=100$):

$\frac{d}{dx}(6x^{100}) = 6 \cdot \frac{d}{dx}(x^{100})$

Using the Power Rule: $\frac{d}{dx}(x^{100}) = 100x^{100-1} = 100x^{99}$.

So, $\frac{d}{dx}(6x^{100}) = 6 \cdot (100x^{99}) = 600x^{99}$.

For the second term, $\frac{d}{dx}(x^{55})$, we use the Power Rule ($n=55$):

$\frac{d}{dx}(x^{55}) = 55x^{55-1} = 55x^{54}$.

For the third term, $\frac{d}{dx}(x)$, we use the Power Rule ($n=1$):

$\frac{d}{dx}(x) = \frac{d}{dx}(x^1) = 1 \cdot x^{1-1} = 1 \cdot x^0$.

Since $x^0 = 1$ (for $x \neq 0$), we have:

$\frac{d}{dx}(x) = 1 \cdot 1 = 1$.

Now, we combine the derivatives of the individual terms:

$f'(x) = (600x^{99}) - (55x^{54}) + (1)$

$f'(x) = 600x^{99} - 55x^{54} + 1$

The derivative of the function $f(x) = 6x^{100} - x^{55} + x$ is $600x^{99} - 55x^{54} + 1$.

Given:

The function is $f(x) = 1 + x + x^2 + x^3 + \dots + x^{50}$.

To Find:

The derivative of $f(x)$ at $x=1$, denoted as $f'(1)$.

Solution:

To find the derivative of the function $f(x)$, we will use the rules of differentiation: the Sum Rule, the Constant Rule, and the Power Rule.

The function is a sum of terms: $f(x) = 1 + x^1 + x^2 + x^3 + \dots + x^{50}$.

The derivative of a sum of functions is the sum of their derivatives:

$f'(x) = \frac{d}{dx}(1 + x + x^2 + x^3 + \dots + x^{50})$

$f'(x) = \frac{d}{dx}(1) + \frac{d}{dx}(x) + \frac{d}{dx}(x^2) + \frac{d}{dx}(x^3) + \dots + \frac{d}{dx}(x^{50})$

We apply the derivative rules to each term:

Using the Constant Rule, $\frac{d}{dx}(c) = 0$:

$\frac{d}{dx}(1) = 0$

Using the Power Rule, $\frac{d}{dx}(x^n) = nx^{n-1}$:

$\frac{d}{dx}(x) = \frac{d}{dx}(x^1) = 1 \cdot x^{1-1} = 1 \cdot x^0 = 1$

$\frac{d}{dx}(x^2) = 2 \cdot x^{2-1} = 2x^1 = 2x$

$\frac{d}{dx}(x^3) = 3 \cdot x^{3-1} = 3x^2$

and so on, up to the last term:

$\frac{d}{dx}(x^{50}) = 50 \cdot x^{50-1} = 50x^{49}$

Combining these derivatives, we get the expression for $f'(x)$:

$f'(x) = 0 + 1 + 2x + 3x^2 + \dots + 50x^{49}$

$f'(x) = 1 + 2x + 3x^2 + \dots + 50x^{49}$

Now, we need to evaluate the derivative at $x = 1$. Substitute $x=1$ into the expression for $f'(x)$:

$f'(1) = 1 + 2(1) + 3(1)^2 + \dots + 50(1)^{49}$

Since any positive integer power of $1$ is $1$, this simplifies to:

$f'(1) = 1 + 2(1) + 3(1) + \dots + 50(1)$

$f'(1) = 1 + 2 + 3 + \dots + 50$

This is the sum of the first 50 positive integers. The formula for the sum of the first $n$ positive integers is $\sum\limits_{k=1}^{n} k = \frac{n(n+1)}{2}$.

Here, $n = 50$.

Sum $= \frac{50(50+1)}{2} = \frac{50 \times 51}{2}$

Sum $= \frac{2550}{2}$

Sum $= 1275$

So, $f'(1) = 1275$.

The derivative of $f(x) = 1 + x + x^2 + x^3 + \dots + x^{50}$ at $x = 1$ is $1275$.

Given:

The function is $f(x) = \frac{x + 1}{x}$.

To Find:

The derivative of $f(x)$, denoted as $f'(x)$ or $\frac{d}{dx}(f(x))$.

Solution (Method 1: Simplification):

We can simplify the function $f(x)$ by dividing each term in the numerator by the denominator:

$f(x) = \frac{x + 1}{x} = \frac{x}{x} + \frac{1}{x}$

For $x \neq 0$, $\frac{x}{x} = 1$.

So, $f(x) = 1 + \frac{1}{x}$.

We can rewrite $\frac{1}{x}$ using negative exponents: $\frac{1}{x} = x^{-1}$.

$f(x) = 1 + x^{-1}$

Now, we find the derivative of $f(x) = 1 + x^{-1}$ using the rules of differentiation (Sum Rule, Constant Rule, Power Rule).

$f'(x) = \frac{d}{dx}(1 + x^{-1})$

Using the Sum Rule:

$f'(x) = \frac{d}{dx}(1) + \frac{d}{dx}(x^{-1})$

Using the Constant Rule, $\frac{d}{dx}(c) = 0$:

$\frac{d}{dx}(1) = 0$

Using the Power Rule, $\frac{d}{dx}(x^n) = nx^{n-1}$ with $n = -1$:

$\frac{d}{dx}(x^{-1}) = (-1) \cdot x^{-1-1} = -1 \cdot x^{-2}$

So, $f'(x) = 0 + (-x^{-2}) = -x^{-2}$.

We can write $x^{-2}$ as $\frac{1}{x^2}$.

$f'(x) = -\frac{1}{x^2}$

Solution (Method 2: Using the Quotient Rule):

The Quotient Rule for differentiation states that if $f(x) = \frac{u(x)}{v(x)}$, then $f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2}$.

In this function, $u(x) = x + 1$ and $v(x) = x$.

First, find the derivatives of $u(x)$ and $v(x)$:

$u'(x) = \frac{d}{dx}(x + 1)$

Using the Sum Rule: $u'(x) = \frac{d}{dx}(x) + \frac{d}{dx}(1) = 1 + 0 = 1$.

$v'(x) = \frac{d}{dx}(x) = 1$.

Now, substitute $u(x), v(x), u'(x), v'(x)$ into the Quotient Rule formula:

$f'(x) = \frac{(1)(x) - (x + 1)(1)}{x^2}$

Simplify the numerator:

$f'(x) = \frac{x - (x + 1)}{x^2}$

$f'(x) = \frac{x - x - 1}{x^2}$

$f'(x) = \frac{-1}{x^2}$

Both methods yield the same result.

The derivative of the function $f(x) = \frac{x+1}{x}$ is $-\frac{1}{x^2}$.

Given:

The function is $f(x) = \sin x$.

To Find:

The derivative of $f(x)$, denoted as $f'(x)$ or $\frac{d}{dx}(\sin x)$.

Solution (using the First Principle):

The derivative of a function $f(x)$ with respect to $x$ is defined by the limit:

$f'(x) = \lim\limits_{h \to 0} \frac{f(x+h) - f(x)}{h}$

First, let's find $f(x+h)$:

$f(x+h) = \sin(x+h)$

The function value $f(x)$ is given as $\sin x$.

Now, form the difference $f(x+h) - f(x)$:

$f(x+h) - f(x) = \sin(x+h) - \sin x$

We use the trigonometric identity for the difference of sines: $\sin A - \sin B = 2 \cos\left(\frac{A+B}{2}\right) \sin\left(\frac{A-B}{2}\right)$.

Here, $A = x+h$ and $B = x$.

$A+B = (x+h) + x = 2x + h$

$\frac{A+B}{2} = \frac{2x+h}{2} = x + \frac{h}{2}$

$A-B = (x+h) - x = h$

$\frac{A-B}{2} = \frac{h}{2}$

So, $\sin(x+h) - \sin x = 2 \cos\left(x + \frac{h}{2}\right) \sin\left(\frac{h}{2}\right)$.

Next, form the difference quotient $\frac{f(x+h) - f(x)}{h}$:

$\frac{f(x+h) - f(x)}{h} = \frac{2 \cos\left(x + \frac{h}{2}\right) \sin\left(\frac{h}{2}\right)}{h}$

We can rearrange the terms to use the standard limit $\lim\limits_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$:

$\frac{f(x+h) - f(x)}{h} = \cos\left(x + \frac{h}{2}\right) \cdot \frac{2 \sin\left(\frac{h}{2}\right)}{h}$

$= \cos\left(x + \frac{h}{2}\right) \cdot \frac{\sin\left(\frac{h}{2}\right)}{\frac{h}{2}}$

Now, evaluate the limit as $h \to 0$:

$f'(x) = \lim\limits_{h \to 0} \left[ \cos\left(x + \frac{h}{2}\right) \cdot \frac{\sin\left(\frac{h}{2}\right)}{\frac{h}{2}} \right]$

Using the property that the limit of a product is the product of the limits:

$f'(x) = \lim\limits_{h \to 0} \cos\left(x + \frac{h}{2}\right) \cdot \lim\limits_{h \to 0} \frac{\sin\left(\frac{h}{2}\right)}{\frac{h}{2}}$

For the first limit, as $h \to 0$, $\frac{h}{2} \to 0$. Since the cosine function is continuous, we can substitute:

$\lim\limits_{h \to 0} \cos\left(x + \frac{h}{2}\right) = \cos(x + 0) = \cos x$

For the second limit, let $\theta = \frac{h}{2}$. As $h \to 0$, $\theta \to 0$. Using the standard limit $\lim\limits_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$:

$\lim\limits_{h \to 0} \frac{\sin\left(\frac{h}{2}\right)}{\frac{h}{2}} = \lim\limits_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$

Combining the limits:

$f'(x) = (\cos x) \cdot (1) = \cos x$

Alternate Solution (Using Derivative Rules):

The standard derivative of the sine function is cosine.

$\frac{d}{dx}(\sin x) = \cos x$

The derivative of the function $f(x) = \sin x$ is $\cos x$.

Given:

The function is $f(x) = \tan x$.

To Find:

The derivative of $f(x)$, denoted as $f'(x)$ or $\frac{d}{dx}(\tan x)$.

Solution (Using Derivative Rules):

We know that $\tan x = \frac{\sin x}{\cos x}$. Thus, $f(x)$ is a quotient of two functions, $u(x) = \sin x$ and $v(x) = \cos x$.

We will use the Quotient Rule for differentiation, which states that if $f(x) = \frac{u(x)}{v(x)}$, then the derivative is given by:

$f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2}$

First, find the derivatives of $u(x) = \sin x$ and $v(x) = \cos x$:

The derivative of $\sin x$ is $\cos x$.

$u'(x) = \frac{d}{dx}(\sin x) = \cos x$

The derivative of $\cos x$ is $-\sin x$.

$v'(x) = \frac{d}{dx}(\cos x) = -\sin x$

Now, substitute $u(x), v(x), u'(x), v'(x)$ into the Quotient Rule formula:

$f'(x) = \frac{(\cos x)(\cos x) - (\sin x)(-\sin x)}{(\cos x)^2}$

Simplify the numerator:

$f'(x) = \frac{\cos^2 x - (-\sin^2 x)}{\cos^2 x}$

$f'(x) = \frac{\cos^2 x + \sin^2 x}{\cos^2 x}$

Using the fundamental trigonometric identity $\sin^2 x + \cos^2 x = 1$:

$f'(x) = \frac{1}{\cos^2 x}$

Recall that $\sec x = \frac{1}{\cos x}$. Therefore, $\frac{1}{\cos^2 x} = \left(\frac{1}{\cos x}\right)^2 = \sec^2 x$.

$f'(x) = \sec^2 x$

The derivative of the function $f(x) = \tan x$ is $\sec^2 x$.

Given:

The function is $f(x) = \sin^2 x$.

To Find:

The derivative of $f(x)$, denoted as $f'(x)$ or $\frac{d}{dx}(\sin^2 x)$.

Solution (Using the Chain Rule):

The function $f(x) = \sin^2 x$ can be written as $f(x) = (\sin x)^2$. This is a composite function of the form $[g(x)]^n$, where $g(x) = \sin x$ and $n=2$.

We use the Chain Rule, which states that if $y = F(g(x))$, then $\frac{dy}{dx} = F'(g(x)) \cdot g'(x)$.

Let $u = g(x) = \sin x$. Then $f(x) = u^2$.

The Chain Rule can be written as $\frac{df}{dx} = \frac{df}{du} \cdot \frac{du}{dx}$.

First, find the derivative of $f$ with respect to $u$:

$f = u^2$

$\frac{df}{du} = \frac{d}{du}(u^2)$

Using the Power Rule $\frac{d}{du}(u^n) = nu^{n-1}$:

$\frac{df}{du} = 2u^{2-1} = 2u$

Next, find the derivative of $u$ with respect to $x$:

$u = \sin x$

$\frac{du}{dx} = \frac{d}{dx}(\sin x)$

The derivative of $\sin x$ is $\cos x$:

$\frac{du}{dx} = \cos x$

Now, apply the Chain Rule $\frac{df}{dx} = \frac{df}{du} \cdot \frac{du}{dx}$:

$f'(x) = (2u) \cdot (\cos x)$

Substitute back $u = \sin x$:

$f'(x) = 2(\sin x) \cdot (\cos x)$

$f'(x) = 2 \sin x \cos x$

We can also use the double angle identity $2 \sin x \cos x = \sin(2x)$.

$f'(x) = \sin(2x)$

The derivative of the function $f(x) = \sin^2 x$ is $2 \sin x \cos x$ or $\sin(2x)$.

Given:

The function is $f(x) = x^2 - 2$.

The point at which the derivative is to be found is $x = 10$.

To Find:

The derivative of $f(x)$ at $x=10$, denoted as $f'(10)$.

Solution:

The derivative of a function $f(x)$ at a specific point $x=c$ is defined using the limit formula:

$f'(c) = \lim\limits_{h \to 0} \frac{f(c+h) - f(c)}{h}$

In this problem, $c = 10$ and $f(x) = x^2 - 2$.

First, let's evaluate the function at $x = c = 10$:

$f(10) = (10)^2 - 2 = 100 - 2 = 98$

Next, let's evaluate the function at $x = c+h = 10+h$:

$f(10+h) = (10+h)^2 - 2$

Expand $(10+h)^2$:

$(10+h)^2 = 10^2 + 2(10)(h) + h^2 = 100 + 20h + h^2$

So, $f(10+h) = (100 + 20h + h^2) - 2 = 98 + 20h + h^2$.

Now, form the difference $f(10+h) - f(10)$:

$f(10+h) - f(10) = (98 + 20h + h^2) - (98)$

$f(10+h) - f(10) = 98 + 20h + h^2 - 98 = 20h + h^2$

Next, form the difference quotient $\frac{f(10+h) - f(10)}{h}$:

$\frac{f(10+h) - f(10)}{h} = \frac{20h + h^2}{h}$

Factor out $h$ from the numerator:

$\frac{20h + h^2}{h} = \frac{h(20 + h)}{h}$

For the limit as $h \to 0$, $h$ is a non-zero value approaching $0$. Thus, we can cancel $h$ from the numerator and denominator:

$\frac{\cancel{h}(20 + h)}{\cancel{h}} = 20 + h$

Finally, evaluate the limit as $h \to 0$:

$f'(10) = \lim\limits_{h \to 0} \frac{f(10+h) - f(10)}{h}$

$f'(10) = \lim\limits_{h \to 0} (20 + h)$

Since $20+h$ is a simple polynomial in $h$, we can evaluate the limit by substituting $h=0$:

$f'(10) = 20 + 0 = 20$

The derivative of $f(x) = x^2 - 2$ at $x = 10$ is $20$.

Given:

The function is $f(x) = x$.

The point at which the derivative is to be found is $x = 1$.

To Find:

The derivative of $f(x)$ at $x=1$, denoted as $f'(1)$.

Solution (using the First Principle):

The derivative of a function $f(x)$ at a point $x=c$ is defined by the limit:

$f'(c) = \lim\limits_{h \to 0} \frac{f(c+h) - f(c)}{h}$

In this problem, $c = 1$ and $f(x) = x$.

First, let's evaluate the function at $x = c = 1$:

$f(1) = 1$

Next, let's evaluate the function at $x = c+h = 1+h$:

$f(1+h) = 1+h$

Now, form the difference $f(1+h) - f(1)$:

$f(1+h) - f(1) = (1+h) - (1)$

$f(1+h) - f(1) = 1 + h - 1 = h$

Next, form the difference quotient $\frac{f(1+h) - f(1)}{h}$:

$\frac{f(1+h) - f(1)}{h} = \frac{h}{h}$

For the limit as $h \to 0$, $h$ is a non-zero value approaching $0$. Thus, we can simplify the fraction:

$\frac{\cancel{h}}{\cancel{h}} = 1$

Finally, evaluate the limit as $h \to 0$:

$f'(1) = \lim\limits_{h \to 0} \frac{f(1+h) - f(1)}{h}$

$f'(1) = \lim\limits_{h \to 0} 1$

The limit of a constant (which is $1$) is the constant itself.

$f'(1) = 1$

Alternate Solution (Using Derivative Rules):

The function is $f(x) = x$. Using the power rule $\frac{d}{dx}(x^n) = nx^{n-1}$ with $n=1$:

$f'(x) = \frac{d}{dx}(x^1) = 1 \cdot x^{1-1} = 1 \cdot x^0 = 1$

The derivative of $f(x) = x$ is $f'(x) = 1$ for all values of $x$.

Therefore, the derivative at $x=1$ is $f'(1) = 1$.

The derivative of $f(x) = x$ at $x=1$ is $1$.

Given:

The function is $f(x) = 99x$.

The point at which the derivative is to be found is $x = 100$.

To Find:

The derivative of $f(x)$ at $x=100$, denoted as $f'(100)$.

Solution (using the First Principle):

The derivative of a function $f(x)$ at a point $x=c$ is defined by the limit:

$f'(c) = \lim\limits_{h \to 0} \frac{f(c+h) - f(c)}{h}$

In this problem, $c = 100$ and $f(x) = 99x$.

First, let's evaluate the function at $x = c = 100$:

$f(100) = 99 \times 100 = 9900$

Next, let's evaluate the function at $x = c+h = 100+h$:

$f(100+h) = 99(100+h)$

$f(100+h) = 99 \times 100 + 99 \times h = 9900 + 99h$

Now, form the difference $f(100+h) - f(100)$:

$f(100+h) - f(100) = (9900 + 99h) - (9900)$

$f(100+h) - f(100) = 9900 + 99h - 9900 = 99h$

Next, form the difference quotient $\frac{f(100+h) - f(100)}{h}$:

$\frac{f(100+h) - f(100)}{h} = \frac{99h}{h}$

For the limit as $h \to 0$, $h$ is a non-zero value approaching $0$. Thus, we can cancel $h$ from the numerator and denominator:

$\frac{99\cancel{h}}{\cancel{h}} = 99$

Finally, evaluate the limit as $h \to 0$:

$f'(100) = \lim\limits_{h \to 0} \frac{f(100+h) - f(100)}{h}$

$f'(100) = \lim\limits_{h \to 0} 99$

The limit of a constant (which is $99$) is the constant itself.

$f'(100) = 99$

Alternate Solution (Using Derivative Rules):

The function is $f(x) = 99x$. Using the constant multiple rule and the power rule $\frac{d}{dx}(x^1) = 1$:

$f'(x) = \frac{d}{dx}(99x) = 99 \cdot \frac{d}{dx}(x) = 99 \cdot 1 = 99$

The derivative of $f(x) = 99x$ is $f'(x) = 99$ for all values of $x$.

Therefore, the derivative at $x=100$ is $f'(100) = 99$.

The derivative of $f(x) = 99x$ at $x = 100$ is $99$.

The derivative of a function $f(x)$ from the first principle is defined by the limit:

$f'(x) = \lim\limits_{h \to 0} \frac{f(x+h) - f(x)}{h}$

Part (i): $f(x) = x^3 - 27$

Given:

The function is $f(x) = x^3 - 27$.

To Find:

The derivative of $f(x)$ using the first principle.

Solution:

$f(x) = x^3 - 27$

$f(x+h) = (x+h)^3 - 27$

Using the binomial expansion $(a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$:

$f(x+h) = (x^3 + 3x^2h + 3xh^2 + h^3) - 27$

Form the difference $f(x+h) - f(x)$:

$f(x+h) - f(x) = (x^3 + 3x^2h + 3xh^2 + h^3 - 27) - (x^3 - 27)$

$f(x+h) - f(x) = x^3 + 3x^2h + 3xh^2 + h^3 - 27 - x^3 + 27$

$f(x+h) - f(x) = 3x^2h + 3xh^2 + h^3$

Form the difference quotient $\frac{f(x+h) - f(x)}{h}$:

$\frac{f(x+h) - f(x)}{h} = \frac{3x^2h + 3xh^2 + h^3}{h}$

Factor out $h$ from the numerator:

$\frac{h(3x^2 + 3xh + h^2)}{h}$

For $h \neq 0$, cancel $h$:

$3x^2 + 3xh + h^2$

Evaluate the limit as $h \to 0$:

$f'(x) = \lim\limits_{h \to 0} (3x^2 + 3xh + h^2)$

$f'(x) = 3x^2 + 3x(0) + (0)^2$

$f'(x) = 3x^2 + 0 + 0 = 3x^2$

The derivative of $f(x) = x^3 - 27$ is $3x^2$.

Part (ii): $f(x) = (x - 1)(x - 2)$

Given:

The function is $f(x) = (x - 1)(x - 2)$.

To Find:

The derivative of $f(x)$ using the first principle.

Solution:

First, expand the function:

$f(x) = x^2 - 2x - x + 2 = x^2 - 3x + 2$

$f(x+h) = (x+h)^2 - 3(x+h) + 2$

$f(x+h) = (x^2 + 2xh + h^2) - (3x + 3h) + 2$

$f(x+h) = x^2 + 2xh + h^2 - 3x - 3h + 2$

Form the difference $f(x+h) - f(x)$:

$f(x+h) - f(x) = (x^2 + 2xh + h^2 - 3x - 3h + 2) - (x^2 - 3x + 2)$

$f(x+h) - f(x) = x^2 + 2xh + h^2 - 3x - 3h + 2 - x^2 + 3x - 2$

Cancel terms:

$f(x+h) - f(x) = 2xh + h^2 - 3h$

Form the difference quotient $\frac{f(x+h) - f(x)}{h}$:

$\frac{f(x+h) - f(x)}{h} = \frac{2xh + h^2 - 3h}{h}$

Factor out $h$ from the numerator:

$\frac{h(2x + h - 3)}{h}$

For $h \neq 0$, cancel $h$:

$2x + h - 3$

Evaluate the limit as $h \to 0$:

$f'(x) = \lim\limits_{h \to 0} (2x + h - 3)$

$f'(x) = 2x + 0 - 3$

$f'(x) = 2x - 3$

The derivative of $f(x) = (x - 1)(x - 2)$ is $2x - 3$.

Part (iii): $f(x) = \frac{1}{x^{2}}$

Given:

The function is $f(x) = \frac{1}{x^2}$.

To Find:

The derivative of $f(x)$ using the first principle.

Solution:

$f(x) = \frac{1}{x^2}$

$f(x+h) = \frac{1}{(x+h)^2}$

Form the difference $f(x+h) - f(x)$:

$f(x+h) - f(x) = \frac{1}{(x+h)^2} - \frac{1}{x^2}$

Find a common denominator $x^2 (x+h)^2$:

$f(x+h) - f(x) = \frac{1 \cdot x^2}{x^2 (x+h)^2} - \frac{1 \cdot (x+h)^2}{x^2 (x+h)^2}$

$= \frac{x^2 - (x+h)^2}{x^2 (x+h)^2}$

Expand $(x+h)^2 = x^2 + 2xh + h^2$:

$= \frac{x^2 - (x^2 + 2xh + h^2)}{x^2 (x+h)^2}$

$= \frac{x^2 - x^2 - 2xh - h^2}{x^2 (x+h)^2}$

$= \frac{-2xh - h^2}{x^2 (x+h)^2}$

Form the difference quotient $\frac{f(x+h) - f(x)}{h}$:

$\frac{f(x+h) - f(x)}{h} = \frac{\frac{-2xh - h^2}{x^2 (x+h)^2}}{h}$

Dividing by $h$ is multiplying by $\frac{1}{h}$:

$= \frac{-2xh - h^2}{x^2 (x+h)^2} \cdot \frac{1}{h}$

Factor out $h$ from the numerator:

$= \frac{h(-2x - h)}{x^2 (x+h)^2} \cdot \frac{1}{h}$

For $h \neq 0$, cancel $h$:

$= \frac{-2x - h}{x^2 (x+h)^2}$

Evaluate the limit as $h \to 0$:

$f'(x) = \lim\limits_{h \to 0} \frac{-2x - h}{x^2 (x+h)^2}$

Substitute $h=0$:

$f'(x) = \frac{-2x - 0}{x^2 (x+0)^2}$

$f'(x) = \frac{-2x}{x^2 (x^2)}$

$f'(x) = \frac{-2x}{x^4}$

Simplify the fraction (cancel $x$):

$f'(x) = \frac{-2}{x^3}$

The derivative of $f(x) = \frac{1}{x^2}$ is $-\frac{2}{x^3}$.

Part (iv): $f(x) = \frac{ x \;+\; 1}{x \;-\; 1}$

Given:

The function is $f(x) = \frac{x + 1}{x - 1}$.

To Find:

The derivative of $f(x)$ using the first principle.

Solution:

$f(x) = \frac{x + 1}{x - 1}$

$f(x+h) = \frac{(x+h) + 1}{(x+h) - 1} = \frac{x+h+1}{x+h-1}$

Form the difference $f(x+h) - f(x)$:

$f(x+h) - f(x) = \frac{x+h+1}{x+h-1} - \frac{x+1}{x-1}$

Find a common denominator $(x+h-1)(x-1)$:

$= \frac{(x+h+1)(x-1) - (x+1)(x+h-1)}{(x+h-1)(x-1)}$

Expand the terms in the numerator:

$(x+h+1)(x-1) = x(x-1) + h(x-1) + 1(x-1) = x^2 - x + hx - h + x - 1 = x^2 + hx - h - 1$

$(x+1)(x+h-1) = x(x+h-1) + 1(x+h-1) = x^2 + xh - x + x + h - 1 = x^2 + xh + h - 1$

Substitute these back into the numerator difference:

Numerator $= (x^2 + hx - h - 1) - (x^2 + xh + h - 1)$

Numerator $= x^2 + hx - h - 1 - x^2 - xh - h + 1$

Numerator $= (x^2 - x^2) + (hx - xh) + (-h - h) + (-1 + 1)$

Numerator $= 0 + 0 - 2h + 0 = -2h$

So, $f(x+h) - f(x) = \frac{-2h}{(x+h-1)(x-1)}$.

Form the difference quotient $\frac{f(x+h) - f(x)}{h}$:

$\frac{f(x+h) - f(x)}{h} = \frac{\frac{-2h}{(x+h-1)(x-1)}}{h}$

Dividing by $h$ is multiplying by $\frac{1}{h}$:

$= \frac{-2h}{(x+h-1)(x-1)} \cdot \frac{1}{h}$

For $h \neq 0$, cancel $h$:

$= \frac{-2}{(x+h-1)(x-1)}$

Evaluate the limit as $h \to 0$:

$f'(x) = \lim\limits_{h \to 0} \frac{-2}{(x+h-1)(x-1)}$

Substitute $h=0$:

$f'(x) = \frac{-2}{(x+0-1)(x-1)}$

$f'(x) = \frac{-2}{(x-1)(x-1)}$

$f'(x) = \frac{-2}{(x-1)^2}$

The derivative of $f(x) = \frac{x+1}{x-1}$ is $-\frac{2}{(x-1)^2}$.

Given:

The function is $f(x) = \frac{x^{100}}{100} + \frac{x^{99}}{99} + \dots + \frac{x^{2}}{2} + x + 1$.

To Prove:

$f'(1) = 100 f'(0)$

Solution:

To find the derivative $f'(x)$, we differentiate each term of the polynomial function $f(x)$. We use the Sum Rule, Constant Multiple Rule, and Power Rule for differentiation.

$f(x) = \frac{1}{100}x^{100} + \frac{1}{99}x^{99} + \dots + \frac{1}{2}x^2 + x^1 + 1$

The derivative $f'(x)$ is given by:

$f'(x) = \frac{d}{dx}\left(\frac{x^{100}}{100}\right) + \frac{d}{dx}\left(\frac{x^{99}}{99}\right) + \dots + \frac{d}{dx}\left(\frac{x^{2}}{2}\right) + \frac{d}{dx}(x) + \frac{d}{dx}(1)$

Applying the rules:

$\frac{d}{dx}\left(\frac{x^{100}}{100}\right) = \frac{1}{100} \cdot 100x^{100-1} = x^{99}$

$\frac{d}{dx}\left(\frac{x^{99}}{99}\right) = \frac{1}{99} \cdot 99x^{99-1} = x^{98}$

This pattern continues for each term down to $x^2$.

$\frac{d}{dx}\left(\frac{x^{k}}{k}\right) = \frac{1}{k} \cdot kx^{k-1} = x^{k-1}$

For the term $\frac{x^2}{2}$ (where $k=2$):

$\frac{d}{dx}\left(\frac{x^{2}}{2}\right) = \frac{1}{2} \cdot 2x^{2-1} = x^{1} = x$

For the term $x$:

$\frac{d}{dx}(x) = 1$

For the constant term $1$:

$\frac{d}{dx}(1) = 0$

Summing these derivatives, we get the expression for $f'(x)$:

$f'(x) = x^{99} + x^{98} + \dots + x^1 + 1 + 0$

$f'(x) = 1 + x + x^2 + \dots + x^{98} + x^{99}$

Now, evaluate $f'(x)$ at $x = 1$:

$f'(1) = 1 + (1) + (1)^2 + \dots + (1)^{98} + (1)^{99}$

Since $(1)^n = 1$ for any positive integer $n$, each term in the sum is $1$.

The sum is $1$ plus $1$ repeated $99$ times (from $x^1$ to $x^{99}$). So there are $1 + 99 = 100$ terms in total.

$f'(1) = \underbrace{1 + 1 + 1 + \dots + 1}_{100 \text{ terms}}$

$f'(1) = 100$

Next, evaluate $f'(x)$ at $x = 0$:

$f'(0) = 1 + (0) + (0)^2 + \dots + (0)^{98} + (0)^{99}$

Any positive integer power of $0$ is $0$.

$f'(0) = 1 + 0 + 0 + \dots + 0 + 0$

$f'(0) = 1$

Now, we check the relation $f'(1) = 100 f'(0)$.

Substitute the calculated values:

$100 = 100 \times 1$

$100 = 100$

The equality holds true.

Therefore, $f'(1) = 100 f'(0)$ is proven.

Given:

The function $f(x) = x^n + ax^{n-1} + a^2x^{n-2} + \dots + a^{n-1}x + a^n$.

Here, $n$ is a positive integer (implied by the polynomial form) and $a$ is a fixed real number (constant).

To Find:

The derivative of $f(x)$, denoted as $f'(x)$ or $\frac{d}{dx}(f(x))$.

Solution:

The given function is a polynomial, which is a sum of several terms. We can find the derivative of the sum of functions by finding the derivative of each term separately and then adding them together. This is based on the **Sum Rule** of differentiation.

For each term of the form $c \cdot x^k$, where $c$ is a constant and $k$ is a power of $x$, we use the **Constant Multiple Rule** ($\frac{d}{dx}(c \cdot g(x)) = c \cdot \frac{d}{dx}(g(x))$) and the **Power Rule** ($\frac{d}{dx}(x^k) = kx^{k-1}$).

Let's differentiate each term of $f(x)$: $f(x) = x^n + ax^{n-1} + a^2x^{n-2} + \dots + a^{n-1}x^1 + a^n$

1. Derivative of the first term, $x^n$:

$\frac{d}{dx}(x^n) = nx^{n-1}$ (using Power Rule with $k=n$)

2. Derivative of the second term, $ax^{n-1}$:

Here, the constant is $a$, and the power is $n-1$.

$\frac{d}{dx}(ax^{n-1}) = a \cdot \frac{d}{dx}(x^{n-1})$ (using Constant Multiple Rule)

$= a \cdot (n-1)x^{(n-1)-1}$ (using Power Rule with $k=n-1$)

$= a(n-1)x^{n-2}$

3. Derivative of the third term, $a^2x^{n-2}$:

Here, the constant is $a^2$, and the power is $n-2$.

$\frac{d}{dx}(a^2x^{n-2}) = a^2 \cdot \frac{d}{dx}(x^{n-2})$ (using Constant Multiple Rule)

$= a^2 \cdot (n-2)x^{(n-2)-1}$ (using Power Rule with $k=n-2$)

$= a^2(n-2)x^{n-3}$

This pattern continues for all terms up to the second-to-last term.

For a general term $a^k x^{n-k}$ (where $k$ goes from $0$ to $n-1$):

$\frac{d}{dx}(a^k x^{n-k}) = a^k \cdot (n-k)x^{n-k-1}$

4. Derivative of the second-to-last term, $a^{n-1}x$ (which is $a^{n-1}x^1$):

Here, the constant is $a^{n-1}$, and the power is 1.

$\frac{d}{dx}(a^{n-1}x^1) = a^{n-1} \cdot \frac{d}{dx}(x^1)$ (using Constant Multiple Rule)

$= a^{n-1} \cdot 1x^{1-1}$ (using Power Rule with $k=1$)

$= a^{n-1}x^0 = a^{n-1} \cdot 1 = a^{n-1}$

5. Derivative of the last term, $a^n$:

Since $a$ is a fixed real number and $n$ is a fixed positive integer, $a^n$ is a constant.

$\frac{d}{dx}(a^n) = 0$ (using Constant Rule)

Now, add the derivatives of all the terms to find $f'(x)$: $f'(x) = nx^{n-1} + a(n-1)x^{n-2} + a^2(n-2)x^{n-3} + \dots + a^{n-1} + 0$

The sum stops at the term $a^{n-1}$ because the derivative of the last term $a^n$ is 0.

Thus, the derivative of $f(x)$ is:

$f'(x) = nx^{n-1} + a(n-1)x^{n-2} + a^2(n-2)x^{n-3} + \dots + a^{n-1}$.

This can also be written using summation notation:

$f'(x) = \sum\limits_{k=0}^{n-1} a^k (n-k) x^{n-k-1}$.

Answer:

The derivative of the given function is $\mathbf{nx^{n-1} + a(n-1)x^{n-2} + a^2(n-2)x^{n-3} + \dots + a^{n-1}}$.

We will find the derivative for each function using standard differentiation rules, assuming $a$ and $b$ are fixed real number constants.

Part (i): $f(x) = (x - a)(x - b)$

Given:

The function is $f(x) = (x - a)(x - b)$.

To Find:

The derivative of $f(x)$, $f'(x)$.

Solution:

First, expand the function $f(x)$:

$f(x) = x(x - b) - a(x - b)$

$f(x) = x^2 - bx - ax + ab$

$f(x) = x^2 - (a+b)x + ab$

Now, differentiate $f(x)$ with respect to $x$ using the Sum/Difference Rule, Constant Multiple Rule, and Power Rule.

$f'(x) = \frac{d}{dx}(x^2) - \frac{d}{dx}((a+b)x) + \frac{d}{dx}(ab)$

Using the Power Rule $\frac{d}{dx}(x^n) = nx^{n-1}$:

$\frac{d}{dx}(x^2) = 2x^{2-1} = 2x$

Using the Constant Multiple Rule and Power Rule:

$\frac{d}{dx}((a+b)x) = (a+b) \frac{d}{dx}(x) = (a+b) \cdot 1 = a+b$

Using the Constant Rule $\frac{d}{dx}(c) = 0$ (since $a$ and $b$ are constants, $ab$ is a constant):

$\frac{d}{dx}(ab) = 0$

Combining the terms:

$f'(x) = 2x - (a+b) + 0$

$f'(x) = 2x - a - b$

The derivative of $f(x) = (x - a)(x - b)$ is $2x - a - b$.

Part (ii): $f(x) = (ax^2 + b)^2$

Given:

The function is $f(x) = (ax^2 + b)^2$.

To Find:

The derivative of $f(x)$, $f'(x)$.

Solution:

First, expand the function $f(x)$:

$f(x) = (ax^2)^2 + 2(ax^2)(b) + b^2$

$f(x) = a^2x^4 + 2abx^2 + b^2$

Now, differentiate $f(x)$ with respect to $x$ using the Sum Rule, Constant Multiple Rule, and Power Rule.

$f'(x) = \frac{d}{dx}(a^2x^4) + \frac{d}{dx}(2abx^2) + \frac{d}{dx}(b^2)$

Using the Constant Multiple Rule and Power Rule:

$\frac{d}{dx}(a^2x^4) = a^2 \frac{d}{dx}(x^4) = a^2 (4x^{4-1}) = 4a^2x^3$

$\frac{d}{dx}(2abx^2) = 2ab \frac{d}{dx}(x^2) = 2ab (2x^{2-1}) = 2ab (2x) = 4abx$

Using the Constant Rule $\frac{d}{dx}(c) = 0$ (since $b$ is a constant, $b^2$ is a constant):

$\frac{d}{dx}(b^2) = 0$

Combining the terms:

$f'(x) = 4a^2x^3 + 4abx + 0$

$f'(x) = 4a^2x^3 + 4abx$

We can factor out $4ax$ from the expression:

$f'(x) = 4ax(ax^2 + b)$

The derivative of $f(x) = (ax^2 + b)^2$ is $4a^2x^3 + 4abx$ or $4ax(ax^2 + b)$.

Part (iii): $f(x) = \frac{ x \;-\; a}{x \;-\;b}$

Given:

The function is $f(x) = \frac{x - a}{x - b}$, for $x \neq b$.

To Find:

The derivative of $f(x)$, $f'(x)$.

Solution:

We use the Quotient Rule for differentiation, which states that if $f(x) = \frac{u(x)}{v(x)}$, then $f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2}$.

Let $u(x) = x - a$ and $v(x) = x - b$.

Find the derivatives of $u(x)$ and $v(x)$ with respect to $x$:

$u'(x) = \frac{d}{dx}(x - a) = \frac{d}{dx}(x) - \frac{d}{dx}(a)$

Using the Power Rule $\frac{d}{dx}(x) = 1$ and the Constant Rule $\frac{d}{dx}(a) = 0$:

$u'(x) = 1 - 0 = 1$

Similarly, for $v(x)$:

$v'(x) = \frac{d}{dx}(x - b) = \frac{d}{dx}(x) - \frac{d}{dx}(b)$

$v'(x) = 1 - 0 = 1$

Now, apply the Quotient Rule formula:

$f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2}$

$f'(x) = \frac{(1)(x - b) - (x - a)(1)}{(x - b)^2}$

Simplify the numerator:

$f'(x) = \frac{x - b - (x - a)}{(x - b)^2}$

$f'(x) = \frac{x - b - x + a}{(x - b)^2}$

$f'(x) = \frac{(x - x) + (a - b)}{(x - b)^2}$

$f'(x) = \frac{0 + a - b}{(x - b)^2}$

$f'(x) = \frac{a - b}{(x - b)^2}$

The derivative of $f(x) = \frac{x - a}{x - b}$ is $\frac{a - b}{(x - b)^2}$.

Given:

Here, $a$ is a constant and $n$ is a fixed number. The function is defined for $x \neq a$.

To Find:

The derivative of $f(x)$, denoted as $f'(x)$ or $\frac{d}{dx}\left(\frac{x^n - a^n}{x - a}\right)$.

Solution:

We need to find the derivative of a quotient of two functions. We can use the **Quotient Rule** of differentiation.

The Quotient Rule states that if $f(x) = \frac{u(x)}{v(x)}$, where $u(x)$ and $v(x)$ are differentiable functions and $v(x) \neq 0$, then the derivative of $f(x)$ is given by:

$f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2}$

In our case, the numerator is $u(x) = x^n - a^n$ and the denominator is $v(x) = x - a$.

First, find the derivatives of $u(x)$ and $v(x)$ with respect to $x$:

1. Find the derivative of the numerator, $u'(x) = \frac{d}{dx}(x^n - a^n)$.

Using the **Difference Rule** and the **Power Rule** ($\frac{d}{dx}(x^k) = kx^{k-1}$) and the **Constant Rule** ($\frac{d}{dx}(\text{constant}) = 0$), we get:

2. Find the derivative of the denominator, $v'(x) = \frac{d}{dx}(x - a)$.

Using the **Difference Rule** and the derivatives of $x$ (which is 1) and a constant (which is 0), we get:

Now, apply the Quotient Rule formula with $u(x) = x^n - a^n$, $v(x) = x - a$, $u'(x) = nx^{n-1}$, and $v'(x) = 1$:

$f'(x) = \frac{(nx^{n-1})(x - a) - (x^n - a^n)(1)}{(x - a)^2}$

Expand the numerator:

Numerator $= nx^{n-1} \times x - nx^{n-1} \times a - (x^n - a^n) \times 1$

Using the exponent rule $x^m \times x^p = x^{m+p}$:

Numerator $= nx^{(n-1)+1} - anx^{n-1} - x^n + a^n$

Numerator $= nx^n - anx^{n-1} - x^n + a^n$

Substitute the expanded numerator back into the derivative formula:

This derivative is valid for all $x$ where the function is defined, i.e., for $x \neq a$.

Answer:

The derivative of $\frac{x^n - a^n}{x - a}$ is $\mathbf{\frac{nx^n - anx^{n-1} - x^n + a^n}{(x - a)^2}}$ for $x \neq a$.

We will find the derivative for each function using the rules of differentiation (Sum/Difference Rule, Constant Multiple Rule, Power Rule, Product Rule, Quotient Rule).

Part (i): $f(x) = 2x - \frac{3}{4}$

Given:

The function is $f(x) = 2x - \frac{3}{4}$.

To Find:

The derivative of $f(x)$, $f'(x)$.

Solution:

Using the Difference Rule and Constant Multiple Rule:

$f'(x) = \frac{d}{dx}(2x - \frac{3}{4}) = \frac{d}{dx}(2x) - \frac{d}{dx}(\frac{3}{4})$

Using the Constant Multiple Rule and Power Rule for the first term:

$\frac{d}{dx}(2x) = 2 \cdot \frac{d}{dx}(x) = 2 \cdot 1 = 2$

Using the Constant Rule for the second term:

$\frac{d}{dx}(\frac{3}{4}) = 0$

Combining the terms:

$f'(x) = 2 - 0 = 2$

The derivative of $f(x) = 2x - \frac{3}{4}$ is $2$.

Part (ii): $f(x) = (5x^3 + 3x - 1) (x - 1)$

Given:

The function is $f(x) = (5x^3 + 3x - 1) (x - 1)$.

To Find:

The derivative of $f(x)$, $f'(x)$.

Solution (Method 1: Expanding first):

Expand the function:

$f(x) = 5x^3(x) + 5x^3(-1) + 3x(x) + 3x(-1) - 1(x) - 1(-1)$

$f(x) = 5x^4 - 5x^3 + 3x^2 - 3x - x + 1$

$f(x) = 5x^4 - 5x^3 + 3x^2 - 4x + 1$

Differentiate term by term using the Sum/Difference Rule, Constant Multiple Rule, and Power Rule:

$f'(x) = \frac{d}{dx}(5x^4) - \frac{d}{dx}(5x^3) + \frac{d}{dx}(3x^2) - \frac{d}{dx}(4x) + \frac{d}{dx}(1)$

$f'(x) = 5(4x^3) - 5(3x^2) + 3(2x^1) - 4(1x^0) + 0$

$f'(x) = 20x^3 - 15x^2 + 6x - 4$

Alternate Solution (Method 2: Using the Product Rule):

Let $u(x) = 5x^3 + 3x - 1$ and $v(x) = x - 1$.

Find the derivatives $u'(x)$ and $v'(x)$:

$u'(x) = \frac{d}{dx}(5x^3 + 3x - 1) = 15x^2 + 3 - 0 = 15x^2 + 3$

$v'(x) = \frac{d}{dx}(x - 1) = 1 - 0 = 1$

Using the Product Rule $(uv)' = u'v + uv'$:

$f'(x) = (15x^2 + 3)(x - 1) + (5x^3 + 3x - 1)(1)$

Expand and simplify:

$f'(x) = (15x^3 - 15x^2 + 3x - 3) + (5x^3 + 3x - 1)$

$f'(x) = 15x^3 + 5x^3 - 15x^2 + 3x + 3x - 3 - 1$

$f'(x) = 20x^3 - 15x^2 + 6x - 4$

Both methods yield the same result.

The derivative of $f(x) = (5x^3 + 3x - 1) (x - 1)$ is $20x^3 - 15x^2 + 6x - 4$.

Part (iii): $f(x) = x^{-3} (5 + 3x)$

Given:

The function is $f(x) = x^{-3} (5 + 3x)$.

To Find:

The derivative of $f(x)$, $f'(x)$.

Solution (Method 1: Expanding first):

Expand the function:

$f(x) = x^{-3} \cdot 5 + x^{-3} \cdot 3x$

$f(x) = 5x^{-3} + 3x^{-3+1}$

$f(x) = 5x^{-3} + 3x^{-2}$

Differentiate term by term using the Sum Rule, Constant Multiple Rule, and Power Rule:

$f'(x) = \frac{d}{dx}(5x^{-3}) + \frac{d}{dx}(3x^{-2})$

$f'(x) = 5(-3x^{-3-1}) + 3(-2x^{-2-1})$

$f'(x) = 5(-3x^{-4}) + 3(-2x^{-3})$

$f'(x) = -15x^{-4} - 6x^{-3}$

Alternate Solution (Method 2: Using the Product Rule):

Let $u(x) = x^{-3}$ and $v(x) = 5 + 3x$.

Find the derivatives $u'(x)$ and $v'(x)$:

$u'(x) = \frac{d}{dx}(x^{-3}) = -3x^{-3-1} = -3x^{-4}$

$v'(x) = \frac{d}{dx}(5 + 3x) = 0 + 3 = 3$

Using the Product Rule $(uv)' = u'v + uv'$:

$f'(x) = (-3x^{-4})(5 + 3x) + (x^{-3})(3)$

Expand:

$f'(x) = -3x^{-4} \cdot 5 + (-3x^{-4}) \cdot 3x + x^{-3} \cdot 3$

$f'(x) = -15x^{-4} - 9x^{-4+1} + 3x^{-3}$

$f'(x) = -15x^{-4} - 9x^{-3} + 3x^{-3}$

Combine terms with $x^{-3}$:

$f'(x) = -15x^{-4} + (-9 + 3)x^{-3}$

$f'(x) = -15x^{-4} - 6x^{-3}$

Both methods yield the same result. We can also write the answer with positive exponents: $f'(x) = -\frac{15}{x^4} - \frac{6}{x^3}$.

The derivative of $f(x) = x^{-3} (5 + 3x)$ is $-15x^{-4} - 6x^{-3}$.

Part (iv): $f(x) = x^{5} (3 - 6x^{-9})$

Given:

The function is $f(x) = x^{5} (3 - 6x^{-9})$.

To Find:

The derivative of $f(x)$, $f'(x)$.

Solution (Expanding first):

Expand the function:

$f(x) = x^5 \cdot 3 - x^5 \cdot 6x^{-9}$

$f(x) = 3x^5 - 6x^{5-9}$

$f(x) = 3x^5 - 6x^{-4}$

Differentiate term by term using the Difference Rule, Constant Multiple Rule, and Power Rule:

$f'(x) = \frac{d}{dx}(3x^5) - \frac{d}{dx}(6x^{-4})$

$f'(x) = 3(5x^{5-1}) - 6(-4x^{-4-1})$

$f'(x) = 3(5x^4) - 6(-4x^{-5})$

$f'(x) = 15x^4 + 24x^{-5}$

The derivative of $f(x) = x^{5} (3 - 6x^{-9})$ is $15x^4 + 24x^{-5}$.

Part (v): $f(x) = x^{-4} (3 - 4x^{-5})$

Given:

The function is $f(x) = x^{-4} (3 - 4x^{-5})$.

To Find:

The derivative of $f(x)$, $f'(x)$.

Solution (Expanding first):

Expand the function:

$f(x) = x^{-4} \cdot 3 - x^{-4} \cdot 4x^{-5}$

$f(x) = 3x^{-4} - 4x^{-4-5}$

$f(x) = 3x^{-4} - 4x^{-9}$

Differentiate term by term using the Difference Rule, Constant Multiple Rule, and Power Rule:

$f'(x) = \frac{d}{dx}(3x^{-4}) - \frac{d}{dx}(4x^{-9})$

$f'(x) = 3(-4x^{-4-1}) - 4(-9x^{-9-1})$

$f'(x) = 3(-4x^{-5}) - 4(-9x^{-10})$

$f'(x) = -12x^{-5} + 36x^{-10}$

The derivative of $f(x) = x^{-4} (3 - 4x^{-5})$ is $-12x^{-5} + 36x^{-10}$.

Part (vi): $f(x) = \frac{2}{x \;+\; 1} - \frac{x^{2}}{3x\;-\;1}$

Given:

The function is $f(x) = \frac{2}{x + 1} - \frac{x^2}{3x - 1}$.

To Find:

The derivative of $f(x)$, $f'(x)$.

Solution:

We use the Difference Rule to differentiate each term separately:

$f'(x) = \frac{d}{dx}\left(\frac{2}{x + 1}\right) - \frac{d}{dx}\left(\frac{x^2}{3x - 1}\right)$

We use the Quotient Rule for each term.

For the first term, let $u(x) = 2$ and $v(x) = x + 1$.

$u'(x) = \frac{d}{dx}(2) = 0$

$v'(x) = \frac{d}{dx}(x + 1) = 1$

The derivative of the first term is:

$\frac{d}{dx}\left(\frac{2}{x + 1}\right) = \frac{u'v - uv'}{v^2} = \frac{(0)(x + 1) - (2)(1)}{(x + 1)^2} = \frac{0 - 2}{(x + 1)^2} = \frac{-2}{(x + 1)^2}$

For the second term, let $p(x) = x^2$ and $q(x) = 3x - 1$.

$p'(x) = \frac{d}{dx}(x^2) = 2x$

$q'(x) = \frac{d}{dx}(3x - 1) = 3$

The derivative of the second term is:

$\frac{d}{dx}\left(\frac{x^2}{3x - 1}\right) = \frac{p'q - pq'}{q^2} = \frac{(2x)(3x - 1) - (x^2)(3)}{(3x - 1)^2}$

$= \frac{6x^2 - 2x - 3x^2}{(3x - 1)^2} = \frac{3x^2 - 2x}{(3x - 1)^2}$

Now, subtract the second derivative from the first one:

$f'(x) = \frac{-2}{(x + 1)^2} - \frac{3x^2 - 2x}{(3x - 1)^2}$

The derivative of $f(x) = \frac{2}{x + 1} - \frac{x^2}{3x - 1}$ is $\frac{-2}{(x + 1)^2} - \frac{3x^2 - 2x}{(3x - 1)^2}$.

Given:

The function is $f(x) = \cos x$.

To Find:

The derivative of $f(x)$ from the first principle, denoted as $f'(x)$.

Solution:

The derivative of a function $f(x)$ from the first principle is defined by the limit:

$f'(x) = \lim\limits_{h \to 0} \frac{f(x+h) - f(x)}{h}$

First, let's find $f(x+h)$:

$f(x+h) = \cos(x+h)$

The function value $f(x)$ is given as $\cos x$.

Now, form the difference $f(x+h) - f(x)$:

$f(x+h) - f(x) = \cos(x+h) - \cos x$

We use the trigonometric identity for the difference of cosines: $\cos A - \cos B = -2 \sin\left(\frac{A+B}{2}\right) \sin\left(\frac{A-B}{2}\right)$.

Here, $A = x+h$ and $B = x$.

$\frac{A+B}{2} = \frac{(x+h) + x}{2} = \frac{2x+h}{2} = x + \frac{h}{2}$

$\frac{A-B}{2} = \frac{(x+h) - x}{2} = \frac{h}{2}$

So, $\cos(x+h) - \cos x = -2 \sin\left(x + \frac{h}{2}\right) \sin\left(\frac{h}{2}\right)$.

Next, form the difference quotient $\frac{f(x+h) - f(x)}{h}$:

$\frac{f(x+h) - f(x)}{h} = \frac{-2 \sin\left(x + \frac{h}{2}\right) \sin\left(\frac{h}{2}\right)}{h}$

We rearrange the terms to use the standard limit $\lim\limits_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$:

$\frac{f(x+h) - f(x)}{h} = -\sin\left(x + \frac{h}{2}\right) \cdot \frac{2 \sin\left(\frac{h}{2}\right)}{h}$

$= -\sin\left(x + \frac{h}{2}\right) \cdot \frac{\sin\left(\frac{h}{2}\right)}{\frac{h}{2}}$

Now, evaluate the limit as $h \to 0$:

$f'(x) = \lim\limits_{h \to 0} \left[ -\sin\left(x + \frac{h}{2}\right) \cdot \frac{\sin\left(\frac{h}{2}\right)}{\frac{h}{2}} \right]$

Using the property that the limit of a product is the product of the limits:

$f'(x) = \lim\limits_{h \to 0} \left( -\sin\left(x + \frac{h}{2}\right) \right) \cdot \lim\limits_{h \to 0} \left( \frac{\sin\left(\frac{h}{2}\right)}{\frac{h}{2}} \right)$

For the first limit, as $h \to 0$, $\frac{h}{2} \to 0$. Since the sine function is continuous, we can substitute:

$\lim\limits_{h \to 0} -\sin\left(x + \frac{h}{2}\right) = -\sin(x + 0) = -\sin x$

For the second limit, let $\theta = \frac{h}{2}$. As $h \to 0$, $\theta \to 0$. Using the standard limit $\lim\limits_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$:

$\lim\limits_{h \to 0} \frac{\sin\left(\frac{h}{2}\right)}{\frac{h}{2}} = \lim\limits_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$

Combining the limits:

$f'(x) = (-\sin x) \cdot (1) = -\sin x$

The derivative of the function $f(x) = \cos x$ from first principle is $-\sin x$.

We will find the derivative for each function using the rules of differentiation.

Part (i): $f(x) = \sin x \cos x$

Given:

The function is $f(x) = \sin x \cos x$.

To Find:

The derivative of $f(x)$, $f'(x)$.

Solution:

We use the Product Rule: If $f(x) = u(x)v(x)$, then $f'(x) = u'(x)v(x) + u(x)v'(x)$.

Let $u(x) = \sin x$ and $v(x) = \cos x$.

The derivatives are $u'(x) = \frac{d}{dx}(\sin x) = \cos x$ and $v'(x) = \frac{d}{dx}(\cos x) = -\sin x$.

Applying the Product Rule:

$f'(x) = (\cos x)(\cos x) + (\sin x)(-\sin x)$

$f'(x) = \cos^2 x - \sin^2 x$

Using the double angle identity $\cos(2x) = \cos^2 x - \sin^2 x$, we can also write the derivative as:

$f'(x) = \cos(2x)$

The derivative of $f(x) = \sin x \cos x$ is $\cos^2 x - \sin^2 x$ or $\cos(2x)$.

Part (ii): $f(x) = \sec x$

Given:

The function is $f(x) = \sec x$.

To Find:

The derivative of $f(x)$, $f'(x)$.

Solution:

We can write $\sec x = \frac{1}{\cos x}$. We use the Quotient Rule: If $f(x) = \frac{u(x)}{v(x)}$, then $f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2}$.

Let $u(x) = 1$ and $v(x) = \cos x$.

The derivatives are $u'(x) = \frac{d}{dx}(1) = 0$ and $v'(x) = \frac{d}{dx}(\cos x) = -\sin x$.

Applying the Quotient Rule:

$f'(x) = \frac{(0)(\cos x) - (1)(-\sin x)}{(\cos x)^2}$

$f'(x) = \frac{0 + \sin x}{\cos^2 x} = \frac{\sin x}{\cos^2 x}$

We can rewrite this as:

$f'(x) = \frac{\sin x}{\cos x} \cdot \frac{1}{\cos x}$

Since $\frac{\sin x}{\cos x} = \tan x$ and $\frac{1}{\cos x} = \sec x$:

$f'(x) = \tan x \sec x$

The derivative of $f(x) = \sec x$ is $\sec x \tan x$.

Part (iii): $f(x) = 5\sec x + 4\cos x$

Given:

The function is $f(x) = 5\sec x + 4\cos x$.

To Find:

The derivative of $f(x)$, $f'(x)$.

Solution:

We use the Sum Rule and Constant Multiple Rule: $\frac{d}{dx}(cu(x) + dv(x)) = c \frac{d}{dx}(u(x)) + d \frac{d}{dx}(v(x))$.

$f'(x) = \frac{d}{dx}(5\sec x) + \frac{d}{dx}(4\cos x)$

$f'(x) = 5 \frac{d}{dx}(\sec x) + 4 \frac{d}{dx}(\cos x)$

Using the known derivatives $\frac{d}{dx}(\sec x) = \sec x \tan x$ (from part ii) and $\frac{d}{dx}(\cos x) = -\sin x$:

$f'(x) = 5 (\sec x \tan x) + 4 (-\sin x)$

$f'(x) = 5\sec x \tan x - 4\sin x$

The derivative of $f(x) = 5\sec x + 4\cos x$ is $5\sec x \tan x - 4\sin x$.

Part (iv): $f(x) = \text{cosec } x$

Given:

The function is $f(x) = \text{cosec } x$.

To Find:

The derivative of $f(x)$, $f'(x)$.

Solution:

We can write $\text{cosec } x = \frac{1}{\sin x}$. We use the Quotient Rule: If $f(x) = \frac{u(x)}{v(x)}$, then $f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2}$.

Let $u(x) = 1$ and $v(x) = \sin x$.

The derivatives are $u'(x) = \frac{d}{dx}(1) = 0$ and $v'(x) = \frac{d}{dx}(\sin x) = \cos x$.

Applying the Quotient Rule:

$f'(x) = \frac{(0)(\sin x) - (1)(\cos x)}{(\sin x)^2}$

$f'(x) = \frac{0 - \cos x}{\sin^2 x} = \frac{-\cos x}{\sin^2 x}$

We can rewrite this as:

$f'(x) = -\frac{\cos x}{\sin x} \cdot \frac{1}{\sin x}$

Since $\frac{\cos x}{\sin x} = \cot x$ and $\frac{1}{\sin x} = \text{cosec } x$:

$f'(x) = -\cot x \text{ cosec } x$

The derivative of $f(x) = \text{cosec } x$ is $-\text{cosec } x \cot x$.

Part (v): $f(x) = 3\cot x + 5\text{cosec } x$

Given:

The function is $f(x) = 3\cot x + 5\text{cosec } x$.

To Find:

The derivative of $f(x)$, $f'(x)$.

Solution:

We use the Sum Rule and Constant Multiple Rule. We need the derivative of $\cot x$.

We can find the derivative of $\cot x = \frac{\cos x}{\sin x}$ using the Quotient Rule:

Let $u(x) = \cos x$ and $v(x) = \sin x$.

$u'(x) = -\sin x$ and $v'(x) = \cos x$.

$\frac{d}{dx}(\cot x) = \frac{(-\sin x)(\sin x) - (\cos x)(\cos x)}{(\sin x)^2} = \frac{-\sin^2 x - \cos^2 x}{\sin^2 x}$

$= \frac{-(\sin^2 x + \cos^2 x)}{\sin^2 x} = \frac{-1}{\sin^2 x}$

Since $\frac{1}{\sin x} = \text{cosec } x$, we have $\frac{d}{dx}(\cot x) = -\text{cosec}^2 x$.

Now, applying the Sum Rule and Constant Multiple Rule to the original function:

$f'(x) = \frac{d}{dx}(3\cot x) + \frac{d}{dx}(5\text{cosec } x)$

$f'(x) = 3 \frac{d}{dx}(\cot x) + 5 \frac{d}{dx}(\text{cosec } x)$

Using the known derivatives $\frac{d}{dx}(\cot x) = -\text{cosec}^2 x$ and $\frac{d}{dx}(\text{cosec } x) = -\text{cosec } x \cot x$ (from part iv):

$f'(x) = 3(-\text{cosec}^2 x) + 5(-\text{cosec } x \cot x)$

$f'(x) = -3\text{cosec}^2 x - 5\text{cosec } x \cot x$

The derivative of $f(x) = 3\cot x + 5\text{cosec } x$ is $-3\text{cosec}^2 x - 5\text{cosec } x \cot x$.

Part (vi): $f(x) = 5\sin x - 6\cos x + 7$

Given:

The function is $f(x) = 5\sin x - 6\cos x + 7$.

To Find:

The derivative of $f(x)$, $f'(x)$.

Solution:

We use the Sum/Difference Rule, Constant Multiple Rule, and Constant Rule.

$f'(x) = \frac{d}{dx}(5\sin x) - \frac{d}{dx}(6\cos x) + \frac{d}{dx}(7)$

$f'(x) = 5 \frac{d}{dx}(\sin x) - 6 \frac{d}{dx}(\cos x) + \frac{d}{dx}(7)$

Using the known derivatives $\frac{d}{dx}(\sin x) = \cos x$, $\frac{d}{dx}(\cos x) = -\sin x$, and $\frac{d}{dx}(7) = 0$:

$f'(x) = 5 (\cos x) - 6 (-\sin x) + 0$

$f'(x) = 5\cos x + 6\sin x$

The derivative of $f(x) = 5\sin x - 6\cos x + 7$ is $5\cos x + 6\sin x$.

Part (vii): $f(x) = 2\tan x - 7\sec x$

Given:

The function is $f(x) = 2\tan x - 7\sec x$.

To Find:

The derivative of $f(x)$, $f'(x)$.

Solution:

We use the Difference Rule and Constant Multiple Rule.

$f'(x) = \frac{d}{dx}(2\tan x) - \frac{d}{dx}(7\sec x)$

$f'(x) = 2 \frac{d}{dx}(\tan x) - 7 \frac{d}{dx}(\sec x)$

Using the known derivatives $\frac{d}{dx}(\tan x) = \sec^2 x$ (from Example 17) and $\frac{d}{dx}(\sec x) = \sec x \tan x$ (from part ii):

$f'(x) = 2 (\sec^2 x) - 7 (\sec x \tan x)$

$f'(x) = 2\sec^2 x - 7\sec x \tan x$

The derivative of $f(x) = 2\tan x - 7\sec x$ is $2\sec^2 x - 7\sec x \tan x$.

We need to find the derivative of the given functions from the first principle using the definition:

$f'(x) = \lim\limits_{h \to 0} \frac{f(x+h) - f(x)}{h}$

Part (i): $f(x) = \frac{2x \;+\; 3}{x \;-\; 2}$

Given:

The function is $f(x) = \frac{2x + 3}{x - 2}$.

To Find:

The derivative of $f(x)$ using the first principle.

Solution:

First, find $f(x+h)$:

$f(x+h) = \frac{2(x+h) + 3}{(x+h) - 2} = \frac{2x + 2h + 3}{x + h - 2}$

Next, find $f(x+h) - f(x)$:

$f(x+h) - f(x) = \frac{2x + 2h + 3}{x + h - 2} - \frac{2x + 3}{x - 2}$

Find a common denominator $(x+h-2)(x-2)$:

$= \frac{(2x + 2h + 3)(x - 2) - (2x + 3)(x + h - 2)}{(x+h-2)(x-2)}$

Expand the numerator:

Numerator $= (2x(x-2) + 2h(x-2) + 3(x-2)) - (2x(x+h-2) + 3(x+h-2))$

$= (2x^2 - 4x + 2hx - 4h + 3x - 6) - (2x^2 + 2xh - 4x + 3x + 3h - 6)$

$= (2x^2 - x + 2hx - 4h - 6) - (2x^2 - x + 2xh + 3h - 6)$

$= 2x^2 - x + 2hx - 4h - 6 - 2x^2 + x - 2xh - 3h + 6$

Combine like terms:

$= (2x^2 - 2x^2) + (-x + x) + (2hx - 2xh) + (-4h - 3h) + (-6 + 6)$

$= 0 + 0 + 0 - 7h + 0 = -7h$

So, $f(x+h) - f(x) = \frac{-7h}{(x+h-2)(x-2)}$.

Form the difference quotient $\frac{f(x+h) - f(x)}{h}$:

$\frac{f(x+h) - f(x)}{h} = \frac{\frac{-7h}{(x+h-2)(x-2)}}{h}$

$= \frac{-7h}{(x+h-2)(x-2)} \cdot \frac{1}{h}$

For $h \neq 0$, cancel $h$:

$= \frac{-7\cancel{h}}{(x+h-2)(x-2)\cancel{h}}$

$= \frac{-7}{(x+h-2)(x-2)}$

Evaluate the limit as $h \to 0$:

$f'(x) = \lim\limits_{h \to 0} \frac{-7}{(x+h-2)(x-2)}$

Substitute $h=0$:

$f'(x) = \frac{-7}{(x+0-2)(x-2)} = \frac{-7}{(x-2)(x-2)} = \frac{-7}{(x-2)^2}$

The derivative of $f(x) = \frac{2x + 3}{x - 2}$ is $-\frac{7}{(x-2)^2}$.

Part (ii): $f(x) = x + \frac{1}{x}$

Given:

The function is $f(x) = x + \frac{1}{x}$.

To Find:

The derivative of $f(x)$ using the first principle.

Solution:

First, find $f(x+h)$:

$f(x+h) = (x+h) + \frac{1}{x+h}$

Next, find $f(x+h) - f(x)$:

$f(x+h) - f(x) = \left(x + h + \frac{1}{x+h}\right) - \left(x + \frac{1}{x}\right)$

$= x + h + \frac{1}{x+h} - x - \frac{1}{x}$

$= h + \frac{1}{x+h} - \frac{1}{x}$

Combine the fractional terms by finding a common denominator $x(x+h)$:

$= h + \left(\frac{x}{x(x+h)} - \frac{x+h}{x(x+h)}\right)$

$= h + \frac{x - (x+h)}{x(x+h)}$

$= h + \frac{x - x - h}{x(x+h)}$

$= h + \frac{-h}{x(x+h)}$

Form the difference quotient $\frac{f(x+h) - f(x)}{h}$:

$\frac{f(x+h) - f(x)}{h} = \frac{h + \frac{-h}{x(x+h)}}{h}$

Split the fraction:

$= \frac{h}{h} + \frac{\frac{-h}{x(x+h)}}{h}$

$= 1 + \frac{-h}{x(x+h)} \cdot \frac{1}{h}$

For $h \neq 0$, cancel $h$:

$= 1 + \frac{-\cancel{h}}{x(x+h)\cancel{h}}$

$= 1 + \frac{-1}{x(x+h)}$

Evaluate the limit as $h \to 0$:

$f'(x) = \lim\limits_{h \to 0} \left(1 + \frac{-1}{x(x+h)}\right)$

Using the sum rule for limits:

$f'(x) = \lim\limits_{h \to 0} 1 + \lim\limits_{h \to 0} \frac{-1}{x(x+h)}$

$f'(x) = 1 + \frac{-1}{x(x+0)}$

$f'(x) = 1 + \frac{-1}{x^2}$

$f'(x) = 1 - \frac{1}{x^2}$

The derivative of $f(x) = x + \frac{1}{x}$ is $1 - \frac{1}{x^2}$.

We need to find the derivative of the given functions from the first principle using the definition:

$f'(x) = \lim\limits_{h \to 0} \frac{f(x+h) - f(x)}{h}$

Part (i): $f(x) = \sin x + \cos x$

Given:

The function is $f(x) = \sin x + \cos x$.

To Find:

The derivative of $f(x)$ using the first principle.

Solution:

First, find $f(x+h)$:

$f(x+h) = \sin(x+h) + \cos(x+h)$

Next, find $f(x+h) - f(x)$:

$f(x+h) - f(x) = (\sin(x+h) + \cos(x+h)) - (\sin x + \cos x)$

$f(x+h) - f(x) = (\sin(x+h) - \sin x) + (\cos(x+h) - \cos x)$

Form the difference quotient $\frac{f(x+h) - f(x)}{h}$:

$\frac{f(x+h) - f(x)}{h} = \frac{(\sin(x+h) - \sin x) + (\cos(x+h) - \cos x)}{h}$

$\frac{f(x+h) - f(x)}{h} = \frac{\sin(x+h) - \sin x}{h} + \frac{\cos(x+h) - \cos x}{h}$

Evaluate the limit as $h \to 0$:

$f'(x) = \lim\limits_{h \to 0} \left( \frac{\sin(x+h) - \sin x}{h} + \frac{\cos(x+h) - \cos x}{h} \right)$

Using the sum rule for limits:

$f'(x) = \lim\limits_{h \to 0} \frac{\sin(x+h) - \sin x}{h} + \lim\limits_{h \to 0} \frac{\cos(x+h) - \cos x}{h}$

The first limit is the definition of the derivative of $\sin x$, which is $\cos x$.

The second limit is the definition of the derivative of $\cos x$, which is $-\sin x$.

Therefore,

$f'(x) = \cos x + (-\sin x)$

$f'(x) = \cos x - \sin x$

The derivative of $f(x) = \sin x + \cos x$ is $\cos x - \sin x$.

Part (ii): $f(x) = x \sin x$

Given:

The function is $f(x) = x \sin x$.

To Find:

The derivative of $f(x)$ using the first principle.

Solution:

First, find $f(x+h)$:

$f(x+h) = (x+h)\sin(x+h)$

Next, find $f(x+h) - f(x)$:

$f(x+h) - f(x) = (x+h)\sin(x+h) - x\sin x$

Expand and rearrange terms to group $\sin x$ and $\cos x$:

Using $\sin(x+h) = \sin x \cos h + \cos x \sin h$:

$f(x+h) - f(x) = (x+h)(\sin x \cos h + \cos x \sin h) - x\sin x$

$= x\sin x \cos h + x\cos x \sin h + h\sin x \cos h + h\cos x \sin h - x\sin x$

Group terms with $\sin x$ and terms with $\cos x$:

$= (x\sin x \cos h - x\sin x) + (x\cos x \sin h) + (h\sin x \cos h) + (h\cos x \sin h)$

Factor:

$= x\sin x (\cos h - 1) + x\cos x \sin h + h\sin x \cos h + h\cos x \sin h$

Form the difference quotient $\frac{f(x+h) - f(x)}{h}$:

$\frac{f(x+h) - f(x)}{h} = \frac{x\sin x (\cos h - 1) + x\cos x \sin h + h\sin x \cos h + h\cos x \sin h}{h}$

Split the fraction:

$= x\sin x \frac{\cos h - 1}{h} + x\cos x \frac{\sin h}{h} + \sin x \cos h + \cos x \sin h$

Evaluate the limit as $h \to 0$:

$f'(x) = \lim\limits_{h \to 0} \left( x\sin x \frac{\cos h - 1}{h} + x\cos x \frac{\sin h}{h} + \sin x \cos h + \cos x \sin h \right)$

Using the properties of limits and standard limits $\lim\limits_{h \to 0} \frac{\cos h - 1}{h} = 0$, $\lim\limits_{h \to 0} \frac{\sin h}{h} = 1$, $\lim\limits_{h \to 0} \cos h = 1$, $\lim\limits_{h \to 0} \sin h = 0$:

$f'(x) = x\sin x \cdot \lim\limits_{h \to 0} \frac{\cos h - 1}{h} + x\cos x \cdot \lim\limits_{h \to 0} \frac{\sin h}{h} + \sin x \cdot \lim\limits_{h \to 0} \cos h + \cos x \cdot \lim\limits_{h \to 0} \sin h$

$f'(x) = x\sin x \cdot (0) + x\cos x \cdot (1) + \sin x \cdot (1) + \cos x \cdot (0)$

$f'(x) = 0 + x\cos x + \sin x + 0$

$f'(x) = x\cos x + \sin x$

The derivative of $f(x) = x \sin x$ is $x\cos x + \sin x$.

We will compute the derivative for each function using standard differentiation rules.

Part (i): $f(x) = \sin 2x$

Given:

The function is $f(x) = \sin 2x$.

To Find:

The derivative of $f(x)$, denoted as $f'(x)$ or $\frac{d}{dx}(\sin 2x)$.

Solution:

This is a composite function, so we use the Chain Rule. The function is of the form $\sin(u)$, where $u = 2x$.

The Chain Rule states that if $y = F(g(x))$, then $\frac{dy}{dx} = F'(g(x)) \cdot g'(x)$.

Let $F(u) = \sin u$ and $u = g(x) = 2x$.

First, find the derivative of the outer function $F(u)$ with respect to $u$:

$F'(u) = \frac{d}{du}(\sin u) = \cos u$

Next, find the derivative of the inner function $u$ with respect to $x$:

$g'(x) = \frac{du}{dx} = \frac{d}{dx}(2x)$

Using the Constant Multiple Rule $\frac{d}{dx}(cx) = c$:

$\frac{du}{dx} = 2$

Now, apply the Chain Rule: $\frac{df}{dx} = F'(u) \cdot \frac{du}{dx}$:

$f'(x) = (\cos u) \cdot (2)$

Substitute back $u = 2x$:

$f'(x) = 2\cos(2x)$

The derivative of $f(x) = \sin 2x$ is $2\cos(2x)$.

Part (ii): $g(x) = \cot x$

Given:

The function is $g(x) = \cot x$.

To Find:

The derivative of $g(x)$, denoted as $g'(x)$ or $\frac{d}{dx}(\cot x)$.

Solution:

We can write $\cot x = \frac{\cos x}{\sin x}$. This is a quotient of two functions. We will use the Quotient Rule.

The Quotient Rule states that if $g(x) = \frac{u(x)}{v(x)}$, then $g'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2}$.

Let $u(x) = \cos x$ and $v(x) = \sin x$.

Find the derivatives $u'(x)$ and $v'(x)$:

$u'(x) = \frac{d}{dx}(\cos x) = -\sin x$

$v'(x) = \frac{d}{dx}(\sin x) = \cos x$

Now, apply the Quotient Rule formula:

$g'(x) = \frac{(-\sin x)(\sin x) - (\cos x)(\cos x)}{(\sin x)^2}$

Simplify the numerator:

$g'(x) = \frac{-\sin^2 x - \cos^2 x}{\sin^2 x}$

Factor out $-1$ from the numerator:

$g'(x) = \frac{-(\sin^2 x + \cos^2 x)}{\sin^2 x}$

Using the fundamental trigonometric identity $\sin^2 x + \cos^2 x = 1$:

$g'(x) = \frac{-1}{\sin^2 x}$

Recall that $\text{cosec } x = \frac{1}{\sin x}$. Therefore, $\frac{1}{\sin^2 x} = \left(\frac{1}{\sin x}\right)^2 = \text{cosec}^2 x$.

$g'(x) = -\text{cosec}^2 x$

The derivative of $g(x) = \cot x$ is $-\text{cosec}^2 x$.

We will use the Quotient Rule to find the derivative of the given functions. The Quotient Rule states that if $f(x) = \frac{u(x)}{v(x)}$, then the derivative $f'(x)$ is given by:

$f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2}$

Part (i): $f(x) = \frac{x^{5}\;-\;\cos\; x}{\sin\; x}$

Given:

The function is $f(x) = \frac{x^5 - \cos x}{\sin x}$, where $\sin x \neq 0$.

To Find:

The derivative of $f(x)$, $f'(x)$.

Solution:

Let $u(x) = x^5 - \cos x$ and $v(x) = \sin x$.

Find the derivatives of $u(x)$ and $v(x)$:

$u'(x) = \frac{d}{dx}(x^5 - \cos x) = \frac{d}{dx}(x^5) - \frac{d}{dx}(\cos x)$

Using the Power Rule and the derivative of $\cos x$:

$u'(x) = 5x^{5-1} - (-\sin x) = 5x^4 + \sin x$

$v'(x) = \frac{d}{dx}(\sin x) = \cos x$

Now, apply the Quotient Rule formula:

$f'(x) = \frac{(5x^4 + \sin x)(\sin x) - (x^5 - \cos x)(\cos x)}{(\sin x)^2}$

Expand the numerator:

Numerator $= 5x^4 \sin x + \sin^2 x - (x^5 \cos x - \cos^2 x)$

Numerator $= 5x^4 \sin x + \sin^2 x - x^5 \cos x + \cos^2 x$

Rearrange terms and use the identity $\sin^2 x + \cos^2 x = 1$:

Numerator $= 5x^4 \sin x - x^5 \cos x + (\sin^2 x + \cos^2 x)$

Numerator $= 5x^4 \sin x - x^5 \cos x + 1$

So, the derivative is:

$f'(x) = \frac{5x^4 \sin x - x^5 \cos x + 1}{\sin^2 x}$

The derivative of $f(x) = \frac{x^{5}\;-\;\cos\; x}{\sin\; x}$ is $\frac{5x^4 \sin x - x^5 \cos x + 1}{\sin^2 x}$.

Part (ii): $g(x) = \frac{x \;+\; \cos\; x}{\tan\; x}$

Given:

The function is $g(x) = \frac{x + \cos x}{\tan x}$, where $\tan x \neq 0$.

To Find:

The derivative of $g(x)$, $g'(x)$.

Solution:

Let $u(x) = x + \cos x$ and $v(x) = \tan x$.

Find the derivatives of $u(x)$ and $v(x)$:

$u'(x) = \frac{d}{dx}(x + \cos x) = \frac{d}{dx}(x) + \frac{d}{dx}(\cos x)$

Using the Power Rule and the derivative of $\cos x$:

$u'(x) = 1 + (-\sin x) = 1 - \sin x$

$v'(x) = \frac{d}{dx}(\tan x) = \sec^2 x$

Now, apply the Quotient Rule formula:

$g'(x) = \frac{(1 - \sin x)(\tan x) - (x + \cos x)(\sec^2 x)}{(\tan x)^2}$

Expand the numerator:

Numerator $= (1)(\tan x) - (\sin x)(\tan x) - [(x)(\sec^2 x) + (\cos x)(\sec^2 x)]$

Numerator $= \tan x - \sin x \tan x - x \sec^2 x - \cos x \sec^2 x$

Rewrite $\tan x$ as $\frac{\sin x}{\cos x}$ and $\sec x$ as $\frac{1}{\cos x}$:

Numerator $= \frac{\sin x}{\cos x} - \sin x \left(\frac{\sin x}{\cos x}\right) - x \left(\frac{1}{\cos x}\right)^2 - \cos x \left(\frac{1}{\cos x}\right)^2$

Numerator $= \frac{\sin x}{\cos x} - \frac{\sin^2 x}{\cos x} - \frac{x}{\cos^2 x} - \frac{\cos x}{\cos^2 x}$

Find a common denominator $\cos^2 x$ for the terms in the numerator:

Numerator $= \frac{\sin x \cos x}{\cos^2 x} - \frac{\sin^2 x \cos x}{\cos^2 x} - \frac{x}{\cos^2 x} - \frac{\cos x}{\cos^2 x}$

Numerator $= \frac{\sin x \cos x - \sin^2 x \cos x - x - \cos x}{\cos^2 x}$

The denominator of the main fraction is $(\tan x)^2 = \left(\frac{\sin x}{\cos x}\right)^2 = \frac{\sin^2 x}{\cos^2 x}$.

So, $g'(x) = \frac{\frac{\sin x \cos x - \sin^2 x \cos x - x - \cos x}{\cos^2 x}}{\frac{\sin^2 x}{\cos^2 x}}$

Multiply the numerator by the reciprocal of the denominator:

$g'(x) = \frac{\sin x \cos x - \sin^2 x \cos x - x - \cos x}{\cos^2 x} \cdot \frac{\cos^2 x}{\sin^2 x}$

Cancel $\cos^2 x$:

$g'(x) = \frac{\sin x \cos x - \sin^2 x \cos x - x - \cos x}{\sin^2 x}$

The derivative of $g(x) = \frac{x + \cos x}{\tan x}$ is $\frac{\sin x \cos x - \sin^2 x \cos x - x - \cos x}{\sin^2 x}$.

To find the derivative of a function $f(x)$ from the first principle, we use the formula:

$f'(x) = \lim\limits_{h \to 0} \frac{f(x+h) - f(x)}{h}$

(i) Let $f(x) = -x$.

We need to find $f'(x)$ using the first principle.

First, find $f(x+h)$:

$f(x+h) = -(x+h) = -x - h$

Next, find the difference $f(x+h) - f(x)$:

$f(x+h) - f(x) = (-x - h) - (-x)$

$f(x+h) - f(x) = -x - h + x$

$f(x+h) - f(x) = -h$

Now, divide the difference by $h$:

$\frac{f(x+h) - f(x)}{h} = \frac{-h}{h} = -1$

Finally, take the limit as $h \to 0$:

$f'(x) = \lim\limits_{h \to 0} (-1)$

$f'(x) = -1$

Thus, the derivative of $-x$ from the first principle is $-1$.

(ii) Let $f(x) = (-x)^{-1} = -\frac{1}{x}$.

We need to find $f'(x)$ using the first principle.

First, find $f(x+h)$:

$f(x+h) = -(x+h)^{-1} = -\frac{1}{x+h}$

Next, find the difference $f(x+h) - f(x)$:

$f(x+h) - f(x) = -\frac{1}{x+h} - \left(-\frac{1}{x}\right)$

$f(x+h) - f(x) = \frac{1}{x} - \frac{1}{x+h}$

Combine the terms by finding a common denominator:

$f(x+h) - f(x) = \frac{1 \cdot (x+h)}{x(x+h)} - \frac{1 \cdot x}{(x+h)x}$

$f(x+h) - f(x) = \frac{x+h - x}{x(x+h)}$

$f(x+h) - f(x) = \frac{h}{x(x+h)}$

Now, divide the difference by $h$:

$\frac{f(x+h) - f(x)}{h} = \frac{\frac{h}{x(x+h)}}{h}$

$\frac{f(x+h) - f(x)}{h} = \frac{h}{h \cdot x(x+h)}$

$\frac{f(x+h) - f(x)}{h} = \frac{1}{x(x+h)}$

Finally, take the limit as $h \to 0$:

$f'(x) = \lim\limits_{h \to 0} \frac{1}{x(x+h)}$

Substitute $h=0$ into the expression:

$f'(x) = \frac{1}{x(x+0)} = \frac{1}{x \cdot x} = \frac{1}{x^2}$

Thus, the derivative of $(-x)^{-1}$ from the first principle is $\frac{1}{x^2}$.

(iii) Let $f(x) = \sin(x+1)$.

We need to find $f'(x)$ using the first principle.

First, find $f(x+h)$:

$f(x+h) = \sin((x+h)+1) = \sin(x+1+h)$

Next, find the difference $f(x+h) - f(x)$:

$f(x+h) - f(x) = \sin(x+1+h) - \sin(x+1)$

Use the trigonometric identity $\sin A - \sin B = 2 \cos\left(\frac{A+B}{2}\right) \sin\left(\frac{A-B}{2}\right)$.

Here, $A = x+1+h$ and $B = x+1$.

$\frac{A+B}{2} = \frac{(x+1+h) + (x+1)}{2} = \frac{2x+2+h}{2} = x+1+\frac{h}{2}$

$\frac{A-B}{2} = \frac{(x+1+h) - (x+1)}{2} = \frac{h}{2}$

So, the difference is:

$f(x+h) - f(x) = 2 \cos\left(x+1+\frac{h}{2}\right) \sin\left(\frac{h}{2}\right)$

Now, divide the difference by $h$:

$\frac{f(x+h) - f(x)}{h} = \frac{2 \cos\left(x+1+\frac{h}{2}\right) \sin\left(\frac{h}{2}\right)}{h}$

We can rewrite this expression:

$\frac{f(x+h) - f(x)}{h} = 2 \cos\left(x+1+\frac{h}{2}\right) \cdot \frac{\sin\left(\frac{h}{2}\right)}{h}$

Recall the standard limit $\lim\limits_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$. We can write $\frac{\sin\left(\frac{h}{2}\right)}{h}$ as $\frac{1}{2} \cdot \frac{\sin\left(\frac{h}{2}\right)}{\frac{h}{2}}$.

As $h \to 0$, $\frac{h}{2} \to 0$. Therefore, $\lim\limits_{h \to 0} \frac{\sin\left(\frac{h}{2}\right)}{\frac{h}{2}} = 1$.

Thus, $\lim\limits_{h \to 0} \frac{\sin\left(\frac{h}{2}\right)}{h} = \lim\limits_{h \to 0} \left(\frac{1}{2} \cdot \frac{\sin\left(\frac{h}{2}\right)}{\frac{h}{2}}\right) = \frac{1}{2} \cdot 1 = \frac{1}{2}$.

Finally, take the limit as $h \to 0$ of the entire expression:

$f'(x) = \lim\limits_{h \to 0} \left( 2 \cos\left(x+1+\frac{h}{2}\right) \cdot \frac{\sin\left(\frac{h}{2}\right)}{h} \right)$

Using limit properties:

$f'(x) = \left(\lim\limits_{h \to 0} 2 \cos\left(x+1+\frac{h}{2}\right)\right) \cdot \left(\lim\limits_{h \to 0} \frac{\sin\left(\frac{h}{2}\right)}{h}\right)$

Substitute $h=0$ into the cosine term:

$f'(x) = \left(2 \cos\left(x+1+0\right)\right) \cdot \left(\frac{1}{2}\right)$

$f'(x) = 2 \cos(x+1) \cdot \frac{1}{2}$

$f'(x) = \cos(x+1)$

Thus, the derivative of $\sin(x+1)$ from the first principle is $\cos(x+1)$.

(iv) Let $f(x) = \cos\left(x - \frac{\pi}{8}\right)$.

We need to find $f'(x)$ using the first principle.

First, find $f(x+h)$:

$f(x+h) = \cos\left((x+h) - \frac{\pi}{8}\right) = \cos\left(x - \frac{\pi}{8} + h\right)$

Next, find the difference $f(x+h) - f(x)$:

$f(x+h) - f(x) = \cos\left(x - \frac{\pi}{8} + h\right) - \cos\left(x - \frac{\pi}{8}\right)$

Use the trigonometric identity $\cos A - \cos B = -2 \sin\left(\frac{A+B}{2}\right) \sin\left(\frac{A-B}{2}\right)$.

Here, $A = x - \frac{\pi}{8} + h$ and $B = x - \frac{\pi}{8}$.

$\frac{A+B}{2} = \frac{\left(x - \frac{\pi}{8} + h\right) + \left(x - \frac{\pi}{8}\right)}{2} = \frac{2x - \frac{2\pi}{8} + h}{2} = \frac{2x - \frac{\pi}{4} + h}{2} = x - \frac{\pi}{8} + \frac{h}{2}$

$\frac{A-B}{2} = \frac{\left(x - \frac{\pi}{8} + h\right) - \left(x - \frac{\pi}{8}\right)}{2} = \frac{h}{2}$

So, the difference is:

$f(x+h) - f(x) = -2 \sin\left(x - \frac{\pi}{8} + \frac{h}{2}\right) \sin\left(\frac{h}{2}\right)$

Now, divide the difference by $h$:

$\frac{f(x+h) - f(x)}{h} = \frac{-2 \sin\left(x - \frac{\pi}{8} + \frac{h}{2}\right) \sin\left(\frac{h}{2}\right)}{h}$

We can rewrite this expression:

$\frac{f(x+h) - f(x)}{h} = -2 \sin\left(x - \frac{\pi}{8} + \frac{h}{2}\right) \cdot \frac{\sin\left(\frac{h}{2}\right)}{h}$

As shown in part (iii), $\lim\limits_{h \to 0} \frac{\sin\left(\frac{h}{2}\right)}{h} = \frac{1}{2}$.

Finally, take the limit as $h \to 0$ of the entire expression:

$f'(x) = \lim\limits_{h \to 0} \left( -2 \sin\left(x - \frac{\pi}{8} + \frac{h}{2}\right) \cdot \frac{\sin\left(\frac{h}{2}\right)}{h} \right)$

Using limit properties:

$f'(x) = \left(\lim\limits_{h \to 0} -2 \sin\left(x - \frac{\pi}{8} + \frac{h}{2}\right)\right) \cdot \left(\lim\limits_{h \to 0} \frac{\sin\left(\frac{h}{2}\right)}{h}\right)$

Substitute $h=0$ into the sine term:

$f'(x) = \left(-2 \sin\left(x - \frac{\pi}{8} + 0\right)\right) \cdot \left(\frac{1}{2}\right)$

$f'(x) = -2 \sin\left(x - \frac{\pi}{8}\right) \cdot \frac{1}{2}$

$f'(x) = -\sin\left(x - \frac{\pi}{8}\right)$

Thus, the derivative of $\cos\left(x - \frac{\pi}{8}\right)$ from the first principle is $-\sin\left(x - \frac{\pi}{8}\right)$.

Let $f(x) = x+a$. We need to find the derivative of $f(x)$ with respect to $x$.

We can use the sum rule for differentiation, which states $\frac{d}{dx}[u(x) + v(x)] = \frac{du}{dx} + \frac{dv}{dx}$.

Here, $u(x) = x$ and $v(x) = a$.

The derivative of $x$ with respect to $x$ is $\frac{d}{dx}(x) = 1$.

The derivative of a constant $a$ with respect to $x$ is $\frac{d}{dx}(a) = 0$, since $a$ is a fixed non-zero constant.

Applying the sum rule:

$f'(x) = \frac{d}{dx}(x+a) = \frac{d}{dx}(x) + \frac{d}{dx}(a)$

$f'(x) = 1 + 0$

$f'(x) = 1$

Thus, the derivative of $(x+a)$ is $1$.

Let $f(x) = (px + q) \left( \frac{r}{x} + s\right)$. We need to find the derivative of $f(x)$ with respect to $x$.

We can find the derivative by first expanding the expression and then differentiating each term.

Expand $f(x)$:

$f(x) = px \left(\frac{r}{x}\right) + px(s) + q\left(\frac{r}{x}\right) + q(s)$

$f(x) = \frac{prx}{x} + psx + \frac{qr}{x} + qs$

For $x \neq 0$, this simplifies to:

$f(x) = pr + psx + qr x^{-1} + qs$

Now, differentiate $f(x)$ with respect to $x$ using the sum rule, constant multiple rule, and power rule $\frac{d}{dx}(x^n) = nx^{n-1}$.

$\frac{d}{dx}[f(x)] = \frac{d}{dx}[pr + psx + qr x^{-1} + qs]$

Apply the sum rule:

$\frac{d}{dx}[f(x)] = \frac{d}{dx}(pr) + \frac{d}{dx}(psx) + \frac{d}{dx}(qr x^{-1}) + \frac{d}{dx}(qs)$

Differentiate each term:

The derivative of a constant is $0$. Since $p$ and $r$ are constants, $\frac{d}{dx}(pr) = 0$.

Using the constant multiple rule, $\frac{d}{dx}(psx) = ps \frac{d}{dx}(x)$. Since $\frac{d}{dx}(x) = 1$, $\frac{d}{dx}(psx) = ps \cdot 1 = ps$.

Using the constant multiple rule and power rule, $\frac{d}{dx}(qr x^{-1}) = qr \frac{d}{dx}(x^{-1})$. Since $\frac{d}{dx}(x^{-1}) = -1 \cdot x^{-1-1} = -x^{-2} = -\frac{1}{x^2}$, $\frac{d}{dx}(qr x^{-1}) = qr \left(-\frac{1}{x^2}\right) = -\frac{qr}{x^2}$.

The derivative of a constant is $0$. Since $q$ and $s$ are constants, $\frac{d}{dx}(qs) = 0$.

Combine the derivatives of the terms:

$f'(x) = 0 + ps + \left(-\frac{qr}{x^2}\right) + 0$

$f'(x) = ps - \frac{qr}{x^2}$

We can write this as a single fraction:

$f'(x) = \frac{ps \cdot x^2}{x^2} - \frac{qr}{x^2} = \frac{psx^2 - qr}{x^2}$

Thus, the derivative of $(px + q) \left( \frac{r}{x} + s\right)$ is $ps - \frac{qr}{x^2}$ or $\frac{psx^2 - qr}{x^2}$.

Given:

Here, $a, b, c,$ and $d$ are fixed non-zero constants.

To Find:

The derivative of $f(x)$, denoted as $f'(x)$ or $\frac{d}{dx}\left((ax + b)(cx + d)^2\right)$.

Solution:

We need to find the derivative of a product of two functions. We can use the **Product Rule** of differentiation.

The Product Rule states that if $f(x) = u(x)v(x)$, where $u(x)$ and $v(x)$ are differentiable functions, then the derivative of $f(x)$ is given by:

$f'(x) = u'(x)v(x) + u(x)v'(x)$

In our case, let $u(x) = ax + b$ and $v(x) = (cx + d)^2$.

First, find the derivative of $u(x)$ with respect to $x$:

$u'(x) = \frac{d}{dx}(ax + b)$

Using the **Sum Rule**, $\frac{d}{dx}(ax + b) = \frac{d}{dx}(ax) + \frac{d}{dx}(b)$.

Using the **Constant Multiple Rule**, $\frac{d}{dx}(ax) = a \frac{d}{dx}(x)$. We know $\frac{d}{dx}(x) = 1$. So, $\frac{d}{dx}(ax) = a \times 1 = a$.

Using the **Constant Rule**, $\frac{d}{dx}(b) = 0$ since $b$ is a constant.

So, $u'(x) = a + 0 = a$.

Next, find the derivative of $v(x) = (cx + d)^2$ with respect to $x$. This is a composite function, so we use the **Chain Rule**. Let $w = cx + d$. Then $v = w^2$. The Chain Rule is $\frac{dv}{dx} = \frac{dv}{dw} \times \frac{dw}{dx}$.

Find $\frac{dv}{dw} = \frac{d}{dw}(w^2)$: Using the **Power Rule**, $\frac{dv}{dw} = 2w^{2-1} = 2w$. Substitute back $w = cx + d$, so $\frac{dv}{dw} = 2(cx + d)$.

Find $\frac{dw}{dx} = \frac{d}{dx}(cx + d)$: Using the **Sum Rule** and **Constant Multiple Rule**, $\frac{d}{dx}(cx + d) = \frac{d}{dx}(cx) + \frac{d}{dx}(d) = c \frac{d}{dx}(x) + 0 = c \times 1 + 0 = c$.

Now, apply the Chain Rule: $v'(x) = \frac{dv}{dw} \times \frac{dw}{dx} = (2(cx + d)) \times c = 2c(cx + d)$.

Now, apply the Product Rule $f'(x) = u'(x)v(x) + u(x)v'(x)$ with $u(x) = ax + b$, $v(x) = (cx + d)^2$, $u'(x) = a$, and $v'(x) = 2c(cx + d)$.

$f'(x) = (a) \cdot (cx + d)^2 + (ax + b) \cdot (2c(cx + d))$

$f'(x) = a(cx + d)^2 + 2c(ax + b)(cx + d)$

We can see that $(cx + d)$ is a common factor in both terms. We can factor it out, but the requested form does not require this. The form provided in the hint matches our result.

Answer:

The derivative of $(ax + b)(cx + d)^2$ is $\mathbf{a(cx + d)^2 + 2c(ax + b)(cx + d)}$ or $\mathbf{2c(ax + b)(cx + d) + a(cx + d)^2}$.

Let $f(x) = \frac{ax + b}{cx + d}$. We need to find the derivative of $f(x)$ with respect to $x$.

We will use the quotient rule for differentiation, which states that if $f(x) = \frac{u(x)}{v(x)}$, then the derivative $f'(x)$ is given by the formula:

$f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2}$

Let the numerator be $u(x) = ax + b$.

Let the denominator be $v(x) = cx + d$.

First, find the derivative of the numerator, $u'(x)$:

$u'(x) = \frac{d}{dx}(ax + b)$

$u'(x) = \frac{d}{dx}(ax) + \frac{d}{dx}(b)$

$u'(x) = a \frac{d}{dx}(x) + 0$

$u'(x) = a \cdot 1 = a$

Next, find the derivative of the denominator, $v'(x)$:

$v'(x) = \frac{d}{dx}(cx + d)$

$v'(x) = \frac{d}{dx}(cx) + \frac{d}{dx}(d)$

$v'(x) = c \frac{d}{dx}(x) + 0$

$v'(x) = c \cdot 1 = c$

Now, apply the quotient rule formula:

$f'(x) = \frac{(a)(cx + d) - (ax + b)(c)}{(cx + d)^2}$

Simplify the numerator:

Numerator $= a(cx + d) - c(ax + b)$

Numerator $= acx + ad - (acx + bc)$

Numerator $= acx + ad - acx - bc$

Numerator $= ad - bc$

Substitute the simplified numerator back into the quotient rule formula:

$f'(x) = \frac{ad - bc}{(cx + d)^2}$

Thus, the derivative of $\frac{ax + b}{cx + d}$ is $\frac{ad - bc}{(cx + d)^2}$.

Let $f(x) = \frac{1 \;+\; \frac{1}{x}}{1 \;-\; \frac{1}{x}}$. We need to find the derivative of $f(x)$ with respect to $x$.

First, we can simplify the expression for $f(x)$. Multiply the numerator and denominator by $x$ (assuming $x \neq 0$):

$f(x) = \frac{x \left(1 \;+\; \frac{1}{x}\right)}{x \left(1 \;-\; \frac{1}{x}\right)}$

$f(x) = \frac{x \cdot 1 \;+\; x \cdot \frac{1}{x}}{x \cdot 1 \;-\; x \cdot \frac{1}{x}}$

$f(x) = \frac{x \;+\; 1}{x \;-\; 1}$

(This simplification is valid for $x \neq 0$ and $x \neq 1$)

Now, we find the derivative of $f(x) = \frac{x+1}{x-1}$ using the quotient rule, which states that if $f(x) = \frac{u(x)}{v(x)}$, then $f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2}$.

Let the numerator be $u(x) = x+1$.

Let the denominator be $v(x) = x-1$.

Find the derivative of the numerator, $u'(x)$:

$u'(x) = \frac{d}{dx}(x+1)$

$u'(x) = \frac{d}{dx}(x) + \frac{d}{dx}(1)$

$u'(x) = 1 + 0 = 1$

Find the derivative of the denominator, $v'(x)$:

$v'(x) = \frac{d}{dx}(x-1)$

$v'(x) = \frac{d}{dx}(x) - \frac{d}{dx}(1)$

$v'(x) = 1 - 0 = 1$

Apply the quotient rule formula:

$f'(x) = \frac{(1)(x - 1) - (x + 1)(1)}{(x - 1)^2}$

Simplify the numerator:

Numerator $= (x - 1) - (x + 1)$

Numerator $= x - 1 - x - 1$

Numerator $= -2$

Substitute the simplified numerator back into the formula:

$f'(x) = \frac{-2}{(x - 1)^2}$

Thus, the derivative of $\frac{1 \;+\; \frac{1}{x}}{1 \;-\; \frac{1}{x}}$ is $\frac{-2}{(x - 1)^2}$.

Let $f(x) = \frac{1}{ax^{2}\;+\;bx \;+\;c}$. We need to find the derivative of $f(x)$ with respect to $x$.

We can use the quotient rule or rewrite the function using a negative exponent and use the chain rule.

Method 1: Using the Quotient Rule

Let $u(x) = 1$ and $v(x) = ax^2 + bx + c$.

Find the derivative of the numerator, $u'(x)$:

$u'(x) = \frac{d}{dx}(1)$

$u'(x) = 0$ (The derivative of a constant is 0)

Find the derivative of the denominator, $v'(x)$:

$v'(x) = \frac{d}{dx}(ax^2 + bx + c)$

$v'(x) = \frac{d}{dx}(ax^2) + \frac{d}{dx}(bx) + \frac{d}{dx}(c)$

$v'(x) = a \frac{d}{dx}(x^2) + b \frac{d}{dx}(x) + 0$

$v'(x) = a(2x) + b(1)$

$v'(x) = 2ax + b$

Apply the quotient rule formula $f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2}$:

$f'(x) = \frac{(0)(ax^2 + bx + c) - (1)(2ax + b)}{(ax^2 + bx + c)^2}$

$f'(x) = \frac{0 - (2ax + b)}{(ax^2 + bx + c)^2}$

$f'(x) = \frac{-(2ax + b)}{(ax^2 + bx + c)^2} = \frac{-2ax - b}{(ax^2 + bx + c)^2}$

Method 2: Using the Chain Rule

Rewrite $f(x)$ using a negative exponent:

$f(x) = (ax^2 + bx + c)^{-1}$

Use the chain rule: $\frac{d}{dx}[g(u(x))] = g'(u(x)) \cdot u'(x)$.

Let $u(x) = ax^2 + bx + c$. Let $g(u) = u^{-1}$.

Find the derivative of $g(u)$ with respect to $u$:

$g'(u) = \frac{d}{du}(u^{-1}) = -1 \cdot u^{-1-1} = -u^{-2}$

Find the derivative of $u(x)$ with respect to $x$:

$u'(x) = \frac{d}{dx}(ax^2 + bx + c) = 2ax + b$ (as found in Method 1)

Apply the chain rule formula $f'(x) = g'(u(x)) \cdot u'(x)$:

$f'(x) = -(ax^2 + bx + c)^{-2} \cdot (2ax + b)$

Rewrite the term with the negative exponent in the denominator:

$f'(x) = \frac{-(2ax + b)}{(ax^2 + bx + c)^2} = \frac{-2ax - b}{(ax^2 + bx + c)^2}$

Both methods yield the same result. Thus, the derivative of $\frac{1}{ax^{2}\;+\;bx \;+\;c}$ is $\frac{-(2ax + b)}{(ax^2 + bx + c)^2}$ or $\frac{-2ax - b}{(ax^2 + bx + c)^2}$.

Let $f(x) = \frac{ax\;+\;b}{px^{2}\;+\;qx\;+\;r}$. We need to find the derivative of $f(x)$ with respect to $x$.

We will use the quotient rule for differentiation, which states that if $f(x) = \frac{u(x)}{v(x)}$, then the derivative $f'(x)$ is given by the formula:

$f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2}$

Let the numerator be $u(x) = ax + b$.

Let the denominator be $v(x) = px^2 + qx + r$.

First, find the derivative of the numerator, $u'(x)$:

$u'(x) = \frac{d}{dx}(ax + b)$

$u'(x) = \frac{d}{dx}(ax) + \frac{d}{dx}(b)$

$u'(x) = a \frac{d}{dx}(x) + 0$

$u'(x) = a \cdot 1 = a$

Next, find the derivative of the denominator, $v'(x)$:

$v'(x) = \frac{d}{dx}(px^2 + qx + r)$

$v'(x) = \frac{d}{dx}(px^2) + \frac{d}{dx}(qx) + \frac{d}{dx}(r)$

$v'(x) = p \frac{d}{dx}(x^2) + q \frac{d}{dx}(x) + 0$

$v'(x) = p(2x) + q(1)$

$v'(x) = 2px + q$

Now, apply the quotient rule formula:

$f'(x) = \frac{(a)(px^2 + qx + r) - (ax + b)(2px + q)}{(px^2 + qx + r)^2}$

Simplify the numerator:

Numerator $= a(px^2 + qx + r) - (ax + b)(2px + q)$

Numerator $= (apx^2 + aqx + ar) - (ax(2px) + ax(q) + b(2px) + b(q))$

Numerator $= apx^2 + aqx + ar - (2apx^2 + aqx + 2bpx + bq)$

Numerator $= apx^2 + aqx + ar - 2apx^2 - aqx - 2bpx - bq$

Combine like terms in the numerator:

Numerator $= (apx^2 - 2apx^2) + (aqx - aqx) - 2bpx + ar - bq$

Numerator $= -apx^2 + 0 - 2bpx + ar - bq$

Numerator $= -apx^2 - 2bpx + ar - bq$

Substitute the simplified numerator back into the quotient rule formula:

$f'(x) = \frac{-apx^2 - 2bpx + ar - bq}{(px^2 + qx + r)^2}$

Thus, the derivative of $\frac{ax\;+\;b}{px^{2}\;+\;qx\;+\;r}$ is $\frac{-apx^2 - 2bpx + ar - bq}{(px^2 + qx + r)^2}$.

Let $f(x) = \frac{px^{2}\;+\;qx\;+\;r}{ax \;+\; b}$. We need to find the derivative of $f(x)$ with respect to $x$.

We will use the quotient rule for differentiation, which states that if $f(x) = \frac{u(x)}{v(x)}$, then the derivative $f'(x)$ is given by the formula:

$f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2}$

Let the numerator be $u(x) = px^2 + qx + r$.

Let the denominator be $v(x) = ax + b$.

First, find the derivative of the numerator, $u'(x)$:

$u'(x) = \frac{d}{dx}(px^2 + qx + r)$

$u'(x) = \frac{d}{dx}(px^2) + \frac{d}{dx}(qx) + \frac{d}{dx}(r)$

$u'(x) = p \frac{d}{dx}(x^2) + q \frac{d}{dx}(x) + 0$

$u'(x) = p(2x) + q(1)$

$u'(x) = 2px + q$

Next, find the derivative of the denominator, $v'(x)$:

$v'(x) = \frac{d}{dx}(ax + b)$

$v'(x) = \frac{d}{dx}(ax) + \frac{d}{dx}(b)$

$v'(x) = a \frac{d}{dx}(x) + 0$

$v'(x) = a \cdot 1 = a$

Now, apply the quotient rule formula:

$f'(x) = \frac{(2px + q)(ax + b) - (px^2 + qx + r)(a)}{(ax + b)^2}$

Simplify the numerator:

Numerator $= (2px + q)(ax + b) - a(px^2 + qx + r)$

Expand the products:

$(2px + q)(ax + b) = (2px)(ax) + (2px)(b) + (q)(ax) + (q)(b) = 2apx^2 + 2bpx + aqx + bq$

$a(px^2 + qx + r) = apx^2 + aqx + ar$

Subtract the second expression from the first:

Numerator $= (2apx^2 + 2bpx + aqx + bq) - (apx^2 + aqx + ar)$

Numerator $= 2apx^2 + 2bpx + aqx + bq - apx^2 - aqx - ar$

Combine like terms:

Numerator $= (2apx^2 - apx^2) + (2bpx) + (aqx - aqx) + (bq - ar)$

Numerator $= apx^2 + 2bpx + 0 + bq - ar$

Numerator $= apx^2 + 2bpx + bq - ar$

Substitute the simplified numerator back into the quotient rule formula:

$f'(x) = \frac{apx^2 + 2bpx + bq - ar}{(ax + b)^2}$

Thus, the derivative of $\frac{px^{2}\;+\;qx\;+\;r}{ax \;+\; b}$ is $\frac{apx^2 + 2bpx + bq - ar}{(ax + b)^2}$.

Let $f(x) = \frac{a}{x^{4}}$ - $\frac{b}{x^{2}}$ + cos x. We need to find the derivative of $f(x)$ with respect to $x$.

We can rewrite the function using negative exponents:

$f(x) = ax^{-4} - bx^{-2} + \cos x$

We will differentiate each term using the sum/difference rule, the constant multiple rule, and the power rule for the terms involving $x^n$. For the cosine term, we use the standard derivative.

Apply the sum/difference rule:

$f'(x) = \frac{d}{dx}(ax^{-4}) - \frac{d}{dx}(bx^{-2}) + \frac{d}{dx}(\cos x)$

Differentiate the first term using the constant multiple rule and power rule ($n = -4$):

$\frac{d}{dx}(ax^{-4}) = a \frac{d}{dx}(x^{-4})$

$= a (-4x^{-4-1})$

$= a (-4x^{-5})$

$= -4ax^{-5} = -\frac{4a}{x^5}$

Differentiate the second term using the constant multiple rule and power rule ($n = -2$):

$\frac{d}{dx}(bx^{-2}) = b \frac{d}{dx}(x^{-2})$

$= b (-2x^{-2-1})$

$= b (-2x^{-3})$

$= -2bx^{-3} = -\frac{2b}{x^3}$

Differentiate the third term (standard derivative):

$\frac{d}{dx}(\cos x) = -\sin x$

Combine the derivatives of the terms:

$f'(x) = \left(-\frac{4a}{x^5}\right) - \left(-\frac{2b}{x^3}\right) + (-\sin x)$

$f'(x) = -\frac{4a}{x^5} + \frac{2b}{x^3} - \sin x$

Thus, the derivative of $\frac{a}{x^{4}}$ - $\frac{b}{x^{2}}$ + cos x is $-\frac{4a}{x^5} + \frac{2b}{x^3} - \sin x$.

Let $f(x) = 4\sqrt{x} - 2$. We need to find the derivative of $f(x)$ with respect to $x$.

We can rewrite the function using a fractional exponent for the square root:

$f(x) = 4x^{1/2} - 2$

We will differentiate each term using the difference rule, the constant multiple rule, and the power rule $\frac{d}{dx}(x^n) = nx^{n-1}$. The derivative of a constant is 0.

Apply the difference rule:

$f'(x) = \frac{d}{dx}(4x^{1/2}) - \frac{d}{dx}(2)$

Differentiate the first term using the constant multiple rule and power rule ($n = 1/2$):

$\frac{d}{dx}(4x^{1/2}) = 4 \frac{d}{dx}(x^{1/2})$

$= 4 \left(\frac{1}{2} x^{1/2 - 1}\right)$

$= 4 \left(\frac{1}{2} x^{-1/2}\right)$

$= 2x^{-1/2}$

Differentiate the second term (derivative of a constant):

$\frac{d}{dx}(2) = 0$

Combine the derivatives of the terms:

$f'(x) = 2x^{-1/2} - 0$

$f'(x) = 2x^{-1/2}$

We can rewrite the result using radical notation:

$f'(x) = \frac{2}{x^{1/2}} = \frac{2}{\sqrt{x}}$

Thus, the derivative of $4\sqrt{x} - 2$ is $\frac{2}{\sqrt{x}}$.

Let $f(x) = (ax + b)^n$. We need to find the derivative of $f(x)$ with respect to $x$.

We will use the chain rule for differentiation. The chain rule states that if $y = g(u)$ where $u = h(x)$, then $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$.

Let $u = ax + b$. Then $f(x) = u^n$.

First, find the derivative of $f(x)$ with respect to $u$:

$\frac{df}{du} = \frac{d}{du}(u^n)$

Using the power rule, $\frac{d}{du}(u^n) = nu^{n-1}$.

So, $\frac{df}{du} = nu^{n-1}$.

Next, find the derivative of $u$ with respect to $x$:

$\frac{du}{dx} = \frac{d}{dx}(ax + b)$

Using the sum rule and constant multiple rule:

$\frac{du}{dx} = \frac{d}{dx}(ax) + \frac{d}{dx}(b)$

$\frac{du}{dx} = a \frac{d}{dx}(x) + 0$

$\frac{du}{dx} = a \cdot 1 = a$

Now, apply the chain rule formula $\frac{df}{dx} = \frac{df}{du} \cdot \frac{du}{dx}$:

$\frac{df}{dx} = (nu^{n-1}) \cdot (a)$

Substitute $u = ax + b$ back into the expression:

$\frac{df}{dx} = n(ax + b)^{n-1} \cdot a$

$\frac{df}{dx} = an(ax + b)^{n-1}$

Thus, the derivative of $(ax + b)^n$ is $an(ax + b)^{n-1}$.

Let $f(x) = (ax + b)^n (cx + d)^m$. We need to find the derivative of $f(x)$ with respect to $x$.

We will use the product rule for differentiation, which states that if $f(x) = u(x)v(x)$, then the derivative $f'(x)$ is given by the formula:

$f'(x) = u'(x)v(x) + u(x)v'(x)$

Let $u(x) = (ax + b)^n$ and $v(x) = (cx + d)^m$.

First, find the derivative of $u(x)$ with respect to $x$. We use the chain rule.

Let $w = ax + b$. Then $u(x) = w^n$.

$u'(x) = \frac{d}{dw}(w^n) \cdot \frac{dw}{dx}$

$u'(x) = nw^{n-1} \cdot \frac{d}{dx}(ax + b)$

$u'(x) = n(ax + b)^{n-1} \cdot (a)$

$u'(x) = an(ax + b)^{n-1}$

Next, find the derivative of $v(x)$ with respect to $x$. We use the chain rule.

Let $y = cx + d$. Then $v(x) = y^m$.

$v'(x) = \frac{d}{dy}(y^m) \cdot \frac{dy}{dx}$

$v'(x) = my^{m-1} \cdot \frac{d}{dx}(cx + d)$

$v'(x) = m(cx + d)^{m-1} \cdot (c)$

$v'(x) = cm(cx + d)^{m-1}$

Now, apply the product rule formula $f'(x) = u'(x)v(x) + u(x)v'(x)$:

$f'(x) = \left[an(ax + b)^{n-1}\right] (cx + d)^m + (ax + b)^n \left[cm(cx + d)^{m-1}\right]$

Simplify the expression by factoring out the common terms $(ax + b)^{n-1}$ and $(cx + d)^{m-1}$:

$f'(x) = (ax + b)^{n-1} (cx + d)^{m-1} \left[an \frac{(cx + d)^m}{(cx + d)^{m-1}} + cm \frac{(ax + b)^n}{(ax + b)^{n-1}}\right]$

$f'(x) = (ax + b)^{n-1} (cx + d)^{m-1} \left[an (cx + d)^{m-(m-1)} + cm (ax + b)^{n-(n-1)}\right]$

$f'(x) = (ax + b)^{n-1} (cx + d)^{m-1} \left[an (cx + d)^1 + cm (ax + b)^1\right]$

$f'(x) = (ax + b)^{n-1} (cx + d)^{m-1} [an(cx + d) + cm(ax + b)]$

Expand the terms inside the square brackets:

$an(cx + d) = ancx + and$

$cm(ax + b) = amcx + bmc$

Add the expanded terms:

$ancx + and + amcx + bmc = (anc + amc)x + and + bmc$

So the expression inside the square brackets is $(ac(n+m))x + (adn + bmc)$.

The derivative is:

$f'(x) = (ax + b)^{n-1} (cx + d)^{m-1} [(ac(n+m))x + adn + bmc]$

Alternatively, keeping the bracketed terms unexpanded is also a valid form:

$f'(x) = (ax + b)^{n-1} (cx + d)^{m-1} [an(cx + d) + cm(ax + b)]$

Thus, the derivative of $(ax + b)^n (cx + d)^m$ is $(ax + b)^{n-1} (cx + d)^{m-1} [an(cx + d) + cm(ax + b)]$.

Let $f(x) = \sin(x+a)$. We need to find the derivative of $f(x)$ with respect to $x$.

We will use the chain rule for differentiation. The chain rule states that if $y = g(u)$ where $u = h(x)$, then the derivative $\frac{dy}{dx}$ is given by:

$\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$

Let $u = x+a$. Then $f(x) = \sin(u)$.

First, find the derivative of $f(x)$ with respect to $u$:

$\frac{df}{du} = \frac{d}{du}(\sin u)$

The derivative of $\sin u$ with respect to $u$ is $\cos u$.

So, $\frac{df}{du} = \cos u$.

Next, find the derivative of $u$ with respect to $x$:

$\frac{du}{dx} = \frac{d}{dx}(x+a)$

Using the sum rule and constant rule:

$\frac{du}{dx} = \frac{d}{dx}(x) + \frac{d}{dx}(a)$

$\frac{du}{dx} = 1 + 0$

$\frac{du}{dx} = 1$

Now, apply the chain rule formula $\frac{df}{dx} = \frac{df}{du} \cdot \frac{du}{dx}$:

$\frac{df}{dx} = (\cos u) \cdot (1)$

Substitute $u = x+a$ back into the expression:

$\frac{df}{dx} = \cos(x+a)$

Thus, the derivative of $\sin(x+a)$ is $\cos(x+a)$.

Let $f(x) = \text{cosec } x \cot x$. We need to find the derivative of $f(x)$ with respect to $x$.

We will use the product rule for differentiation, which states that if $f(x) = u(x)v(x)$, then the derivative $f'(x)$ is given by the formula:

$f'(x) = u'(x)v(x) + u(x)v'(x)$

Let $u(x) = \text{cosec } x$ and $v(x) = \cot x$.

First, find the derivative of $u(x)$ with respect to $x$:

$u'(x) = \frac{d}{dx}(\text{cosec } x)$

The derivative of $\text{cosec } x$ is $-\text{cosec } x \cot x$.

$u'(x) = -\text{cosec } x \cot x$

Next, find the derivative of $v(x)$ with respect to $x$:

$v'(x) = \frac{d}{dx}(\cot x)$

The derivative of $\cot x$ is $-\text{cosec}^2 x$.

$v'(x) = -\text{cosec}^2 x$

Now, apply the product rule formula $f'(x) = u'(x)v(x) + u(x)v'(x)$:

$f'(x) = (-\text{cosec } x \cot x)(\cot x) + (\text{cosec } x)(-\text{cosec}^2 x)$

Simplify the expression:

$f'(x) = -\text{cosec } x \cot^2 x - \text{cosec } x \cdot \text{cosec}^2 x$

$f'(x) = -\text{cosec } x \cot^2 x - \text{cosec}^3 x$

We can factor out $-\text{cosec } x$:

$f'(x) = -\text{cosec } x (\cot^2 x + \text{cosec}^2 x)$

Alternatively, we can use the identity $\cot^2 x = \text{cosec}^2 x - 1$ to express the result only in terms of $\text{cosec } x$:

$f'(x) = -\text{cosec } x ((\text{cosec}^2 x - 1) + \text{cosec}^2 x)$

$f'(x) = -\text{cosec } x (2\text{cosec}^2 x - 1)$

$f'(x) = -2\text{cosec}^3 x + \text{cosec } x$

Both simplified forms are correct.

Thus, the derivative of $\text{cosec } x \cot x$ is $-\text{cosec } x (\cot^2 x + \text{cosec}^2 x)$ or $-2\text{cosec}^3 x + \text{cosec } x$.

Let $f(x) = \frac{\cos x}{1 + \sin x}$. We need to find the derivative of $f(x)$ with respect to $x$.

We will use the quotient rule for differentiation, which states that if $f(x) = \frac{u(x)}{v(x)}$, then the derivative $f'(x)$ is given by the formula:

$f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2}$

Let the numerator be $u(x) = \cos x$.

Let the denominator be $v(x) = 1 + \sin x$.

First, find the derivative of the numerator, $u'(x)$:

$u'(x) = \frac{d}{dx}(\cos x)$

$u'(x) = -\sin x$

Next, find the derivative of the denominator, $v'(x)$:

$v'(x) = \frac{d}{dx}(1 + \sin x)$

$v'(x) = \frac{d}{dx}(1) + \frac{d}{dx}(\sin x)$

$v'(x) = 0 + \cos x$

$v'(x) = \cos x$

Now, apply the quotient rule formula:

$f'(x) = \frac{(-\sin x)(1 + \sin x) - (\cos x)(\cos x)}{(1 + \sin x)^2}$

Simplify the numerator:

Numerator $= (-\sin x)(1) + (-\sin x)(\sin x) - \cos x \cdot \cos x$

Numerator $= -\sin x - \sin^2 x - \cos^2 x$

Factor out $-1$ from the last two terms:

Numerator $= -\sin x - (\sin^2 x + \cos^2 x)$

Use the Pythagorean identity $\sin^2 x + \cos^2 x = 1$:

Numerator $= -\sin x - (1)$

Numerator $= -1 - \sin x$

We can factor out $-1$ from the numerator:

Numerator $= -(1 + \sin x)$

Substitute the simplified numerator back into the quotient rule formula:

$f'(x) = \frac{-(1 + \sin x)}{(1 + \sin x)^2}$

Cancel out one factor of $(1 + \sin x)$ from the numerator and denominator (assuming $1 + \sin x \neq 0$):

$f'(x) = \frac{\cancel{-(1 + \sin x)}}{\cancel{(1 + \sin x)}^2}$

$f'(x) = \frac{-1}{1 + \sin x}$

Thus, the derivative of $\frac{\cos x}{1 + \sin x}$ is $\frac{-1}{1 + \sin x}$.

Given:

The function is defined when $\sin x - \cos x \neq 0$, i.e., $\sin x \neq \cos x$, which means $\tan x \neq 1$.

To Find:

The derivative of $f(x)$, denoted as $f'(x)$ or $\frac{d}{dx}\left(\frac{\sin x + \cos x}{\sin x - \cos x}\right)$.

Solution:

We will use the **Quotient Rule** of differentiation to find the derivative of this function. The Quotient Rule states that if $f(x) = \frac{u(x)}{v(x)}$, where $u(x)$ and $v(x)$ are differentiable functions and $v(x) \neq 0$, then the derivative of $f(x)$ is given by:

$f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2}$

In our case, the numerator is $u(x) = \sin x + \cos x$ and the denominator is $v(x) = \sin x - \cos x$.

First, we find the derivatives of $u(x)$ and $v(x)$ with respect to $x$.

1. Find the derivative of the numerator, $u'(x) = \frac{d}{dx}(\sin x + \cos x)$.

Using the properties of derivatives: $\frac{d}{dx}(\sin x + \cos x) = \frac{d}{dx}(\sin x) + \frac{d}{dx}(\cos x)$.

We know that $\frac{d}{dx}(\sin x) = \cos x$ and $\frac{d}{dx}(\cos x) = -\sin x$.

So, $u'(x) = \cos x + (-\sin x) = \cos x - \sin x$.

2. Find the derivative of the denominator, $v'(x) = \frac{d}{dx}(\sin x - \cos x)$.

Using the properties of derivatives: $\frac{d}{dx}(\sin x - \cos x) = \frac{d}{dx}(\sin x) - \frac{d}{dx}(\cos x)$.

So, $v'(x) = \cos x - (-\sin x) = \cos x + \sin x$.

Now, substitute $u(x), v(x), u'(x),$ and $v'(x)$ into the Quotient Rule formula:

$f'(x) = \frac{(\cos x - \sin x)(\sin x - \cos x) - (\sin x + \cos x)(\cos x + \sin x)}{(\sin x - \cos x)^2}$

Simplify the numerator:

Numerator $= (\cos x - \sin x)(-(\cos x - \sin x)) - (\sin x + \cos x)(\sin x + \cos x)$

Numerator $= -(\cos x - \sin x)^2 - (\sin x + \cos x)^2$

Expand the squared terms using the identities $(a-b)^2 = a^2 - 2ab + b^2$ and $(a+b)^2 = a^2 + 2ab + b^2$:

$(\cos x - \sin x)^2 = \cos^2 x - 2 \sin x \cos x + \sin^2 x$

$(\sin x + \cos x)^2 = \sin^2 x + 2 \sin x \cos x + \cos^2 x$}

Using the identity $\sin^2 x + \cos^2 x = 1$, these become:

$(\cos x - \sin x)^2 = (\cos^2 x + \sin^2 x) - 2 \sin x \cos x = 1 - 2 \sin x \cos x$

$(\sin x + \cos x)^2 = (\sin^2 x + \cos^2 x) + 2 \sin x \cos x = 1 + 2 \sin x \cos x$

Now substitute these back into the numerator expression:

Numerator $= -(1 - 2 \sin x \cos x) - (1 + 2 \sin x \cos x)$

Numerator $= -1 + 2 \sin x \cos x - 1 - 2 \sin x \cos x$

Numerator $= -1 - 1 = -2$

The denominator is $(\sin x - \cos x)^2$.

Substitute the simplified numerator and the denominator back into the derivative formula:

This derivative is valid wherever the original function is defined, i.e., when $\sin x - \cos x \neq 0$.

Alternate Approach (Simplification using tan x):

We can simplify the original function by dividing the numerator and denominator by $\cos x$ (assuming $\cos x \neq 0$):

$f(x) = \frac{(\sin x + \cos x)/\cos x}{(\sin x - \cos x)/\cos x} = \frac{\frac{\sin x}{\cos x} + \frac{\cos x}{\cos x}}{\frac{\sin x}{\cos x} - \frac{\cos x}{\cos x}} = \frac{\tan x + 1}{\tan x - 1}$

We know the tangent addition formula: $\tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$. Using $\tan(\frac{\pi}{4}) = 1$, we have $\tan(\frac{\pi}{4} + x) = \frac{\tan(\frac{\pi}{4}) + \tan x}{1 - \tan(\frac{\pi}{4})\tan x} = \frac{1 + \tan x}{1 - \tan x}$.

Notice that $\frac{\tan x + 1}{\tan x - 1} = \frac{-(1 + \tan x)}{-(1 - \tan x)} = - \frac{1 + \tan x}{1 - \tan x}$.

So, $f(x) = - \tan\left(\frac{\pi}{4} + x\right)$.

Now differentiate $f(x) = - \tan\left(\frac{\pi}{4} + x\right)$ using the Chain Rule.

$f'(x) = \frac{d}{dx}\left(- \tan\left(\frac{\pi}{4} + x\right)\right) = - \frac{d}{dx}\left(\tan\left(\frac{\pi}{4} + x\right)\right)$

Using $\frac{d}{du}(\tan u) = \sec^2 u$ and $\frac{d}{dx}(\frac{\pi}{4} + x) = 1$:

$f'(x) = - \sec^2\left(\frac{\pi}{4} + x\right) \times \frac{d}{dx}\left(\frac{\pi}{4} + x\right) = - \sec^2\left(\frac{\pi}{4} + x\right) \times 1 = - \sec^2\left(\frac{\pi}{4} + x\right)$.

Let's verify this matches the Quotient Rule result. $\sec\left(\frac{\pi}{4} + x\right) = \frac{1}{\cos\left(\frac{\pi}{4} + x\right)}$.

Using $\cos(A+B) = \cos A \cos B - \sin A \sin B$:

$\cos\left(\frac{\pi}{4} + x\right) = \cos\left(\frac{\pi}{4}\right)\cos x - \sin\left(\frac{\pi}{4}\right)\sin x = \frac{1}{\sqrt{2}}\cos x - \frac{1}{\sqrt{2}}\sin x = \frac{1}{\sqrt{2}}(\cos x - \sin x)$.

So, $\cos^2\left(\frac{\pi}{4} + x\right) = \left(\frac{1}{\sqrt{2}}(\cos x - \sin x)\right)^2 = \frac{1}{2}(\cos x - \sin x)^2 = \frac{1}{2}(\sin x - \cos x)^2$.

Therefore, $f'(x) = - \sec^2\left(\frac{\pi}{4} + x\right) = \frac{-1}{\cos^2\left(\frac{\pi}{4} + x\right)} = \frac{-1}{\frac{1}{2}(\sin x - \cos x)^2} = \frac{-2}{(\sin x - \cos x)^2}$.

Both methods give the same result.

Answer:

The derivative of the function is $\mathbf{\frac{-2}{(\sin x - \cos x)^2}}$.

Let $f(x) = \frac{\sec x - 1}{\sec x + 1}$. We need to find the derivative of $f(x)$ with respect to $x$.

Before applying the quotient rule, we can simplify the expression by writing $\sec x$ in terms of $\cos x$, since $\sec x = \frac{1}{\cos x}$.

$f(x) = \frac{\frac{1}{\cos x} - 1}{\frac{1}{\cos x} + 1}$

Multiply the numerator and denominator by $\cos x$ (assuming $\cos x \neq 0$):

$f(x) = \frac{\cos x \left(\frac{1}{\cos x} - 1\right)}{\cos x \left(\frac{1}{\cos x} + 1\right)}$

$f(x) = \frac{\cos x \cdot \frac{1}{\cos x} - \cos x \cdot 1}{\cos x \cdot \frac{1}{\cos x} + \cos x \cdot 1}$

$f(x) = \frac{1 - \cos x}{1 + \cos x}$

Now, we find the derivative of $f(x) = \frac{1 - \cos x}{1 + \cos x}$ using the quotient rule, which states that if $f(x) = \frac{u(x)}{v(x)}$, then $f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2}$.

Let the numerator be $u(x) = 1 - \cos x$.

Let the denominator be $v(x) = 1 + \cos x$.

First, find the derivative of the numerator, $u'(x)$:

$u'(x) = \frac{d}{dx}(1 - \cos x)$

$u'(x) = \frac{d}{dx}(1) - \frac{d}{dx}(\cos x)$

$u'(x) = 0 - (-\sin x)$

$u'(x) = \sin x$

Next, find the derivative of the denominator, $v'(x)$:

$v'(x) = \frac{d}{dx}(1 + \cos x)$

$v'(x) = \frac{d}{dx}(1) + \frac{d}{dx}(\cos x)$

$v'(x) = 0 + (-\sin x)$

$v'(x) = -\sin x$

Now, apply the quotient rule formula:

$f'(x) = \frac{(\sin x)(1 + \cos x) - (1 - \cos x)(-\sin x)}{(1 + \cos x)^2}$

Simplify the numerator:

Numerator $= (\sin x)(1) + (\sin x)(\cos x) - [(1)(-\sin x) - (\cos x)(-\sin x)]$

Numerator $= \sin x + \sin x \cos x - [-\sin x + \sin x \cos x]$

Numerator $= \sin x + \sin x \cos x + \sin x - \sin x \cos x$

Combine the terms:

Numerator $= (\sin x + \sin x) + (\sin x \cos x - \sin x \cos x)$

Numerator $= 2 \sin x + 0$

Numerator $= 2 \sin x$

Substitute the simplified numerator back into the quotient rule formula:

$f'(x) = \frac{2 \sin x}{(1 + \cos x)^2}$

We can also express the result using half-angle identities. Recall that $1 + \cos x = 2 \cos^2 \left(\frac{x}{2}\right)$ and $\sin x = 2 \sin \left(\frac{x}{2}\right) \cos \left(\frac{x}{2}\right)$.

$f'(x) = \frac{2 \left(2 \sin \left(\frac{x}{2}\right) \cos \left(\frac{x}{2}\right)\right)}{\left(2 \cos^2 \left(\frac{x}{2}\right)\right)^2}$

$f'(x) = \frac{4 \sin \left(\frac{x}{2}\right) \cos \left(\frac{x}{2}\right)}{4 \cos^4 \left(\frac{x}{2}\right)}$

Cancel out $4$ and one factor of $\cos \left(\frac{x}{2}\right)$ (assuming $\cos \left(\frac{x}{2}\right) \neq 0$):

$f'(x) = \frac{\cancel{4} \sin \left(\frac{x}{2}\right) \cancel{\cos \left(\frac{x}{2}\right)}}{\cancel{4} \cos^{\cancel{4}3} \left(\frac{x}{2}\right)}$

$f'(x) = \frac{\sin \left(\frac{x}{2}\right)}{\cos^3 \left(\frac{x}{2}\right)}$

$f'(x) = \frac{\sin \left(\frac{x}{2}\right)}{\cos \left(\frac{x}{2}\right) \cos^2 \left(\frac{x}{2}\right)}$

$f'(x) = \tan \left(\frac{x}{2}\right) \sec^2 \left(\frac{x}{2}\right)$

Both forms $\frac{2 \sin x}{(1 + \cos x)^2}$ and $\tan \left(\frac{x}{2}\right) \sec^2 \left(\frac{x}{2}\right)$ are correct.

Thus, the derivative of $\frac{\sec x - 1}{\sec x + 1}$ is $\frac{2 \sin x}{(1 + \cos x)^2}$ or $\tan \left(\frac{x}{2}\right) \sec^2 \left(\frac{x}{2}\right)$.

Let $f(x) = \sin^n x$. This can be written as $f(x) = (\sin x)^n$. We need to find the derivative of $f(x)$ with respect to $x$.

We will use the chain rule for differentiation. The chain rule states that if $y = g(u)$ where $u = h(x)$, then the derivative $\frac{dy}{dx}$ is given by:

$\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$

Let the inner function be $u = \sin x$.

Let the outer function be $g(u) = u^n$. Then $f(x) = g(u(x)) = (\sin x)^n = u^n$.

First, find the derivative of the outer function $g(u)$ with respect to $u$:

$\frac{dg}{du} = \frac{d}{du}(u^n)$

Using the power rule, $\frac{d}{du}(u^n) = nu^{n-1}$.

So, $\frac{dg}{du} = nu^{n-1}$.

Next, find the derivative of the inner function $u(x)$ with respect to $x$:

$\frac{du}{dx} = \frac{d}{dx}(\sin x)$

The derivative of $\sin x$ is $\cos x$.

So, $\frac{du}{dx} = \cos x$.

Now, apply the chain rule formula $\frac{df}{dx} = \frac{dg}{du} \cdot \frac{du}{dx}$:

$\frac{df}{dx} = (nu^{n-1}) \cdot (\cos x)$

Substitute $u = \sin x$ back into the expression:

$\frac{df}{dx} = n(\sin x)^{n-1} \cdot \cos x$

$\frac{df}{dx} = n \sin^{n-1} x \cos x$

Thus, the derivative of $\sin^n x$ is $n \sin^{n-1} x \cos x$.

Given:

Here, $a, b, c,$ and $d$ are fixed non-zero constants.

The function is defined for values of $x$ where the denominator $c + d \cos x \neq 0$.

To Find:

The derivative of $f(x)$, denoted as $f'(x)$ or $\frac{d}{dx}\left(\frac{a + b \sin x}{c + d \cos x}\right)$.

Solution:

We need to find the derivative of a quotient of two functions. We will use the **Quotient Rule** of differentiation.

The Quotient Rule states that if $f(x) = \frac{u(x)}{v(x)}$, where $u(x)$ and $v(x)$ are differentiable functions and $v(x) \neq 0$, then the derivative of $f(x)$ is given by:

$f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2}$

In our case, the numerator is $u(x) = a + b \sin x$ and the denominator is $v(x) = c + d \cos x$.

First, find the derivatives of $u(x)$ and $v(x)$ with respect to $x$.

1. Find the derivative of the numerator, $u'(x) = \frac{d}{dx}(a + b \sin x)$.

Using the **Sum Rule** and **Constant Multiple Rule**:

$u'(x) = \frac{d}{dx}(a) + \frac{d}{dx}(b \sin x) = 0 + b \frac{d}{dx}(\sin x)$

We know that $\frac{d}{dx}(\sin x) = \cos x$.

2. Find the derivative of the denominator, $v'(x) = \frac{d}{dx}(c + d \cos x)$.

Using the **Sum Rule** and **Constant Multiple Rule**:

$v'(x) = \frac{d}{dx}(c) + \frac{d}{dx}(d \cos x) = 0 + d \frac{d}{dx}(\cos x)$

We know that $\frac{d}{dx}(\cos x) = -\sin x$.

Now, substitute $u(x), v(x), u'(x),$ and $v'(x)$ into the Quotient Rule formula:

$f'(x) = \frac{(b \cos x)(c + d \cos x) - (a + b \sin x)(-d \sin x)}{(c + d \cos x)^2}$

Expand the terms in the numerator:

Numerator $= (b \cos x \times c) + (b \cos x \times d \cos x) - [(a \times -d \sin x) + (b \sin x \times -d \sin x)]$

Numerator $= bc \cos x + bd \cos^2 x - [-ad \sin x - bd \sin^2 x]$

Numerator $= bc \cos x + bd \cos^2 x + ad \sin x + bd \sin^2 x$

Group the terms involving $bd$:

Numerator $= ad \sin x + bc \cos x + bd (\cos^2 x + \sin^2 x)$

Using the fundamental trigonometric identity $\sin^2 x + \cos^2 x = 1$:

Numerator $= ad \sin x + bc \cos x + bd (1)$

Numerator $= ad \sin x + bc \cos x + bd$

Substitute the simplified numerator back into the derivative formula:

This derivative is valid for all $x$ where the original function is defined, i.e., when $c + d \cos x \neq 0$.

Answer:

The derivative of the function is $\mathbf{\frac{ad \sin x + bc \cos x + bd}{(c + d \cos x)^2}}$.

Let $f(x) = \frac{\sin (x \;+\; a)}{\cos x}$. We need to find the derivative of $f(x)$ with respect to $x$.

We will use the quotient rule for differentiation, which states that if $f(x) = \frac{u(x)}{v(x)}$, then the derivative $f'(x)$ is given by the formula:

$f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2}$

Let the numerator be $u(x) = \sin (x + a)$.

Let the denominator be $v(x) = \cos x$.

First, find the derivative of the numerator, $u'(x)$:

$u'(x) = \frac{d}{dx}(\sin (x + a))$

Using the chain rule, let $w = x+a$. Then $\frac{d}{dx}(\sin (x+a)) = \frac{d}{dw}(\sin w) \cdot \frac{dw}{dx}$.

$\frac{d}{dw}(\sin w) = \cos w$

$\frac{dw}{dx} = \frac{d}{dx}(x+a) = 1+0 = 1$

So, $u'(x) = \cos(x+a) \cdot 1 = \cos(x+a)$.

Next, find the derivative of the denominator, $v'(x)$:

$v'(x) = \frac{d}{dx}(\cos x)$

$v'(x) = -\sin x$

Now, apply the quotient rule formula:

$f'(x) = \frac{(\cos (x + a))(\cos x) - (\sin (x + a))(-\sin x)}{(\cos x)^2}$

Simplify the numerator:

Numerator $= \cos (x + a) \cos x + \sin (x + a) \sin x$

Recall the trigonometric identity $\cos(A-B) = \cos A \cos B + \sin A \sin B$.

Let $A = x+a$ and $B = x$.

Numerator $= \cos((x+a) - x) = \cos(a)$

Substitute the simplified numerator back into the quotient rule formula:

$f'(x) = \frac{\cos a}{(\cos x)^2} = \frac{\cos a}{\cos^2 x}$

Thus, the derivative of $\frac{\sin (x + a)}{\cos x}$ is $\frac{\cos a}{\cos^2 x}$.

Let $f(x) = x^4 (5 \sin x - 3 \cos x)$. We need to find the derivative of $f(x)$ with respect to $x$.

We will use the product rule for differentiation, which states that if $f(x) = u(x)v(x)$, then the derivative $f'(x)$ is given by the formula:

$f'(x) = u'(x)v(x) + u(x)v'(x)$

Let $u(x) = x^4$ and $v(x) = 5 \sin x - 3 \cos x$.

First, find the derivative of $u(x)$ with respect to $x$:

$u'(x) = \frac{d}{dx}(x^4)$

Using the power rule, $u'(x) = 4x^{4-1} = 4x^3$.

$u'(x) = 4x^3$

Next, find the derivative of $v(x)$ with respect to $x$:

$v'(x) = \frac{d}{dx}(5 \sin x - 3 \cos x)$

Using the difference rule and constant multiple rule:

$v'(x) = \frac{d}{dx}(5 \sin x) - \frac{d}{dx}(3 \cos x)$

$v'(x) = 5 \frac{d}{dx}(\sin x) - 3 \frac{d}{dx}(\cos x)$

$v'(x) = 5 (\cos x) - 3 (-\sin x)$

$v'(x) = 5 \cos x + 3 \sin x$

Now, apply the product rule formula $f'(x) = u'(x)v(x) + u(x)v'(x)$:

$f'(x) = (4x^3)(5 \sin x - 3 \cos x) + (x^4)(5 \cos x + 3 \sin x)$

Expand the terms:

$f'(x) = 4x^3 \cdot 5 \sin x - 4x^3 \cdot 3 \cos x + x^4 \cdot 5 \cos x + x^4 \cdot 3 \sin x$

$f'(x) = 20x^3 \sin x - 12x^3 \cos x + 5x^4 \cos x + 3x^4 \sin x$

Group terms with $\sin x$ and $\cos x$:

$f'(x) = (20x^3 \sin x + 3x^4 \sin x) + (-12x^3 \cos x + 5x^4 \cos x)$

Factor out $\sin x$ and $\cos x$:

$f'(x) = x^3(20 + 3x) \sin x + x^3(-12 + 5x) \cos x$

Or, factor out the common term $x^3$ from the entire expression:

$f'(x) = x^3 [ (20 \sin x - 12 \cos x) + x(5 \cos x + 3 \sin x) ]$

$f'(x) = x^3 [ 20 \sin x - 12 \cos x + 5x \cos x + 3x \sin x ]$

Group terms inside the bracket by $\sin x$ and $\cos x$:

$f'(x) = x^3 [ (20 + 3x) \sin x + (5x - 12) \cos x ]$

Thus, the derivative of $x^4 (5 \sin x - 3 \cos x)$ is $x^3 [ (20 + 3x) \sin x + (5x - 12) \cos x ]$.

Let $f(x) = (x^2 + 1) \cos x$. We need to find the derivative of $f(x)$ with respect to $x$.

We will use the product rule for differentiation, which states that if $f(x) = u(x)v(x)$, then the derivative $f'(x)$ is given by the formula:

$f'(x) = u'(x)v(x) + u(x)v'(x)$

Let $u(x) = x^2 + 1$ and $v(x) = \cos x$.

First, find the derivative of $u(x)$ with respect to $x$:

$u'(x) = \frac{d}{dx}(x^2 + 1)$

Using the sum rule and power rule:

$u'(x) = \frac{d}{dx}(x^2) + \frac{d}{dx}(1)$

$u'(x) = 2x^{2-1} + 0$

$u'(x) = 2x$

Next, find the derivative of $v(x)$ with respect to $x$:

$v'(x) = \frac{d}{dx}(\cos x)$

The derivative of $\cos x$ is $-\sin x$.

$v'(x) = -\sin x$

Now, apply the product rule formula $f'(x) = u'(x)v(x) + u(x)v'(x)$:

$f'(x) = (2x)(\cos x) + (x^2 + 1)(-\sin x)$

Simplify the expression:

$f'(x) = 2x \cos x - (x^2 + 1) \sin x$

$f'(x) = 2x \cos x - x^2 \sin x - \sin x$

Thus, the derivative of $(x^2 + 1) \cos x$ is $2x \cos x - (x^2 + 1) \sin x$.

Given:

Here, $a, p,$ and $q$ are fixed non-zero constants.

To Find:

The derivative of $f(x)$, denoted as $f'(x)$ or $\frac{d}{dx}((ax^2 + \sin x)(p + q \cos x))$.

Solution:

We need to find the derivative of a product of two functions. We will use the **Product Rule** of differentiation.

The Product Rule states that if $f(x) = u(x)v(x)$, where $u(x)$ and $v(x)$ are differentiable functions, then the derivative of $f(x)$ is given by:

$f'(x) = u'(x)v(x) + u(x)v'(x)$

In our case, let $u(x) = ax^2 + \sin x$ and $v(x) = p + q \cos x$.

First, find the derivative of $u(x)$ with respect to $x$: $u'(x) = \frac{d}{dx}(ax^2 + \sin x)$.

Using the **Sum Rule**: $u'(x) = \frac{d}{dx}(ax^2) + \frac{d}{dx}(\sin x)$.

Using the **Constant Multiple Rule** and **Power Rule**: $\frac{d}{dx}(ax^2) = a \frac{d}{dx}(x^2) = a(2x) = 2ax$.

Using the derivative of $\sin x$: $\frac{d}{dx}(\sin x) = \cos x$.

So, $u'(x) = 2ax + \cos x$.

Next, find the derivative of $v(x)$ with respect to $x$: $v'(x) = \frac{d}{dx}(p + q \cos x)$.

Using the **Sum Rule**: $v'(x) = \frac{d}{dx}(p) + \frac{d}{dx}(q \cos x)$.

Using the **Constant Rule**: $\frac{d}{dx}(p) = 0$ (since $p$ is a constant).

Using the **Constant Multiple Rule**: $\frac{d}{dx}(q \cos x) = q \frac{d}{dx}(\cos x)$.

Using the derivative of $\cos x$: $\frac{d}{dx}(\cos x) = -\sin x$.

So, $v'(x) = 0 + q(-\sin x) = -q \sin x$.

Now, apply the Product Rule formula $f'(x) = u'(x)v(x) + u(x)v'(x)$ with $u(x) = ax^2 + \sin x$, $v(x) = p + q \cos x$, $u'(x) = 2ax + \cos x$, and $v'(x) = -q \sin x$.

$f'(x) = (2ax + \cos x)(p + q \cos x) + (ax^2 + \sin x)(-q \sin x)$

Rearrange the terms to match the structure of the given answer:

$f'(x) = (ax^2 + \sin x)(-q \sin x) + (2ax + \cos x)(p + q \cos x)$

$f'(x) = -q \sin x (ax^2 + \sin x) + (p + q \cos x)(2ax + \cos x)$

Answer:

The derivative of $(ax^2 + \sin x)(p + q \cos x)$ is $\mathbf{-q \sin x (ax^2 + \sin x) + (p + q \cos x)(2ax + \cos x)}$.

Given:

The function is defined for values of $x$ where $\tan x$ is defined, i.e., $x \neq \frac{\pi}{2} + n\pi$, where $n$ is an integer.

To Find:

The derivative of $f(x)$, denoted as $f'(x)$ or $\frac{d}{dx}((x + \cos x)(x - \tan x))$.

Solution:

We need to find the derivative of a product of two functions. We will use the **Product Rule** of differentiation.

The Product Rule states that if $f(x) = u(x)v(x)$, where $u(x)$ and $v(x)$ are differentiable functions, then the derivative of $f(x)$ is given by:

$f'(x) = u'(x)v(x) + u(x)v'(x)$

In our case, let $u(x) = x + \cos x$ and $v(x) = x - \tan x$.

First, find the derivative of $u(x)$ with respect to $x$: $u'(x) = \frac{d}{dx}(x + \cos x)$.

Using the **Sum Rule**: $u'(x) = \frac{d}{dx}(x) + \frac{d}{dx}(\cos x)$.

We know that $\frac{d}{dx}(x) = 1$ and $\frac{d}{dx}(\cos x) = -\sin x$.

So, $u'(x) = 1 + (-\sin x) = 1 - \sin x$.

Next, find the derivative of $v(x)$ with respect to $x$: $v'(x) = \frac{d}{dx}(x - \tan x)$.

Using the **Difference Rule**: $v'(x) = \frac{d}{dx}(x) - \frac{d}{dx}(\tan x)$.

We know that $\frac{d}{dx}(x) = 1$ and $\frac{d}{dx}(\tan x) = \sec^2 x$.

So, $v'(x) = 1 - \sec^2 x$.

Recall the trigonometric identity $1 + \tan^2 x = \sec^2 x$. Rearranging this, we get $1 - \sec^2 x = -\tan^2 x$.

So, $v'(x) = -\tan^2 x$.

Now, apply the Product Rule formula $f'(x) = u'(x)v(x) + u(x)v'(x)$ with $u(x) = x + \cos x$, $v(x) = x - \tan x$, $u'(x) = 1 - \sin x$, and $v'(x) = 1 - \sec^2 x$ (or $-\tan^2 x$).

$f'(x) = (1 - \sin x)(x - \tan x) + (x + \cos x)(1 - \sec^2 x)$

Using $1 - \sec^2 x = -\tan^2 x$:

$f'(x) = (1 - \sin x)(x - \tan x) + (x + \cos x)(-\tan^2 x)$

$f'(x) = (1 - \sin x)(x - \tan x) - \tan^2 x (x + \cos x)$

The result can be written in different equivalent forms. The provided "correct answer" format is unusual as it includes the original function in the derivative expression, which is generally incorrect unless obtained through specific manipulations or identities not directly from standard derivative rules.

The derivative calculated using the standard Product Rule is:

or

Expanding the terms:

$f'(x) = x - \tan x - x \sin x + \sin x \tan x - x \tan^2 x - \cos x \tan^2 x$

$f'(x) = x - \tan x - x \sin x + \sin x \frac{\sin x}{\cos x} - x \tan^2 x - \cos x \frac{\sin^2 x}{\cos^2 x}$

$f'(x) = x - \tan x - x \sin x + \frac{\sin^2 x}{\cos x} - x \tan^2 x - \frac{\sin^2 x}{\cos x}$

$f'(x) = x - \tan x - x \sin x - x \tan^2 x$

Factoring out $x$ from the terms involving $x$:

$f'(x) = x(1 - \sin x - \tan^2 x) - \tan x$

However, the most direct form using the product rule is usually preferred unless a specific simplified form is required.

Answer:

The derivative of the function is $\mathbf{(1 - \sin x)(x - \tan x) + (x + \cos x)(1 - \sec^2 x)}$ or $\mathbf{(1 - \sin x)(x - \tan x) - \tan^2 x (x + \cos x)}$.

Let $f(x) = \frac{4x \;+\; 5 \sin x}{3x \;+\; 7 \cos x}$. We need to find the derivative of $f(x)$ with respect to $x$.

We will use the quotient rule for differentiation, which states that if $f(x) = \frac{u(x)}{v(x)}$, then the derivative $f'(x)$ is given by the formula:

$f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2}$

Let the numerator be $u(x) = 4x + 5 \sin x$.

Let the denominator be $v(x) = 3x + 7 \cos x$.

First, find the derivative of the numerator, $u'(x)$:

$u'(x) = \frac{d}{dx}(4x + 5 \sin x)$

$u'(x) = \frac{d}{dx}(4x) + \frac{d}{dx}(5 \sin x)$

$u'(x) = 4 \frac{d}{dx}(x) + 5 \frac{d}{dx}(\sin x)$

$u'(x) = 4(1) + 5(\cos x)$

$u'(x) = 4 + 5 \cos x$

Next, find the derivative of the denominator, $v'(x)$:

$v'(x) = \frac{d}{dx}(3x + 7 \cos x)$

$v'(x) = \frac{d}{dx}(3x) + \frac{d}{dx}(7 \cos x)$

$v'(x) = 3 \frac{d}{dx}(x) + 7 \frac{d}{dx}(\cos x)$

$v'(x) = 3(1) + 7(-\sin x)$

$v'(x) = 3 - 7 \sin x$

Now, apply the quotient rule formula:

$f'(x) = \frac{(4 + 5 \cos x)(3x + 7 \cos x) - (4x + 5 \sin x)(3 - 7 \sin x)}{(3x + 7 \cos x)^2}$

Simplify the numerator:

Numerator $= (4 + 5 \cos x)(3x + 7 \cos x) - (4x + 5 \sin x)(3 - 7 \sin x)$

Expand the first product:

$(4)(3x) + (4)(7 \cos x) + (5 \cos x)(3x) + (5 \cos x)(7 \cos x)$

$= 12x + 28 \cos x + 15x \cos x + 35 \cos^2 x$

Expand the second product:

$(4x)(3) + (4x)(-7 \sin x) + (5 \sin x)(3) + (5 \sin x)(-7 \sin x)$

$= 12x - 28x \sin x + 15 \sin x - 35 \sin^2 x$

Subtract the second expanded result from the first:

Numerator $= (12x + 28 \cos x + 15x \cos x + 35 \cos^2 x) - (12x - 28x \sin x + 15 \sin x - 35 \sin^2 x)$

Numerator $= 12x + 28 \cos x + 15x \cos x + 35 \cos^2 x - 12x + 28x \sin x - 15 \sin x + 35 \sin^2 x$

Combine like terms:

Numerator $= (12x - 12x) + (28 \cos x + 15x \cos x) + (28x \sin x - 15 \sin x) + (35 \cos^2 x + 35 \sin^2 x)$

Numerator $= 0 + (28 + 15x) \cos x + (28x - 15) \sin x + 35 (\cos^2 x + \sin^2 x)$

Using the identity $\cos^2 x + \sin^2 x = 1$:

Numerator $= (15x + 28) \cos x + (28x - 15) \sin x + 35(1)$

Numerator $= 35 + (15x + 28) \cos x + (28x - 15) \sin x$

Substitute the simplified numerator back into the quotient rule formula:

$f'(x) = \frac{35 + (15x + 28) \cos x + (28x - 15) \sin x}{(3x + 7 \cos x)^2}$

Thus, the derivative of $\frac{4x \;+\; 5 \sin x}{3x \;+\; 7 \cos x}$ is $\frac{35 + (15x + 28) \cos x + (28x - 15) \sin x}{(3x + 7 \cos x)^2}$.

Let $f(x) = \frac{x^{2}\cos\left( \frac{\pi}{4} \right)}{\sin x}$. We need to find the derivative of $f(x)$ with respect to $x$.

Note that $\cos\left(\frac{\pi}{4}\right)$ is a constant, equal to $\frac{\sqrt{2}}{2}$.

We will use the quotient rule for differentiation, which states that if $f(x) = \frac{u(x)}{v(x)}$, then the derivative $f'(x)$ is given by the formula:

$f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2}$

Let the numerator be $u(x) = x^2 \cos\left( \frac{\pi}{4} \right)$.

Let the denominator be $v(x) = \sin x$.

First, find the derivative of the numerator, $u'(x)$. Since $\cos\left(\frac{\pi}{4}\right)$ is a constant, we use the constant multiple rule:

$u'(x) = \frac{d}{dx}\left( x^2 \cos\left( \frac{\pi}{4} \right) \right)$

$u'(x) = \cos\left( \frac{\pi}{4} \right) \frac{d}{dx}(x^2)$

Using the power rule, $\frac{d}{dx}(x^2) = 2x$.

$u'(x) = \cos\left( \frac{\pi}{4} \right) (2x)$

$u'(x) = 2x \cos\left( \frac{\pi}{4} \right)$

Next, find the derivative of the denominator, $v'(x)$:

$v'(x) = \frac{d}{dx}(\sin x)$

$v'(x) = \cos x$

Now, apply the quotient rule formula:

$f'(x) = \frac{\left(2x \cos\left( \frac{\pi}{4} \right)\right)(\sin x) - \left(x^2 \cos\left( \frac{\pi}{4} \right)\right)(\cos x)}{(\sin x)^2}$

Simplify the numerator:

Numerator $= 2x \cos\left( \frac{\pi}{4} \right) \sin x - x^2 \cos\left( \frac{\pi}{4} \right) \cos x$

Factor out the common term $\cos\left( \frac{\pi}{4} \right)$:

Numerator $= \cos\left( \frac{\pi}{4} \right) (2x \sin x - x^2 \cos x)$

Substitute the simplified numerator back into the quotient rule formula:

$f'(x) = \frac{\cos\left( \frac{\pi}{4} \right) (2x \sin x - x^2 \cos x)}{\sin^2 x}$

We can also substitute the value $\cos\left( \frac{\pi}{4} \right) = \frac{\sqrt{2}}{2}$:

$f'(x) = \frac{\frac{\sqrt{2}}{2} (2x \sin x - x^2 \cos x)}{\sin^2 x}$

$f'(x) = \frac{\sqrt{2} (2x \sin x - x^2 \cos x)}{2 \sin^2 x}$

Thus, the derivative of $\frac{x^{2}\cos\left( \frac{\pi}{4} \right)}{\sin x}$ is $\frac{\cos\left( \frac{\pi}{4} \right) (2x \sin x - x^2 \cos x)}{\sin^2 x}$ or $\frac{\sqrt{2} (2x \sin x - x^2 \cos x)}{2 \sin^2 x}$.

Let $f(x) = \frac{x}{1 \;+\; \tan x}$. We need to find the derivative of $f(x)$ with respect to $x$.

We will use the quotient rule for differentiation, which states that if $f(x) = \frac{u(x)}{v(x)}$, then the derivative $f'(x)$ is given by the formula:

$f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2}$

Let the numerator be $u(x) = x$.

Let the denominator be $v(x) = 1 + \tan x$.

First, find the derivative of the numerator, $u'(x)$:

$u'(x) = \frac{d}{dx}(x)$

$u'(x) = 1$

Next, find the derivative of the denominator, $v'(x)$:

$v'(x) = \frac{d}{dx}(1 + \tan x)$

$v'(x) = \frac{d}{dx}(1) + \frac{d}{dx}(\tan x)$

$v'(x) = 0 + \sec^2 x$

$v'(x) = \sec^2 x$

Now, apply the quotient rule formula:

$f'(x) = \frac{(1)(1 + \tan x) - (x)(\sec^2 x)}{(1 + \tan x)^2}$

Simplify the numerator:

Numerator $= 1 + \tan x - x \sec^2 x$

Substitute the simplified numerator back into the quotient rule formula:

$f'(x) = \frac{1 + \tan x - x \sec^2 x}{(1 + \tan x)^2}$

Thus, the derivative of $\frac{x}{1 \;+\; \tan x}$ is $\frac{1 + \tan x - x \sec^2 x}{(1 + \tan x)^2}$.

Let $f(x) = (x + \sec x) (x - \tan x)$. We need to find the derivative of $f(x)$ with respect to $x$.

We will use the product rule for differentiation, which states that if $f(x) = u(x)v(x)$, then the derivative $f'(x)$ is given by the formula:

$f'(x) = u'(x)v(x) + u(x)v'(x)$

Let $u(x) = x + \sec x$ and $v(x) = x - \tan x$.

First, find the derivative of $u(x)$ with respect to $x$:

$u'(x) = \frac{d}{dx}(x + \sec x)$

Using the sum rule and standard derivatives:

$u'(x) = \frac{d}{dx}(x) + \frac{d}{dx}(\sec x)$

$u'(x) = 1 + \sec x \tan x$

Next, find the derivative of $v(x)$ with respect to $x$:

$v'(x) = \frac{d}{dx}(x - \tan x)$

Using the difference rule and standard derivatives:

$v'(x) = \frac{d}{dx}(x) - \frac{d}{dx}(\tan x)$

$v'(x) = 1 - \sec^2 x$

Now, apply the product rule formula $f'(x) = u'(x)v(x) + u(x)v'(x)$:

$f'(x) = (1 + \sec x \tan x)(x - \tan x) + (x + \sec x)(1 - \sec^2 x)$

Expand the terms:

$(1 + \sec x \tan x)(x - \tan x) = 1(x) - 1(\tan x) + (\sec x \tan x)(x) - (\sec x \tan x)(\tan x)$

$= x - \tan x + x \sec x \tan x - \sec x \tan^2 x$

$(x + \sec x)(1 - \sec^2 x) = x(1) - x(\sec^2 x) + (\sec x)(1) - (\sec x)(\sec^2 x)$

$= x - x \sec^2 x + \sec x - \sec^3 x$

Add the expanded terms:

$f'(x) = (x - \tan x + x \sec x \tan x - \sec x \tan^2 x) + (x - x \sec^2 x + \sec x - \sec^3 x)$

$f'(x) = x - \tan x + x \sec x \tan x - \sec x \tan^2 x + x - x \sec^2 x + \sec x - \sec^3 x$

Combine like terms:

$f'(x) = 2x - \tan x + \sec x + x \sec x \tan x - \sec x \tan^2 x - x \sec^2 x - \sec^3 x$

Thus, the derivative of $(x + \sec x) (x - \tan x)$ is $2x - \tan x + \sec x + x \sec x \tan x - \sec x \tan^2 x - x \sec^2 x - \sec^3 x$.

Let $f(x) = \frac{x}{\sin^{n}\;x}$. We need to find the derivative of $f(x)$ with respect to $x$.

We will use the quotient rule for differentiation, which states that if $f(x) = \frac{u(x)}{v(x)}$, then the derivative $f'(x)$ is given by the formula:

$f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2}$

Let the numerator be $u(x) = x$.

Let the denominator be $v(x) = \sin^n x = (\sin x)^n$.

First, find the derivative of the numerator, $u'(x)$:

$u'(x) = \frac{d}{dx}(x)$

$u'(x) = 1$

Next, find the derivative of the denominator, $v'(x)$. We use the chain rule.

Let $w = \sin x$. Then $v(x) = w^n$.

The derivative of $v(x)$ with respect to $x$ is $\frac{dv}{dx} = \frac{dv}{dw} \cdot \frac{dw}{dx}$.

$\frac{dv}{dw} = \frac{d}{du}(w^n) = nw^{n-1} = n(\sin x)^{n-1} = n \sin^{n-1} x$

$\frac{dw}{dx} = \frac{d}{dx}(\sin x) = \cos x$

So, $v'(x) = (n \sin^{n-1} x) \cdot (\cos x) = n \sin^{n-1} x \cos x$.

Now, apply the quotient rule formula:

$f'(x) = \frac{(1)(\sin^n x) - (x)(n \sin^{n-1} x \cos x)}{(\sin^n x)^2}$

Simplify the expression:

$f'(x) = \frac{\sin^n x - nx \sin^{n-1} x \cos x}{\sin^{2n} x}$

Factor out the common term $\sin^{n-1} x$ from the numerator:

$f'(x) = \frac{\sin^{n-1} x (\sin x - nx \cos x)}{\sin^{2n} x}$

Cancel out $\sin^{n-1} x$ from the numerator and denominator (assuming $\sin x \neq 0$):

The exponent in the denominator becomes $2n - (n-1) = 2n - n + 1 = n+1$.

$f'(x) = \frac{\cancel{\sin^{n-1} x} (\sin x - nx \cos x)}{\sin^{\cancel{2n} n+1} x}$

$f'(x) = \frac{\sin x - nx \cos x}{\sin^{n+1} x}$

Thus, the derivative of $\frac{x}{\sin^{n}\;x}$ is $\frac{\sin x - nx \cos x}{\sin^{n+1} x}$.