Chapter 13 Limits And Derivatives

To find the limit of a polynomial function as $x$ approaches a certain value, we can directly substitute the value of $x$ into the function.

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

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

Substitute $x = 1$ into the function:

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

$= 1 - 1 + 1$

$= 1$

Thus, $\lim\limits_{x \to 1} [x^3 - x^2 + 1] = 1$.

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

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

Substitute $x = 3$ into the function:

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

$= 3(4)$

$= 12$

Thus, $\lim\limits_{x \to 3} [x(x + 1)] = 12$.

(iii) Find $\lim\limits_{x \to -1} [1 + x + x^2 + … + x^{10}]$.

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

Substitute $x = -1$ into the function:

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

$= 1 - 1 + 1 - 1 + 1 - 1 + 1 - 1 + 1 - 1 + 1$

The terms are $1, -1, 1, -1, \dots, 1$. There are 11 terms in total (from $x^0$ to $x^{10}$).

Pairing consecutive terms (except the last one): $(1 - 1) + (1 - 1) + (1 - 1) + (1 - 1) + (1 - 1) + 1$

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

$= 1$

Thus, $\lim\limits_{x \to -1} [1 + x + x^2 + … + x^{10}] = 1$.

To find the limit of a rational function $\frac{f(x)}{g(x)}$ as $x \to c$, we first check the value of the denominator at $x=c$. If $g(c) \neq 0$, we can substitute $x=c$ directly. If $g(c) = 0$ and $f(c) = 0$, it is an indeterminate form $\frac{0}{0}$, and we need to simplify the expression by factoring or other methods. If $g(c) = 0$ and $f(c) \neq 0$, the limit is usually $\pm\infty$ or does not exist.

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

The denominator at $x = 1$ is $1 + 100 = 101$, which is not equal to 0. We can substitute $x = 1$ directly into the expression.

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

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

$= \frac{2}{101}$

Thus, $\lim\limits_{x \to 1} \left[ \frac{x^2 \;+ \; 1}{x \;+\; 100} \right] = \frac{2}{101}$.

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

At $x = 2$, the numerator is $(2)^3 - 4(2)^2 + 4(2) = 8 - 16 + 8 = 0$.

The denominator is $(2)^2 - 4 = 4 - 4 = 0$.

This is the indeterminate form $\frac{0}{0}$. We need to factorize the numerator and the denominator.

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

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

So, the expression becomes $\frac{x(x - 2)^2}{(x - 2)(x + 2)}$. For $x \neq 2$, we can cancel one factor of $(x - 2)$ from the numerator and denominator.

$\lim\limits_{x \to 2} \frac{x(x - 2)\cancel{^2}}{\cancel{(x - 2)}(x + 2)} = \lim\limits_{x \to 2} \frac{x(x - 2)}{x + 2}$

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

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

Thus, $\lim\limits_{x \to 2} \left[ \frac{x^{3}\;-\;4x^{2}\;+\;4x}{x^{2}\;-\;4} \right] = 0$.

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

At $x = 2$, the numerator is $(2)^2 - 4 = 0$.

The denominator is $(2)^3 - 4(2)^2 + 4(2) = 0$.

This is the indeterminate form $\frac{0}{0}$. We factorize the numerator and the denominator.

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

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

So, the expression becomes $\frac{(x - 2)(x + 2)}{x(x - 2)^2}$. For $x \neq 2$, we can cancel one factor of $(x - 2)$.

$\lim\limits_{x \to 2} \frac{\cancel{(x - 2)}(x + 2)}{x(x - 2)\cancel{^2}} = \lim\limits_{x \to 2} \frac{x + 2}{x(x - 2)}$

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

Numerator approaches $2 + 2 = 4$.

Denominator approaches $2(2 - 2) = 2(0) = 0$.

Since the numerator approaches a non-zero value and the denominator approaches 0, the limit does not exist. To be more precise, we can examine the one-sided limits.

As $x \to 2^+$, $x > 2$, so $x - 2 > 0$. $x(x - 2)$ is positive. $\lim\limits_{x \to 2^+} \frac{x + 2}{x(x - 2)} = \frac{4}{0^+} = +\infty$.

As $x \to 2^-$, $x < 2$, so $x - 2 < 0$. $x(x - 2)$ is negative. $\lim\limits_{x \to 2^-} \frac{x + 2}{x(x - 2)} = \frac{4}{0^-} = -\infty$.

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

Thus, $\lim\limits_{x \to 2} \left[ \frac{x^{2}\;-\;4}{x^{3}\;-\;4x^{2}\;+\;4x} \right]$ does not exist.

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

At $x = 2$, the numerator is $(2)^3 - 2(2)^2 = 8 - 8 = 0$.

The denominator is $(2)^2 - 5(2) + 6 = 4 - 10 + 6 = 0$.

This is the indeterminate form $\frac{0}{0}$. We need to factorize the numerator and the denominator.

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

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

So, the expression becomes $\frac{x^2(x - 2)}{(x - 2)(x - 3)}$. For $x \neq 2$, we can cancel the factor $(x - 2)$.

$\lim\limits_{x \to 2} \frac{x^2\cancel{(x - 2)}}{\cancel{(x - 2)}(x - 3)} = \lim\limits_{x \to 2} \frac{x^2}{x - 3}$

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

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

Thus, $\lim\limits_{x \to 2} \left[ \frac{x^{3}\;-\;2x^{2}}{x^{2}\;-\;5x\;+\;6} \right] = -4$.

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

First, simplify the expression by combining the two fractions.

Factor the denominators:

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

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

The common denominator is $x(x - 1)(x - 2)$.

Combine the fractions:

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

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

Expand and simplify the numerator:

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

Factor the numerator: $x^2 - 4x + 3 = (x - 1)(x - 3)$

The expression becomes $\frac{(x - 1)(x - 3)}{x(x - 1)(x - 2)}$.

Now, find the limit of this simplified expression as $x \to 1$:

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

At $x = 1$, the numerator is $(1 - 1)(1 - 3) = 0(-2) = 0$.

The denominator is $1(1 - 1)(1 - 2) = 1(0)(-1) = 0$.

This is the indeterminate form $\frac{0}{0}$. For $x \neq 1$, we can cancel the factor $(x - 1)$ from the numerator and denominator.

$\lim\limits_{x \to 1} \frac{\cancel{(x - 1)}(x - 3)}{x\cancel{(x - 1)}(x - 2)} = \lim\limits_{x \to 1} \frac{x - 3}{x(x - 2)}$

Now, substitute $x = 1$ into the simplified expression:

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

Thus, $\lim\limits_{x \to 1} \left[ \frac{x\;-\;2}{x^{2}\;-\;x}-\frac{1}{x^3\;-\;3x^2\;+\;2x} \right] = 2$.

We will use some standard limit formulas and algebraic manipulation to evaluate the given limits.

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

At $x = 1$, the numerator is $1^{15} - 1 = 1 - 1 = 0$.

The denominator is $1^{10} - 1 = 1 - 1 = 0$.

This is the indeterminate form $\frac{0}{0}$. We can use the standard limit formula: $\lim\limits_{x \to a} \frac{x^n - a^n}{x - a} = n a^{n-1}$.

We can rewrite the expression by dividing the numerator and denominator by $(x - 1)$, since $x \to 1$ implies $x \neq 1$.

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

Apply the limit formula to both the numerator and the denominator, 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 $15 \times 1^{15-1} = 15 \times 1^{14} = 15 \times 1 = 15$.

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

So, the limit of the given expression is the ratio of these limits:

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

Simplify the fraction:

$\frac{15}{10} = \frac{3 \times 5}{2 \times 5} = \frac{3}{2}$

Thus, $\lim\limits_{x \to 1} \frac{x^{15}\;-\;1}{x^{10}\;-\;1} = \frac{3}{2}$.

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

At $x = 0$, the numerator is $\sqrt{1 + 0} - 1 = \sqrt{1} - 1 = 1 - 1 = 0$.

The denominator is $0$.

This is the indeterminate form $\frac{0}{0}$. We can rationalize the numerator.

Multiply the numerator and the denominator by the conjugate of the numerator, which is $\sqrt{1 + x} + 1$:

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

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

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

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

For $x \neq 0$, we can cancel the factor $x$ from the numerator and denominator.

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

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

Now, substitute $x = 0$ into the simplified expression:

$= \frac{1}{\sqrt{1\;+\;0}\;+\;1} = \frac{1}{\sqrt{1}\;+\;1} = \frac{1}{1\;+\;1} = \frac{1}{2}$

Thus, $\lim\limits_{x \to 0} \frac{\sqrt{1\;+\;x}\;-\;1}{x} = \frac{1}{2}$.

Alternate Method for (ii):

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

Let $y = 1 + x$. As $x \to 0$, $y \to 1$. The expression becomes:

$\lim\limits_{y \to 1} \frac{\sqrt{y}\;-\;1}{y\;-\;1} = \lim\limits_{y \to 1} \frac{y^{1/2}\;-\;1^{1/2}}{y\;-\;1}$

Here, $y$ is the variable, $a = 1$, and $n = \frac{1}{2}$.

Using the formula, the limit is $\frac{1}{2} \times 1^{\frac{1}{2} - 1} = \frac{1}{2} \times 1^{-\frac{1}{2}} = \frac{1}{2} \times 1 = \frac{1}{2}$.

This confirms the result obtained by rationalization.

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

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

At $x = 0$, the numerator is $\sin(4 \times 0) = \sin(0) = 0$.

The denominator is $\sin(2 \times 0) = \sin(0) = 0$.

This is the indeterminate form $\frac{0}{0}$. We can rewrite the expression to use the standard limit formula.

Divide the numerator and the denominator by $x$:

$\lim\limits_{x \to 0} \frac{\frac{\sin4x}{x}}{\frac{\sin2x}{x}}$

Now, multiply and divide the numerator by 4 and the denominator by 2 to match the arguments of the sine functions:

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

As $x \to 0$, $4x \to 0$ and $2x \to 0$. We can use the property $\lim\limits_{x \to a} \frac{f(x)}{g(x)} = \frac{\lim\limits_{x \to a} f(x)}{\lim\limits_{x \to a} g(x)}$ and the standard limit $\lim\limits_{\theta \to 0} \frac{\sin\theta}{\theta} = 1$.

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

$= \frac{1 \times 4}{1 \times 2}$

$= \frac{4}{2} = 2$

Thus, $\lim\limits_{x \to 0} \frac{\sin4x}{\sin2x} = 2$.

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

At $x = 0$, the numerator is $\tan(0) = 0$.

The denominator is $0$.

This is the indeterminate form $\frac{0}{0}$. We can use the definition of $\tan x$ and the standard limit for $\sin x$.

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

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

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

Using the limit properties $\lim\limits_{x \to a} (f(x) \times g(x)) = \lim\limits_{x \to a} f(x) \times \lim\limits_{x \to a} g(x)$:

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

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

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

So, the limit is $1 \times 1 = 1$.

Thus, $\lim\limits_{x \to 0} \frac{\tan x}{x} = 1$.

Note: The limit $\lim\limits_{\theta \to 0} \frac{\tan\theta}{\theta} = 1$ is a standard limit result, so you could directly state the result if it's allowed in your context.

To evaluate the limit of a polynomial function as $x$ approaches a specific value, we can directly substitute the value of $x$ into the function.

Evaluate $\lim\limits_{x \to 3} (x + 3)$.

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

Substitute $x = 3$ into the function:

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

$= 6$

Thus, $\lim\limits_{x \to 3} x + 3 = 6$.

To evaluate the limit of a linear function as $x$ approaches a specific value, we can directly substitute the value of $x$ into the function.

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

The function is $f(x) = x - \frac{22}{7}$.

Substitute $x = \pi$ into the function:

$\lim\limits_{x \to \pi}\left( x - \frac{22}{7} \right) = \pi - \frac{22}{7}$

Thus, $\lim\limits_{x \to \pi}\left( x - \frac{22}{7} \right) = \pi - \frac{22}{7}$.

To evaluate the limit of a power function (which is a polynomial) as the variable approaches a specific value, we can directly substitute the value into the function. Here, the variable is $r$, and the constant is $\pi$.

Evaluate $\lim\limits_{r \to 1} \pi r^2$.

The function is $f(r) = \pi r^2$.

Substitute $r = 1$ into the function:

$\lim\limits_{r \to 1} \pi r^2 = \pi (1)^2$

$= \pi (1)$

$= \pi$

Thus, $\lim\limits_{r \to 1} \pi r^2 = \pi$.

To evaluate the limit of a rational function $\frac{f(x)}{g(x)}$ as $x$ approaches a specific value, we first check the value of the denominator at that value. If the denominator is non-zero, we can substitute the value of $x$ directly into the function.

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

The function is $f(x) = \frac{4x + 3}{x - 2}$.

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

Since the denominator is 2 (which is not 0), we can substitute $x = 4$ directly into the expression:

$\lim\limits_{x \to 4} \frac{4x + 3}{x - 2} = \frac{4(4) + 3}{4 - 2}$

$= \frac{16 + 3}{2}$

$= \frac{19}{2}$

Thus, $\lim\limits_{x \to 4}\frac{4x \;+\; 3}{x \;-\; 2} = \frac{19}{2}$.

To evaluate the limit of a rational function $\frac{f(x)}{g(x)}$ as $x$ approaches a specific value, we first check the value of the denominator at that value. If the denominator is non-zero, we can substitute the value of $x$ directly into the function.

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

The function is $f(x) = \frac{x^{10} + x^5 + 1}{x - 1}$.

Check the denominator at $x = -1$: $x - 1 = -1 - 1 = -2$.

Since the denominator is -2 (which is not 0), we can substitute $x = -1$ directly into the expression:

$\lim\limits_{x \to -1} \frac{x^{10} + x^5 + 1}{x - 1} = \frac{(-1)^{10} + (-1)^5 + 1}{-1 - 1}$

Recall that an even power of -1 is 1, and an odd power of -1 is -1.

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

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

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

Thus, $\lim\limits_{x \to -1}\frac{x^{10}\;+\;x^{5}\;+\;1}{x\;-\;1} = -\frac{1}{2}$.

To evaluate this limit, we can use the standard limit formula: $\lim\limits_{z \to a} \frac{z^n - a^n}{z - a} = n a^{n-1}$.

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

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

Also, from $z = x + 1$, we have $x = z - 1$.

Substitute these into the limit expression:

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

This matches the standard limit formula with $z$ as the variable, $a = 1$, and $n = 5$.

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

Thus, $\lim\limits_{x \to 0}\frac{(x \;+\; 1)^{5} \;-\; 1}{x} = 5$.

Alternate Method (using binomial expansion):

Expand $(x + 1)^5$ using the binomial theorem: $(x + 1)^5 = 1^5 + \binom{5}{1}1^4 x^1 + \binom{5}{2}1^3 x^2 + \binom{5}{3}1^2 x^3 + \binom{5}{4}1^1 x^4 + \binom{5}{5} x^5$

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

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

The expression becomes $\frac{5x + 10x^2 + 10x^3 + 5x^4 + x^5}{x}$.

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

$\frac{x(5 + 10x + 10x^2 + 5x^3 + x^4)}{x} = 5 + 10x + 10x^2 + 5x^3 + x^4$

Now, evaluate the limit as $x \to 0$ for this polynomial:

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

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

This confirms the result.

To evaluate this limit, we first check the values of the numerator and denominator at $x = 2$.

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

At $x = 2$:

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

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

This is the indeterminate form $\frac{0}{0}$. We need to factorize the numerator and the denominator to cancel the common factor $(x - 2)$.

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

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

Factorize the numerator $3x^2 - x - 10$. We can find the roots of $3x^2 - x - 10 = 0$ using the quadratic formula or by splitting the middle term.

By splitting the middle term: We need 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, rewrite the expression with the factored numerator and denominator:

$\lim\limits_{x \to 2}\frac{(x \;-\; 2)(3x \;+\; 5)}{(x \;-\; 2)(x \;+\; 2)}$

For $x \neq 2$, we can cancel the common factor $(x - 2)$:

$\lim\limits_{x \to 2}\frac{\cancel{(x \;-\; 2)}(3x \;+\; 5)}{\cancel{(x \;-\; 2)}(x \;+\; 2)} = \lim\limits_{x \to 2}\frac{3x \;+\; 5}{x \;+\; 2}$

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

$\frac{3(2) \;+\; 5}{2 \;+\; 2} = \frac{6 \;+\; 5}{4} = \frac{11}{4}$

Thus, $\lim\limits_{x \to 2}\frac{3x^2 \;-\; x \;-\; 10}{x^2 \;-\; 4} = \frac{11}{4}$.

To evaluate this limit, we first check the values of the numerator and denominator at $x = 3$.

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

At $x = 3$:

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

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

This is the indeterminate form $\frac{0}{0}$. We need to factorize the numerator and the denominator to cancel the common factor $(x - 3)$.

Factorize the numerator using the difference of squares formula twice:

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

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

Factorize the denominator $2x^2 - 5x - 3$. We can find the roots of $2x^2 - 5x - 3 = 0$ using the quadratic formula or by splitting the middle term.

By splitting the middle term: We need 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, rewrite the expression with the factored numerator and denominator:

$\lim\limits_{x \to 3}\frac{(x \;-\; 3)(x \;+\; 3)(x^2 \;+\; 9)}{(x \;-\; 3)(2x \;+\; 1)}$

For $x \neq 3$, we can cancel the common factor $(x - 3)$:

$\lim\limits_{x \to 3}\frac{\cancel{(x \;-\; 3)}(x \;+\; 3)(x^2 \;+\; 9)}{\cancel{(x \;-\; 3)}(2x \;+\; 1)} = \lim\limits_{x \to 3}\frac{(x \;+\; 3)(x^2 \;+\; 9)}{2x \;+\; 1}$

Now, substitute $x = 3$ into the simplified expression:

$\frac{(3 \;+\; 3)(3^2 \;+\; 9)}{2(3) \;+\; 1} = \frac{(6)(9 \;+\; 9)}{6 \;+\; 1}$

$= \frac{6 \times 18}{7} = \frac{108}{7}$

Thus, $\lim\limits_{x \to 3}\frac{x^4 \;-\; 81}{2x^2 \;-\; 5x \;-\; 3} = \frac{108}{7}$.

To evaluate the limit of a rational function $\frac{f(x)}{g(x)}$ as $x$ approaches a specific value, we first check the value of the denominator at that value. If the denominator is non-zero, we can substitute the value of $x$ directly into the function.

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

The function is $f(x) = \frac{ax + b}{cx + 1}$.

Check the denominator at $x = 0$: $cx + 1 = c(0) + 1 = 0 + 1 = 1$.

Since the denominator is 1 (which is not 0), we can substitute $x = 0$ directly into the expression:

$\lim\limits_{x \to 0} \frac{ax + b}{cx + 1} = \frac{a(0) + b}{c(0) + 1}$

$= \frac{0 + b}{0 + 1}$

$= \frac{b}{1} = b$

Thus, $\lim\limits_{x \to 0} \frac{ax \;+\; b}{cx \;+\; 1} = b$.

To evaluate this limit, we first check the values of the numerator and denominator at $z = 1$.

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

At $z = 1$:

Numerator: $1^{\frac{1}{3}} - 1 = 1 - 1 = 0$.

Denominator: $1^{\frac{1}{6}} - 1 = 1 - 1 = 0$.

This is the indeterminate form $\frac{0}{0}$. We can use the standard limit formula: $\lim\limits_{x \to a} \frac{x^n - a^n}{x - a} = n a^{n-1}$.

We can rewrite the expression by dividing the numerator and denominator by $(z - 1)$, since $z \to 1$ implies $z \neq 1$.

$\lim\limits_{z \to 1} \frac{z^{\frac{1}{3}} \;-\; 1}{z^{\frac{1}{6}} \;-\; 1} = \lim\limits_{z \to 1} \frac{\frac{z^{\frac{1}{3}} \;-\; 1}{z\;-\;1}}{\frac{z^{\frac{1}{6}} \;-\; 1}{z\;-\;1}}$

Apply the limit formula to both the numerator and the denominator, with $a = 1$.

For the numerator: $\lim\limits_{z \to 1} \frac{z^{\frac{1}{3}} \;-\; 1^{1/3}}{z\;-\;1}$. 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: $\lim\limits_{z \to 1} \frac{z^{\frac{1}{6}} \;-\; 1^{1/6}}{z\;-\;1}$. Here, $n = \frac{1}{6}$ and $a = 1$. The limit is $\frac{1}{6} \times 1^{\frac{1}{6}-1} = \frac{1}{6} \times 1^{-\frac{5}{6}} = \frac{1}{6} \times 1 = \frac{1}{6}$.

So, the limit of the given expression is the ratio of these limits:

$\lim\limits_{z \to 1} \frac{z^{\frac{1}{3}} \;-\; 1}{z^{\frac{1}{6}} \;-\; 1} = \frac{\frac{1}{3}}{\frac{1}{6}}$

$= \frac{1}{3} \times \frac{6}{1} = \frac{6}{3} = 2$

Thus, $\lim\limits_{z \to 1} \frac{z^{\frac{1}{3}} \;-\; 1}{z^{\frac{1}{6}} \;-\; 1} = 2$.

Alternate Method (using substitution):

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

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

The limit expression becomes:

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

The numerator can be factored as a difference of squares: $x^2 - 1 = (x - 1)(x + 1)$.

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

For $x \neq 1$, we can cancel the common factor $(x - 1)$:

$\lim\limits_{x \to 1} (x + 1)$

Now, substitute $x = 1$:

$1 + 1 = 2$

This confirms the result.

To evaluate the limit of a rational function $\frac{f(x)}{g(x)}$ as $x$ approaches a specific value, we first check the value of the denominator at that value. If the denominator is non-zero, we can substitute the value of $x$ directly into the function.

Evaluate $\lim\limits_{x \to 1} \frac{ax^2 \;+\; bx \;+\; c}{cx^2 \;+\; bx \;+\; a}$. Note: The question provided $ax^2 + bx + x$ in the numerator, but the common form of such problems and the context suggest it should be $ax^2 + bx + c$. Assuming the standard form $ax^2 + bx + c$ for the numerator.

Let the function be $f(x) = \frac{ax^2 + bx + c}{cx^2 + bx + a}$.

Check the denominator at $x = 1$: $cx^2 + bx + a = c(1)^2 + b(1) + a = c + b + a$.

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

Since the denominator is $a + b + c$ (which is not 0), we can substitute $x = 1$ directly into the expression:

$\lim\limits_{x \to 1} \frac{ax^2 + bx + c}{cx^2 + bx + a} = \frac{a(1)^2 + b(1) + c}{c(1)^2 + b(1) + a}$

$= \frac{a + b + c}{c + b + a}$

Since $a + b + c \neq 0$, we can cancel the numerator and denominator.

$= 1$

Thus, $\lim\limits_{x \to 1} \frac{ax^2 \;+\; bx \;+\; c}{cx^2 \;+\; bx \;+\; a} = 1$, assuming the numerator is $ax^2 + bx + c$.

Considering the numerator as given in the question:

Let the function be $f(x) = \frac{ax^2 + bx + x}{cx^2 + bx + a} = \frac{ax^2 + (b+1)x}{cx^2 + bx + a}$.

Check the denominator at $x = 1$: $c(1)^2 + b(1) + a = c + b + a$.

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

Since the denominator is $a + b + c$ (which is not 0), we can substitute $x = 1$ directly into the expression:

$\lim\limits_{x \to 1} \frac{ax^2 + bx + x}{cx^2 + bx + a} = \frac{a(1)^2 + b(1) + 1}{c(1)^2 + b(1) + a}$

$= \frac{a + b + 1}{c + b + a}$

Thus, if the numerator is $ax^2 + bx + x$, then $\lim\limits_{x \to 1} \frac{ax^2 \;+\; bx \;+\; x}{cx^2 \;+\; bx \;+\; a} = \frac{a + b + 1}{a + b + c}$.

Since the standard form for such a problem usually involves $c$ in the numerator, the first interpretation is more likely the intended question. However, based strictly on the input, the second result is correct.

To evaluate this limit, we first check the values of the numerator and denominator at $x = -2$.

Evaluate $\lim\limits_{x \to -2} \frac{\frac{1}{x}+\frac{1}{2}}{x \;+\; 2}$.

At $x = -2$:

Numerator: $\frac{1}{-2} + \frac{1}{2} = -\frac{1}{2} + \frac{1}{2} = 0$.

Denominator: $-2 + 2 = 0$.

This is the indeterminate form $\frac{0}{0}$. We need to simplify the expression by combining the fractions in the numerator.

Combine the fractions in the numerator:

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

Now, rewrite the entire expression:

$\frac{\frac{2 \;+\; x}{2x}}{x \;+\; 2}$

Divide the numerator by the denominator (which is multiplying the numerator by the reciprocal of the denominator):

$\frac{2 \;+\; x}{2x} \times \frac{1}{x \;+\; 2} = \frac{x \;+\; 2}{2x(x \;+\; 2)}$

Now, find the limit of this simplified expression as $x \to -2$:

$\lim\limits_{x \to -2} \frac{x \;+\; 2}{2x(x \;+\; 2)}$

For $x \neq -2$, we can cancel the common factor $(x + 2)$:

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

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

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

Thus, $\lim\limits_{x \to -2} \frac{\frac{1}{x}+\frac{1}{2}}{x \;+\; 2} = -\frac{1}{4}$.

To evaluate this limit, we will use the standard trigonometric limit formula: $\lim\limits_{\theta \to 0} \frac{\sin\theta}{\theta} = 1$.

Evaluate $\lim\limits_{x \to 0} \frac{\sin ax}{bx}$.

At $x = 0$, the numerator is $\sin(a \times 0) = \sin(0) = 0$.

The denominator is $b \times 0 = 0$.

This is the indeterminate form $\frac{0}{0}$ (assuming $b \neq 0$).

We can rewrite the expression to use the standard limit formula. Multiply and divide by $ax$ in the numerator:

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

For $x \neq 0$ and $b \neq 0$, we can simplify the second term:

$\frac{ax}{bx} = \frac{a}{b}$

So, the expression becomes:

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

Using the limit property $\lim\limits_{x \to a} (f(x) \times g(x)) = \lim\limits_{x \to a} f(x) \times \lim\limits_{x \to a} g(x)$:

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

As $x \to 0$, $ax \to 0$. Using the standard limit $\lim\limits_{\theta \to 0} \frac{\sin\theta}{\theta} = 1$ (where $\theta = ax$):

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

The second limit is the limit of a constant, which is the constant itself:

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

So, the limit is $1 \times \frac{a}{b} = \frac{a}{b}$.

Thus, $\lim\limits_{x \to 0} \frac{\sin ax}{bx} = \frac{a}{b}$, provided $b \neq 0$.

If $b = 0$, the original expression is $\lim\limits_{x \to 0} \frac{\sin ax}{0}$.

If $a \neq 0$, as $x \to 0$, $\sin ax \to \sin 0 = 0$. The expression is of the form $\frac{0}{0}$ if $a \neq 0$, but the denominator is identically zero, which is not a standard indeterminate form. If $b=0$, the function is undefined for all $x$. Thus, the limit does not exist when $b=0$ unless $a=0$. If $a=0$, the numerator is $\sin(0) = 0$, so the expression is $\frac{0}{0}$ for all $x$. In that case, if $b \neq 0$, the limit is 0. If $a=0$ and $b=0$, the expression is $\frac{0}{0}$, which is undefined.

Assuming the question implies the case where the limit exists in the $\frac{0}{0}$ form, we assume $b \neq 0$.

To evaluate this limit, we will use the standard trigonometric limit formula: $\lim\limits_{\theta \to 0} \frac{\sin\theta}{\theta} = 1$. We are given that $a \neq 0$ and $b \neq 0$.

Evaluate $\lim\limits_{x \to 0} \frac{\sin ax}{\sin bx}$.

At $x = 0$, the numerator is $\sin(a \times 0) = \sin(0) = 0$.

The denominator is $\sin(b \times 0) = \sin(0) = 0$.

This is the indeterminate form $\frac{0}{0}$. We can rewrite the expression to use the standard limit formula.

Divide the numerator and the denominator by $x$:

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

Now, multiply and divide the numerator by $a$ and the denominator by $b$ to match the arguments of the sine functions:

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

As $x \to 0$, $ax \to 0$ and $bx \to 0$. We can use the property $\lim\limits_{x \to a} \frac{f(x)}{g(x)} = \frac{\lim\limits_{x \to a} f(x)}{\lim\limits_{x \to a} g(x)}$ and the standard limit $\lim\limits_{\theta \to 0} \frac{\sin\theta}{\theta} = 1$.

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

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

$= \frac{a}{b}$

Thus, $\lim\limits_{x \to 0} \frac{\sin ax}{\sin bx} = \frac{a}{b}$, provided $a \neq 0$ and $b \neq 0$.

To evaluate this limit, we will use the standard trigonometric limit formula: $\lim\limits_{\theta \to 0} \frac{\sin\theta}{\theta} = 1$.

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

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

The expression involves $\sin(\pi - x)$ and $\pi - x$. Notice that the argument of the sine function is $(\pi - x)$, and a factor of $(\pi - x)$ is in the denominator. Also, there is a constant factor of $\pi$ in the denominator.

We can rewrite the limit in terms of $\theta$:

$\lim\limits_{\theta \to 0} \frac{\sin \theta}{\pi \theta}$

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

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

Now, we use the standard limit formula $\lim\limits_{\theta \to 0} \frac{\sin\theta}{\theta} = 1$.

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

$= \frac{1}{\pi}$

Thus, $\lim\limits_{x \to \pi} \frac{\sin \;(\pi \;-\; x)}{\pi \;(\pi \;-\; x)} = \frac{1}{\pi}$.

To evaluate the limit of a rational function where the numerator and denominator are well-behaved functions (like trigonometric and polynomial functions), we first check the value of the denominator at the limit point. If the denominator is non-zero, we can substitute the value of $x$ directly into the function.

Evaluate $\lim\limits_{x \to 0} \frac{\cos x}{\pi \;-\; x}$.

The function is $f(x) = \frac{\cos x}{\pi - x}$.

Check the denominator at $x = 0$: $\pi - x = \pi - 0 = \pi$.

Since the denominator is $\pi$ (which is not 0), we can substitute $x = 0$ directly into the expression:

$\lim\limits_{x \to 0} \frac{\cos x}{\pi - x} = \frac{\cos(0)}{\pi - 0}$

We know that $\cos(0) = 1$.

$= \frac{1}{\pi}$

Thus, $\lim\limits_{x \to 0} \frac{\cos x}{\pi \;-\; x} = \frac{1}{\pi}$.

To evaluate this limit, we first check the values of the numerator and denominator at $x = 0$.

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

At $x = 0$:

Numerator: $\cos(2 \times 0) - 1 = \cos(0) - 1 = 1 - 1 = 0$.

Denominator: $\cos(0) - 1 = 1 - 1 = 0$.

This is the indeterminate form $\frac{0}{0}$. We can use trigonometric identities to simplify the expression.

We use the identity $1 - \cos 2\theta = 2\sin^2\theta$ and $1 - \cos \theta = 2\sin^2(\frac{\theta}{2})$.

The numerator is $\cos 2x - 1 = -(1 - \cos 2x) = -2\sin^2 x$.

The denominator is $\cos x - 1 = -(1 - \cos x) = -2\sin^2(\frac{x}{2})$.

So, the expression becomes:

$\lim\limits_{x \to 0} \frac{-2\sin^2 x}{-2\sin^2(\frac{x}{2})}$

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

$= \lim\limits_{x \to 0} \left( \frac{\sin x}{\sin(\frac{x}{2})} \right)^2$

We can rewrite the term inside the square to use the standard limit $\lim\limits_{\theta \to 0} \frac{\sin\theta}{\theta} = 1$.

$\frac{\sin x}{\sin(\frac{x}{2})} = \frac{\frac{\sin x}{x} \times x}{\frac{\sin(\frac{x}{2})}{\frac{x}{2}} \times \frac{x}{2}}$

$= \frac{\frac{\sin x}{x}}{\frac{\sin(\frac{x}{2})}{\frac{x}{2}}} \times \frac{x}{\frac{x}{2}}$

For $x \neq 0$, $\frac{x}{\frac{x}{2}} = x \times \frac{2}{x} = 2$.

So, $\lim\limits_{x \to 0} \frac{\sin x}{\sin(\frac{x}{2})} = \lim\limits_{x \to 0} \left( \frac{\frac{\sin x}{x}}{\frac{\sin(\frac{x}{2})}{\frac{x}{2}}} \times 2 \right)$

As $x \to 0$, $\frac{x}{2} \to 0$. Using the standard limit:

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

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

Now, substitute this back into the original limit expression, which was squared:

$\lim\limits_{x \to 0} \left( \frac{\sin x}{\sin(\frac{x}{2})} \right)^2 = (2)^2 = 4$

Thus, $\lim\limits_{x \to 0} \frac{\cos 2x \;-\; 1}{\cos x \;-\; 1} = 4$.

Alternate Method (using L'Hopital's Rule, if applicable):

If L'Hopital's Rule is allowed, we can take the derivative of the numerator and the denominator.

Derivative of numerator: $\frac{d}{dx}(\cos 2x - 1) = -2\sin 2x$

Derivative of denominator: $\frac{d}{dx}(\cos x - 1) = -\sin x$

The limit becomes $\lim\limits_{x \to 0} \frac{-2\sin 2x}{-\sin x} = \lim\limits_{x \to 0} \frac{2\sin 2x}{\sin x}$.

At $x = 0$, this is still $\frac{0}{0}$. Apply L'Hopital's Rule again.

Derivative of new numerator: $\frac{d}{dx}(2\sin 2x) = 2(2\cos 2x) = 4\cos 2x$

Derivative of new denominator: $\frac{d}{dx}(\sin x) = \cos x$

The limit becomes $\lim\limits_{x \to 0} \frac{4\cos 2x}{\cos x}$.

Now, substitute $x = 0$:

$\frac{4\cos(2 \times 0)}{\cos(0)} = \frac{4\cos(0)}{\cos(0)} = \frac{4 \times 1}{1} = 4$

This confirms the result.

To evaluate this limit, we first check the values of the numerator and denominator at $x = 0$.

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

At $x = 0$:

Numerator: $a(0) + 0 \times \cos(0) = 0 + 0 \times 1 = 0$.

Denominator: $b \sin(0) = b \times 0 = 0$.

This is the indeterminate form $\frac{0}{0}$ (assuming $b \neq 0$). We can simplify the expression by factoring out $x$ from the numerator and rewriting the denominator.

Numerator: $ax + x\cos x = x(a + \cos x)$

Denominator: $b\sin x$

The expression becomes $\frac{x(a \;+\; \cos x)}{b\sin x}$.

We can rewrite this as $\frac{a \;+\; \cos x}{b} \times \frac{x}{\sin x}$.

Now, find the limit as $x \to 0$:

$\lim\limits_{x \to 0} \left( \frac{a \;+\; \cos x}{b} \times \frac{x}{\sin x} \right)$

Using the limit property for product:

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

For the first limit, substitute $x = 0$ (assuming $b \neq 0$):

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

For the second limit, we know the standard limit $\lim\limits_{x \to 0} \frac{\sin x}{x} = 1$. The reciprocal of this limit is also 1:

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

So, the limit of the expression is the product of these limits:

$= \frac{a \;+\; 1}{b} \times 1 = \frac{a \;+\; 1}{b}$

Thus, $\lim\limits_{x \to 0} \frac{ax \;+\; x\cos x}{b\sin x} = \frac{a \;+\; 1}{b}$, provided $b \neq 0$.

If $b = 0$, the denominator is always 0 (except where $\sin x = 0$). If $a \neq -1$, the numerator approaches $a$, so the limit would be $\pm \infty$ or not exist. If $a = -1$ and $b = 0$, the expression is $\frac{-x + x\cos x}{0}$, which requires further analysis.

Assuming the problem is set up such that the limit exists in the $\frac{0}{0}$ form, we assume $b \neq 0$.

To evaluate this limit, we can rewrite $\sec x$ in terms of $\cos x$ and then substitute the limit value.

Evaluate $\lim\limits_{x \to 0} x\sec x$.

Recall that $\sec x = \frac{1}{\cos x}$.

So, the expression is $x \times \frac{1}{\cos x} = \frac{x}{\cos x}$.

Now, find the limit of this expression as $x \to 0$:

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

Check the denominator at $x = 0$: $\cos(0) = 1$.

Since the denominator is 1 (which is not 0), we can substitute $x = 0$ directly into the expression:

$\frac{0}{\cos(0)} = \frac{0}{1} = 0$

Thus, $\lim\limits_{x \to 0} x\sec x = 0$.

To evaluate this limit, we first check the values of the numerator and denominator at $x = 0$.

Evaluate $\lim\limits_{x \to 0} \frac{\sin ax\;+\;bx}{ax\;+\;\sin bx}$.

At $x = 0$:

Numerator: $\sin(a \times 0) + b(0) = \sin(0) + 0 = 0 + 0 = 0$.

Denominator: $a(0) + \sin(b \times 0) = 0 + \sin(0) = 0 + 0 = 0$.

This is the indeterminate form $\frac{0}{0}$. We can use the standard trigonometric limit formula $\lim\limits_{\theta \to 0} \frac{\sin\theta}{\theta} = 1$.

Divide the numerator and the denominator by $x$ (since $x \to 0$, we consider $x \neq 0$):

$\lim\limits_{x \to 0} \frac{\frac{\sin ax\;+\;bx}{x}}{\frac{ax\;+\;\sin bx}{x}}$

Split the terms in the numerator and denominator:

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

Simplify the terms $\frac{bx}{x} = b$ and $\frac{ax}{x} = a$ (for $x \neq 0$):

$= \lim\limits_{x \to 0} \frac{\frac{\sin ax}{x} + b}{a + \frac{\sin bx}{x}}$

To use the standard limit for the sine terms, multiply and divide by $a$ in the first numerator term and by $b$ in the second denominator term:

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

As $x \to 0$, $ax \to 0$ and $bx \to 0$. Use the limit properties and the standard limit $\lim\limits_{\theta \to 0} \frac{\sin\theta}{\theta} = 1$.

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

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

$= \frac{a + b}{a + b}$

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

$= 1$

Thus, $\lim\limits_{x \to 0} \frac{\sin ax\;+\;bx}{ax\;+\;\sin bx} = 1$, provided $a, b, a+b \neq 0$.

To evaluate this limit, we first examine the behavior of $\text{cosec}\;x$ and $\cot x$ as $x \to 0$.

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

Recall that $\text{cosec}\;x = \frac{1}{\sin x}$ and $\cot x = \frac{\cos x}{\sin x}$.

As $x \to 0$, $\sin x \to 0$ and $\cos x \to 1$.

So, $\text{cosec}\;x \to \pm\infty$ and $\cot x \to \pm\infty$. This is the indeterminate form $\infty - \infty$.

We can rewrite the expression by combining the terms into a single fraction:

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

Now, find the limit of this expression as $x \to 0$:

$\lim\limits_{x \to 0} \frac{1 \;-\; \cos x}{\sin x}$

At $x = 0$:

Numerator: $1 - \cos(0) = 1 - 1 = 0$.

Denominator: $\sin(0) = 0$.

This is the indeterminate form $\frac{0}{0}$. We can use trigonometric identities or standard limits.

We use the identity $1 - \cos x = 2\sin^2(\frac{x}{2})$ and $\sin x = 2\sin(\frac{x}{2})\cos(\frac{x}{2})$.

The expression becomes:

$\lim\limits_{x \to 0} \frac{2\sin^2(\frac{x}{2})}{2\sin(\frac{x}{2})\cos(\frac{x}{2})}$

For $x \neq 0$, $\sin(\frac{x}{2}) \neq 0$, so we can cancel a factor of $2\sin(\frac{x}{2})$:

$= \lim\limits_{x \to 0} \frac{\cancel{2\sin(\frac{x}{2})}\sin(\frac{x}{2})}{\cancel{2\sin(\frac{x}{2})}\cos(\frac{x}{2})}$

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

Now, substitute $x = 0$:

$= \frac{\sin(\frac{0}{2})}{\cos(\frac{0}{2})} = \frac{\sin(0)}{\cos(0)} = \frac{0}{1} = 0$

Thus, $\lim\limits_{x \to 0} (\text{cosec}\;x-\cot x) = 0$.

Alternate Method (using standard limits):

We had the expression $\lim\limits_{x \to 0} \frac{1 \;-\; \cos x}{\sin x}$.

Divide the numerator and the denominator by $x$:

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

We know the standard limits: $\lim\limits_{x \to 0} \frac{\sin x}{x} = 1$ and $\lim\limits_{x \to 0} \frac{1 \;-\; \cos x}{x} = 0$.

Using the property for the limit of a quotient:

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

This confirms the result.

Alternate Method (using L'Hopital's Rule):

We had the expression $\lim\limits_{x \to 0} \frac{1 \;-\; \cos x}{\sin x}$. At $x=0$, this is $\frac{0}{0}$.

Apply L'Hopital's Rule. Take the derivative of the numerator and denominator:

Derivative of numerator: $\frac{d}{dx}(1 - \cos x) = -(-\sin x) = \sin x$

Derivative of denominator: $\frac{d}{dx}(\sin x) = \cos x$

The limit becomes $\lim\limits_{x \to 0} \frac{\sin x}{\cos x}$.

Now, substitute $x = 0$:

$\frac{\sin(0)}{\cos(0)} = \frac{0}{1} = 0$

This also confirms the result.

To evaluate this limit, we first check the values of the numerator and denominator as $x \to \frac{\pi}{2}$.

Evaluate $\lim\limits_{x \to \frac{\pi}{2}} \frac{\tan2x}{x\;-\;\frac{\pi}{2}}$.

As $x \to \frac{\pi}{2}$:

Numerator: $\tan(2x) \to \tan\left(2 \times \frac{\pi}{2}\right) = \tan(\pi)$. Since $\tan(\pi) = 0$, the numerator approaches 0.

Denominator: $x - \frac{\pi}{2} \to \frac{\pi}{2} - \frac{\pi}{2} = 0$. The denominator approaches 0.

This is the indeterminate form $\frac{0}{0}$. We can use a substitution to simplify the limit.

Let $y = x - \frac{\pi}{2}$. As $x \to \frac{\pi}{2}$, $y \to 0$.

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

The argument of the tangent function is $2x = 2\left(y + \frac{\pi}{2}\right) = 2y + \pi$.

Using the trigonometric identity $\tan(\pi + \theta) = \tan \theta$, we have $\tan(2y + \pi) = \tan(2y)$.

Substitute these into the limit expression:

$\lim\limits_{x \to \frac{\pi}{2}} \frac{\tan2x}{x\;-\;\frac{\pi}{2}} = \lim\limits_{y \to 0} \frac{\tan(2y)}{y}$

We can rewrite the expression to use the standard trigonometric limit $\lim\limits_{\theta \to 0} \frac{\tan\theta}{\theta} = 1$.

Multiply and divide the expression by 2:

$= \lim\limits_{y \to 0} \frac{\tan(2y)}{2y} \times 2$

As $y \to 0$, $2y \to 0$. Let $\theta = 2y$. The first part of the expression approaches the standard limit.

$= \lim\limits_{y \to 0} \frac{\tan(2y)}{2y} \times \lim\limits_{y \to 0} 2$

$= 1 \times 2$

$= 2$

Thus, $\lim\limits_{x \to \frac{\pi}{2}} \frac{\tan2x}{x\;-\;\frac{\pi}{2}} = 2$.

Alternate Method (using L'Hopital's Rule):

We have the indeterminate form $\frac{0}{0}$ at $x = \frac{\pi}{2}$. We can apply L'Hopital's Rule.

Take the derivative of the numerator and the denominator with respect to $x$.

Derivative of the numerator: $\frac{d}{dx}(\tan(2x)) = \sec^2(2x) \times \frac{d}{dx}(2x) = 2\sec^2(2x)$.

Derivative of the denominator: $\frac{d}{dx}(x - \frac{\pi}{2}) = 1$.

The limit becomes:

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

Now, substitute $x = \frac{\pi}{2}$ into the simplified expression:

$2\sec^2\left(2 \times \frac{\pi}{2}\right) = 2\sec^2(\pi)$

Recall that $\sec(\pi) = \frac{1}{\cos(\pi)} = \frac{1}{-1} = -1$.

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

This confirms the result obtained by substitution.

We need to find the limit of the piecewise function $f(x)$ as $x$ approaches 0 and as $x$ approaches 1.

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

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

The function definition changes at $x = 0$. Therefore, we need to evaluate the left-hand limit (LHL) and the right-hand limit (RHL) at $x = 0$.

LHL: $\lim\limits_{x \to 0^-} f(x)$. For $x < 0$, $f(x) = 2x + 3$.

$\lim\limits_{x \to 0^-} f(x) = \lim\limits_{x \to 0^-} (2x + 3)$

Substitute $x = 0$:

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

RHL: $\lim\limits_{x \to 0^+} f(x)$. For $x > 0$, $f(x) = 3(x + 1)$.

$\lim\limits_{x \to 0^+} f(x) = \lim\limits_{x \to 0^+} 3(x + 1)$

Substitute $x = 0$:

$= 3(0 + 1) = 3(1) = 3$

Since the LHL and RHL at $x = 0$ are equal (both are 3), the limit exists at $x = 0$ and is equal to 3.

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

We need to evaluate the limit as $x$ approaches 1. For values of $x$ close to 1 (both less than 1 and greater than 1), the function is defined by $f(x) = 3(x + 1)$, because 1 > 0. The function definition does not change its form around $x = 1$.

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

Substitute $x = 1$ into the function:

$= 3(1 + 1)$

$= 3(2)$

$= 6$

The limits are $\lim\limits_{x \to 0} f(x) = 3$ and $\lim\limits_{x \to 1} f(x) = 6$.

Given:

The function $f(x)$ is defined as:

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

To Find:

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

Solution:

To determine if $\lim\limits_{x \to 1} f(x)$ exists, we need to evaluate the left-hand limit and the right-hand limit of the function as $x$ approaches $1$. The limit exists if and only if the left-hand limit is equal to the right-hand limit.

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

As $x$ approaches $1$ from the left side, $x < 1$. For this case, the function is defined as $f(x) = x^2 - 1$.

So, we evaluate the limit of $x^2 - 1$ as $x \to 1^-$.

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

Substituting $x=1$ into the expression, we get:

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

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

As $x$ approaches $1$ from the right side, $x > 1$. For this case, the function is defined as $f(x) = -x^2 - 1$.

So, we evaluate the limit of $-x^2 - 1$ as $x \to 1^+$.

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

Substituting $x=1$ into the expression, we get:

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

Conclusion:

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

We found that the left-hand limit is $0$ and the right-hand limit is $-2$.

Since $\lim\limits_{x \to 1^-} f(x) \neq \lim\limits_{x \to 1^+} f(x)$, which means $0 \neq -2$, the limit of the function as $x$ approaches $1$ does not exist.

Therefore, $\lim\limits_{x \to 1} f(x)$ does not exist.

Given:

The function $f(x)$ is defined as:

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

To Evaluate:

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

Solution:

To evaluate $\lim\limits_{x \to 0} f(x)$, we need to check the left-hand limit and the right-hand limit as $x$ approaches $0$. The limit exists if and only if both limits are equal.

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

As $x$ approaches $0$ from the left side, $x$ is negative ($x < 0$) but close to $0$.

For $x < 0$, the definition of absolute value is $|x| = -x$.

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

For $x \neq 0$, we can cancel $x$ in the numerator and denominator:

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

Now, we find the limit as $x \to 0^-$:

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

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

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

As $x$ approaches $0$ from the right side, $x$ is positive ($x > 0$) but close to $0$.

For $x > 0$, the definition of absolute value is $|x| = x$.

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

For $x \neq 0$, we can cancel $x$ in the numerator and denominator:

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

Now, we find the limit as $x \to 0^+$:

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

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

Conclusion:

For the limit $\lim\limits_{x \to 0} f(x)$ to exist, the left-hand limit must be equal to the right-hand limit.

We have found that the left-hand limit is $-1$ and the right-hand limit is $1$.

Since $\lim\limits_{x \to 0^-} f(x) \neq \lim\limits_{x \to 0^+} f(x)$, which means $-1 \neq 1$, the limit of the function as $x$ approaches $0$ does not exist.

The value of the function at $x=0$, $f(0)=0$, is not relevant for the existence of the limit as $x \to 0$.

Therefore, $\lim\limits_{x \to 0} f(x)$ does not exist.

Given:

The function $f(x)$ is defined as:

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

To Find:

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

Solution:

To find the limit of $f(x)$ as $x$ approaches $0$, we need to evaluate the left-hand limit and the right-hand limit. The limit exists if and only if these two limits are equal.

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

As $x \to 0^-$ (meaning $x$ approaches $0$ from values less than $0$), $x < 0$.

By the definition of the absolute value, for $x < 0$, $|x| = -x$.

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

For $x \neq 0$, we can simplify the expression:

$f(x) = \frac{x}{-x} = -1$, for $x < 0$.

Now, we evaluate the limit:

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

Since the limit of a constant is the constant itself:

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

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

As $x \to 0^+$ (meaning $x$ approaches $0$ from values greater than $0$), $x > 0$.

By the definition of the absolute value, for $x > 0$, $|x| = x$.

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

For $x \neq 0$, we can simplify the expression:

$f(x) = \frac{x}{x} = 1$, for $x > 0$.

Now, we evaluate the limit:

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

Since the limit of a constant is the constant itself:

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

Conclusion:

For the limit $\lim\limits_{x \to 0} f(x)$ to exist, the left-hand limit must be equal to the right-hand limit.

We found that $\lim\limits_{x \to 0^-} f(x) = -1$ and $\lim\limits_{x \to 0^+} f(x) = 1$.

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

The value of the function at $x=0$, which is $f(0)=0$, does not affect the existence or value of the limit as $x$ approaches $0$.

Therefore, $\lim\limits_{x \to 0} f(x)$ does not exist.

Given:

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

To Find:

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

Solution:

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

The absolute value function $|x|$ is defined as:

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

We are interested in the limit as $x$ approaches $5$. When $x$ is close to $5$, $x$ is positive (specifically, $x > 0$).

For values of $x$ near $5$, $|x| = x$.

Therefore, for $x$ close to $5$, the function $f(x)$ can be written as:

$f(x) = x - 5$

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

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

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

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

The limit of $x$ as $x \to a$ is $a$, and the limit of a constant $c$ as $x \to a$ is $c$.

So,

$\lim\limits_{x \to 5} x = 5$

$\lim\limits_{x \to 5} 5 = 5$

Therefore,

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

Alternatively, we can consider the left-hand limit and the right-hand limit.

Left-hand limit ($\lim\limits_{x \to 5^-} f(x)$):

As $x \to 5^-$ (values of $x$ less than $5$ but close to $5$), $x$ is positive, so $|x| = x$.

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

Right-hand limit ($\lim\limits_{x \to 5^+} f(x)$):

As $x \to 5^+$ (values of $x$ greater than $5$ but close to $5$), $x$ is positive, so $|x| = x$.

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

Conclusion:

Since the left-hand limit is equal to the right-hand limit ($\lim\limits_{x \to 5^-} f(x) = \lim\limits_{x \to 5^+} f(x) = 0$), the limit of the function as $x$ approaches $5$ exists and is equal to $0$.

Therefore, $\lim\limits_{x \to 5} f(x) = 0$.

Given:

The function $f(x)$ is defined as:

$f(x) = \begin{cases} a + bx , & x < 1 \\ 4 , & x = 0 \\ b - ax , & x > 1 \end{cases}$

The condition $\lim\limits_{x \to 1} f(x) = f(1)$ is given.

To Find:

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

Interpretation Note:

The given definition of $f(x)$ includes $f(x)=4$ at $x=0$. However, the condition $\lim\limits_{x \to 1} f(x) = f(1)$ involves the function value at $x=1$. In the context of such problems, it is typical for the middle part of a piecewise definition to specify the function value at the point where the function definition changes. Therefore, we assume there might be a typo and that the condition $\lim\limits_{x \to 1} f(x) = f(1)$ implies that $f(1)$ is the value defined for $x=1$. Based on the structure, it is highly probable that the value $4$ is intended for $x=1$, i.e., $f(1) = 4$. We will proceed with this assumption.

Solution:

The condition $\lim\limits_{x \to 1} f(x) = f(1)$ means that the limit of $f(x)$ as $x$ approaches $1$ exists and is equal to the function value at $x=1$. For the limit to exist, the left-hand limit and the right-hand limit at $x=1$ must be equal.

So, we must have $\lim\limits_{x \to 1^-} f(x) = \lim\limits_{x \to 1^+} f(x) = f(1)$.

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

When $x < 1$, the function is $f(x) = a + bx$.

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

Substituting $x=1$ into the expression for the limit from the left:

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

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

When $x > 1$, the function is $f(x) = b - ax$.

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

Substituting $x=1$ into the expression for the limit from the right:

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

Function value at $x=1$ ($f(1)$):

Based on our interpretation of the likely intent of the question, we assume $f(1) = 4$.

$f(1) = 4$

Applying the condition $\lim\limits_{x \to 1} f(x) = f(1)$:

The condition requires $\lim\limits_{x \to 1^-} f(x) = \lim\limits_{x \to 1^+} f(x) = f(1)$.

Equating the left-hand limit to $f(1)$:

Equating the right-hand limit to $f(1)$:

Solving the system of linear equations:

We have the system:

$a + b = 4 \quad (1)$

$-a + b = 4 \quad (2)$

Add equation (1) and equation (2):

$(a + b) + (-a + b) = 4 + 4$

$a + b - a + b = 8$

$2b = 8$

$b = \frac{8}{2}$

$b = 4$

Substitute the value of $b$ into equation (1):

$a + 4 = 4$

$a = 4 - 4$

$a = 0$

Conclusion:

Based on the likely intent of the question, where $f(1) = 4$ is assumed, the values of $a$ and $b$ that satisfy the condition $\lim\limits_{x \to 1} f(x) = f(1)$ are $a=0$ and $b=4$. This condition means the function is continuous at $x=1$.

Therefore, the possible values are $a=0$ and $b=4$.

Given:

A function $f(x)$ defined as the product of $n$ linear factors:

$f(x) = (x − a_1) (x − a_2)...(x − a_n)$, where $a_1, a_2, \dots, a_n$ are fixed real numbers.

To Find:

1. $\lim\limits_{x \to a_1} f(x)$

2. $\lim\limits_{x \to a} f(x)$, for some $a \neq a_1, a_2, \dots, a_n$.

Solution:

The function $f(x) = (x − a_1) (x − a_2)...(x − a_n)$ is a polynomial function. When the product of $n$ linear factors $(x-c_i)$ is expanded, the result is a polynomial of degree $n$.

A key property of polynomial functions is that they are continuous for all real numbers. For a continuous function $P(x)$ and any real number $c$, the limit of the function as $x$ approaches $c$ is equal to the function evaluated at $c$. That is, $\lim\limits_{x \to c} P(x) = P(c)$.

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

Since $f(x)$ is a polynomial, we can find the limit as $x \to a_1$ by evaluating the function at $x = a_1$.

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

Substitute $x = a_1$ into the expression for $f(x)$:

$f(a_1) = (a_1 − a_1) (a_1 − a_2)...(a_1 − a_n)$

$f(a_1) = (0) (a_1 − a_2)...(a_1 − a_n)$

$f(a_1) = 0$

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

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

Again, since $f(x)$ is a polynomial and continuous everywhere, we can find the limit as $x \to a$ by evaluating the function at $x = a$, given that $a$ is a real number.

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

Substitute $x = a$ into the expression for $f(x)$:

$f(a) = (a − a_1) (a − a_2)...(a − a_n)$

Since it is given that $a \neq a_1, a_2, \dots, a_n$, none of the factors $(a - a_i)$ are equal to zero.

Therefore, the limit is simply the value of the function at $x=a$.

$\lim\limits_{x \to a} f(x) = (a − a_1) (a − a_2)...(a − a_n)$

Summary of Results:

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

2. For some $a \neq a_1, a_2, \dots, a_n$, $\lim\limits_{x \to a} f(x) = (a − a_1) (a − a_2)...(a − a_n)$

Given:

The function $f(x)$ is defined as:

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

To Find:

The value(s) of $a$ for which $\lim\limits_{x \to a} f(x)$ exists.

Solution:

For the limit $\lim\limits_{x \to a} f(x)$ to exist, the left-hand limit and the right-hand limit of the function as $x$ approaches $a$ must be equal.

We need to consider different cases for the value of $a$ based on the piecewise definition of $f(x)$.

Case 1: $a < 0$

If $a < 0$, then for values of $x$ sufficiently close to $a$ (both from the left and right), $x$ will be negative ($x < 0$).

For $x < 0$, the function is $f(x) = |x| + 1$. Since $x < 0$, $|x| = -x$.

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

This is a polynomial function, which is continuous everywhere. Therefore, for any $a < 0$, the limit exists and is equal to $f(a)$.

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

The limit exists for all $a < 0$.

Case 2: $a = 0$

We need to evaluate the left-hand limit and the right-hand limit at $x = 0$.

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

As $x \to 0^-$ (i.e., $x$ approaches $0$ from the left), $x < 0$. The function is $f(x) = |x| + 1 = -x + 1$ for $x < 0$.

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

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

As $x \to 0^+$ (i.e., $x$ approaches $0$ from the right), $x > 0$. The function is $f(x) = |x| - 1 = x - 1$ for $x > 0$.

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

Since the left-hand limit ($1$) is not equal to the right-hand limit ($-1$) at $x=0$, the limit $\lim\limits_{x \to 0} f(x)$ does not exist.

Case 3: $a > 0$

If $a > 0$, then for values of $x$ sufficiently close to $a$ (both from the left and right), $x$ will be positive ($x > 0$).

For $x > 0$, the function is $f(x) = |x| - 1$. Since $x > 0$, $|x| = x$.

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

This is a polynomial function, which is continuous everywhere. Therefore, for any $a > 0$, the limit exists and is equal to $f(a)$.

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

The limit exists for all $a > 0$.

Conclusion:

Based on the three cases, the limit $\lim\limits_{x \to a} f(x)$ exists when $a < 0$ and when $a > 0$. The limit does not exist when $a = 0$.

Therefore, the limit exists for all real values of $a$ except $0$.

The limit $\lim\limits_{x \to a} f(x)$ exists for all values of $a$ such that $a \neq 0$.

Given:

The function $f(x)$ satisfies the condition $\lim\limits_{x \to 1} \frac{f(x) \;-\; 2}{x^2 \;-\; 1} = \pi$.

To Evaluate:

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

Solution:

We are given that the limit of the expression $\frac{f(x) \;-\; 2}{x^2 \;-\; 1}$ as $x$ approaches $1$ exists and is equal to $\pi$, which is a finite non-zero value.

Let's consider the limit of the denominator as $x$ approaches $1$:

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

For the limit of a fraction $\frac{N(x)}{D(x)}$ to be a finite number $\pi$ (which is not zero), when the limit of the denominator $D(x)$ is $0$, the limit of the numerator $N(x)$ must also be $0$. This is a necessary condition for the limit to be of the indeterminate form $\frac{0}{0}$, which can evaluate to a finite non-zero value.

In this case, the numerator is $f(x) - 2$. Therefore, the limit of the numerator as $x$ approaches $1$ must be $0$.

So, we must have:

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

Using the property that the limit of a difference is the difference of the limits (assuming $\lim\limits_{x \to 1} f(x)$ exists):

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

The limit of a constant is the constant itself, so $\lim\limits_{x \to 1} 2 = 2$.

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

Adding $2$ to both sides of the equation:

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

We can verify this by substituting $\lim\limits_{x \to 1} f(x) = 2$ back into the original limit expression. If $\lim\limits_{x \to 1} f(x) = 2$, then $\lim\limits_{x \to 1} (f(x) - 2) = \lim\limits_{x \to 1} f(x) - 2 = 2 - 2 = 0$. This gives the indeterminate form $\frac{0}{0}$, which is consistent with the given finite limit $\pi$.

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

Given:

The function $f(x)$ is defined as:

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

To Find:

The integer values of $m$ and $n$ for which both $\lim\limits_{x \to 0} f(x)$ and $\lim\limits_{x \to 1} f(x)$ exist.

Solution:

For the limit of a function to exist at a point $a$, the left-hand limit and the right-hand limit at that point must be equal.

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

For the limit $\lim\limits_{x \to 0} f(x)$ to exist, the left-hand limit as $x \to 0^-$ must be equal to the right-hand limit as $x \to 0^+$.

The left-hand limit as $x \to 0^-$, where $x < 0$, is given by the first piece of the function definition:

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

Since $mx^2 + n$ is a polynomial, its limit as $x$ approaches $0$ is obtained by direct substitution:

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

The right-hand limit as $x \to 0^+$, where $0 \leq x \leq 1$ (specifically $x > 0$ but close to $0$), is given by the second piece of the function definition:

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

Since $nx + m$ is a polynomial, its limit as $x$ approaches $0$ is obtained by direct substitution:

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

For the limit $\lim\limits_{x \to 0} f(x)$ to exist, the left-hand limit must equal the right-hand limit:

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

For the limit $\lim\limits_{x \to 1} f(x)$ to exist, the left-hand limit as $x \to 1^-$ must be equal to the right-hand limit as $x \to 1^+$.

The left-hand limit as $x \to 1^-$, where $0 \leq x \leq 1$ (specifically $x < 1$ but close to $1$), is given by the second piece of the function definition:

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

Since $nx + m$ is a polynomial, its limit as $x$ approaches $1$ is obtained by direct substitution:

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

The right-hand limit as $x \to 1^+$, where $x > 1$, is given by the third piece of the function definition:

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

Since $nx^3 + m$ is a polynomial, its limit as $x$ approaches $1$ is obtained by direct substitution:

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

For the limit $\lim\limits_{x \to 1} f(x)$ to exist, the left-hand limit must equal the right-hand limit:

This equation is always true for any values of $m$ and $n$. Therefore, the existence of the limit at $x=1$ does not impose any constraints on the values of $m$ and $n$.

Conclusion:

For both $\lim\limits_{x \to 0} f(x)$ and $\lim\limits_{x \to 1} f(x)$ to exist, the conditions for the existence of each limit must be met simultaneously.

The condition for $\lim\limits_{x \to 0} f(x)$ to exist is $n = m$.

The condition for $\lim\limits_{x \to 1} f(x)$ to exist is $n + m = n + m$, which is always true.

Therefore, both limits exist if and only if the condition $n = m$ is satisfied.

The question asks for integer values of $m$ and $n$. The condition $n=m$ means that $m$ and $n$ must be the same integer.

Both $\lim\limits_{x \to 0} f(x)$ and $\lim\limits_{x \to 1} f(x)$ exist if and only if $m$ and $n$ are integers such that $m = n$.

In other words, $m$ and $n$ must be equal integers.

Given:

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

To Find:

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

Solution:

We can find the derivative of a function $f(x)$ at a specific point $x=c$ using the limit definition of the derivative at a point:

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

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

First, we find $f(c)$, which is $f(2)$:

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

Next, we find $f(c+h)$, which is $f(2+h)$:

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

Now, we substitute these into the limit definition:

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

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

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 we are evaluating the limit as $h \to 0$, $h$ is approaching $0$ but is not equal to $0$. Therefore, we can cancel the $h$ terms:

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

The limit of a constant as $h$ approaches any value is the constant itself:

$f'(2) = 3$

Alternatively, we can find the general derivative $f'(x)$ first and then evaluate it at $x=2$.

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

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

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

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

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

$f'(x) = 3$

Now, evaluate $f'(x)$ at $x=2$:

$f'(2) = 3$

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

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

Given:

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

To Find:

1. The derivative of $f(x)$ at $x = -1$, i.e., $f'(-1)$.

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

Solution:

We use the definition of the derivative of a function $f(x)$ at a point $c$, which is given by:

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

Part 1: Find $f'(-1)$

Here, $c = -1$. We need to find $f(-1)$ and $f(-1+h)$.

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

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

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

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

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

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

Now, substitute these into the limit definition for $f'(-1)$:

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

$f'(-1) = \lim\limits_{h \to 0} \frac{(2h^2 - h - 6) - (-6)}{h}$

$f'(-1) = \lim\limits_{h \to 0} \frac{2h^2 - h - 6 + 6}{h}$

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

Factor out $h$ from the numerator:

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

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

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

Substitute $h = 0$ to evaluate the limit:

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

So, the derivative at $x = -1$ is $\mathbf{f'(-1) = -1}$.

Part 2: Prove $f'(0) + 3f'(-1) = 0$

We already found $f'(-1) = -1$. We now need to find $f'(0)$.

Here, $c = 0$. We need to find $f(0)$ and $f(0+h) = f(h)$.

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

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

Now, substitute these into the limit definition for $f'(0)$:

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

$f'(0) = \lim\limits_{h \to 0} \frac{(2h^2 + 3h - 5) - (-5)}{h}$

$f'(0) = \lim\limits_{h \to 0} \frac{2h^2 + 3h - 5 + 5}{h}$

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

Factor out $h$ from the numerator:

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

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

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

Substitute $h = 0$ to evaluate the limit:

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

So, the derivative at $x = 0$ is $\mathbf{f'(0) = 3}$.

Now, we prove the given statement $f'(0) + 3f'(-1) = 0$ by substituting the values we found:

Left Hand Side (LHS) = $f'(0) + 3f'(-1)$

LHS = $(3) + 3(-1)$

LHS = $3 - 3$

LHS = $0$

Right Hand Side (RHS) = $0$

Since LHS = RHS, the statement is proven.

Therefore, $f'(-1) = -1$ and it is proven that $f'(0) + 3f'(-1) = 0$.

Given:

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

To Find:

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

Solution:

We use the limit definition of the derivative of a function $f(x)$ at a point $c$:

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

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

First, we find $f(c)$, which is $f(0)$:

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

Next, we find $f(c+h)$, which is $f(0+h) = f(h)$:

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

Now, substitute these into the limit definition for $f'(0)$:

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

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

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

This is a standard trigonometric limit:

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

Using this standard limit, where $x$ is replaced by $h$:

$f'(0) = 1$

Alternatively, we can find the general derivative $f'(x)$ first and then evaluate it at $x=0$. The derivative of $\sin x$ is $\cos x$.

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

Now, evaluate $f'(x)$ at $x=0$:

$f'(0) = \cos(0)$

The value of $\cos(0)$ is $1$.

$f'(0) = 1$

Therefore, the derivative of $\sin x$ at $x = 0$ is $1$.

The derivative at $x=0$ is $f'(0) = 1$.

Given:

The function $f(x) = 3$.

To Find:

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

Solution:

The function $f(x) = 3$ is a constant function. The derivative of a constant function is always $0$ at any point.

We can show this using the limit definition of the derivative of a function $f(x)$ at a point $c$:

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

Find the derivative at $x = 0$ ($c=0$):

We need to find $f(0+h) = f(h)$ and $f(0)$.

Since $f(x) = 3$ for all values of $x$, we have:

$f(h) = 3$

$f(0) = 3$

Now, substitute these into the limit definition for $f'(0)$:

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

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

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

For $h \neq 0$, $\frac{0}{h} = 0$. So, we have:

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

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

$f'(0) = 0$

Find the derivative at $x = 3$ ($c=3$):

We need to find $f(3+h)$ and $f(3)$.

Since $f(x) = 3$ for all values of $x$, we have:

$f(3+h) = 3$

$f(3) = 3$

Now, substitute these into the limit definition for $f'(3)$:

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

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

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

For $h \neq 0$, $\frac{0}{h} = 0$. So, we have:

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

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

$f'(3) = 0$

General Derivative:

Using the power rule for differentiation, the derivative of a constant $c$ is $0$. For $f(x) = 3$, $f'(x) = \frac{d}{dx}(3) = 0$. Since the derivative is $0$ for all values of $x$, it is $0$ specifically at $x=0$ and $x=3$.

Therefore, the derivative of $f(x) = 3$ at $x = 0$ is $f'(0) = 0$ and at $x = 3$ is $f'(3) = 0$.

Given:

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

To Find:

The derivative of $f(x)$, denoted as $f'(x)$ or $\frac{d}{dx}(f(x))$.

Solution:

We can find the derivative of the function $f(x) = 10x$ using the first principle of differentiation, which is given by the limit definition:

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

First, we find $f(x+h)$ by substituting $x+h$ into the function $f(x) = 10x$:

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

Next, we find the difference $f(x+h) - f(x)$:

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

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

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

Now, substitute this into the limit definition of the derivative:

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

Since we are taking the limit as $h \to 0$, $h$ is approaching $0$ but is not equal to $0$. Therefore, we can cancel the $h$ terms in the numerator and denominator:

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

The limit of a constant as $h$ approaches any value is the constant itself:

$f'(x) = 10$

Alternatively, using the power rule for differentiation, which states that the derivative of $cx^n$ is $cnx^{n-1}$, where $c$ is a constant and $n$ is a real number.

For the function $f(x) = 10x$, we can write it as $f(x) = 10x^1$. Here, $c=10$ and $n=1$.

Using the power rule:

$f'(x) = \frac{d}{dx}(10x^1) = 10 \times 1 \times x^{1-1} = 10 \times x^0 = 10 \times 1 = 10$

So, $f'(x) = 10$.

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

The derivative is $f'(x) = 10$.

Given:

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

To Find:

The derivative of $f(x)$, denoted as $f'(x)$ or $\frac{d}{dx}(f(x))$.

Solution:

We will find the derivative of the function $f(x) = x^2$ using the definition of the derivative (first principle):

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

First, find $f(x+h)$ by substituting $(x+h)$ into the function $f(x) = x^2$:

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

Expand $(x+h)^2$ using the formula $(a+b)^2 = a^2 + 2ab + b^2$:

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

Next, find the difference $f(x+h) - f(x)$:

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

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

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

Now, substitute this into the limit definition of the derivative:

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

Factor out $h$ from the numerator:

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

Since we are evaluating the limit as $h \to 0$, $h$ is approaching $0$ but is not equal to $0$. Therefore, we can cancel the $h$ terms in the numerator and denominator:

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

Now, substitute $h = 0$ to evaluate the limit:

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

$f'(x) = 2x$

Alternate Solution (Using Power Rule):

The power rule for differentiation states that if $f(x) = x^n$, then $f'(x) = nx^{n-1}$.

For the function $f(x) = x^2$, we have $n=2$.

Using the power rule:

$f'(x) = 2 \times x^{2-1}$

$f'(x) = 2x^1$

$f'(x) = 2x$

This matches the result obtained from the first principle.

Therefore, the derivative of the function $f(x) = x^2$ is $2x$.

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

Given:

The function $f(x) = a$, where $a$ is a fixed real number.

To Find:

The derivative of $f(x)$, denoted as $f'(x)$ or $\frac{d}{dx}(f(x))$.

Solution:

We can find the derivative of the constant function $f(x) = a$ using the definition of the derivative (first principle):

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

First, find $f(x+h)$ by substituting $(x+h)$ into the function $f(x) = a$. Since $f(x)$ is a constant function, the value of the function is always $a$, regardless of the input $x$.

$f(x+h) = a$

The value of $f(x)$ is:

$f(x) = a$

Next, find the difference $f(x+h) - f(x)$:

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

$f(x+h) - f(x) = 0$

Now, substitute this into the limit definition of the derivative:

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

Since we are evaluating the limit as $h \to 0$, $h$ is approaching $0$ but is not equal to $0$. Therefore, $\frac{0}{h} = 0$ for all $h \neq 0$.

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

The limit of a constant (which is $0$) as $h$ approaches any value is the constant itself:

$f'(x) = 0$

Alternate Explanation:

Geometrically, the graph of a constant function $f(x) = a$ is a horizontal line. The derivative of a function at a point represents the slope of the tangent line to the graph at that point. For a horizontal line, the slope is always $0$ at every point.

Using the power rule, a constant function $f(x) = a$ can be written as $f(x) = ax^0$ (for $x \neq 0$). While this specific form with $x^0$ isn't necessary as $a$ is a constant, the rule for the derivative of a constant is a fundamental rule derived from the limit definition: $\frac{d}{dx}(c) = 0$ for any constant $c$. In this case, the constant is $a$.

$f'(x) = \frac{d}{dx}(a) = 0$

Therefore, the derivative of the constant function $f(x) = a$ is $0$.

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

Given:

The function $f(x) = \frac{1}{x}$, where $x \neq 0$.

To Find:

The derivative of $f(x)$, denoted as $f'(x)$ or $\frac{d}{dx}(f(x))$.

Solution:

We will find the derivative of the function $f(x) = \frac{1}{x}$ using the definition of the derivative (first principle):

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

First, find $f(x+h)$ by substituting $(x+h)$ into the function $f(x) = \frac{1}{x}$:

$f(x+h) = \frac{1}{x+h}$

Next, find the difference $f(x+h) - f(x)$:

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

To subtract these fractions, find a common denominator, which is $x(x+h)$:

$f(x+h) - f(x) = \frac{1 \times x}{x(x+h)} - \frac{1 \times (x+h)}{x(x+h)}$

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

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

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

Now, substitute this into the limit definition of the derivative:

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

Dividing by $h$ is the same as multiplying by $\frac{1}{h}$:

$f'(x) = \lim\limits_{h \to 0} \frac{-h}{x(x+h)} \times \frac{1}{h}$

For $h \neq 0$, we can cancel the $h$ in the numerator and denominator:

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

Now, substitute $h = 0$ into the expression:

$f'(x) = \frac{-1}{x(x+0)}$

$f'(x) = \frac{-1}{x(x)}$

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

Alternate Solution (Using Power Rule):

The function $f(x) = \frac{1}{x}$ can be written as $f(x) = x^{-1}$.

Using the power rule for differentiation, which states that if $f(x) = x^n$, then $f'(x) = nx^{n-1}$. Here, $n = -1$.

Applying the power rule:

$f'(x) = (-1) \times x^{-1-1}$

$f'(x) = -1 \times x^{-2}$

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

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

This matches the result obtained from the first principle.

Therefore, the derivative of the function $f(x) = \frac{1}{x}$ is $-\frac{1}{x^2}$.

The derivative is $f'(x) = -\frac{1}{x^2}$.

Given:

The function $f(x) = 6x^{100} - x^{55} + x$.

To Compute:

The derivative of $f(x)$, denoted as $f'(x)$ or $\frac{d}{dx}(f(x))$.

Solution:

We can compute the derivative using the properties of derivatives, specifically the sum/difference rule and the power rule.

The sum/difference rule states that the derivative of a sum or difference of functions is the sum or difference of their derivatives: $\frac{d}{dx}[g(x) \pm h(x)] = \frac{d}{dx}[g(x)] \pm \frac{d}{dx}[h(x)]$.

The power rule states that the derivative of $x^n$ is $nx^{n-1}$, i.e., $\frac{d}{dx}(x^n) = nx^{n-1}$. Also, for a constant $c$, $\frac{d}{dx}(cx^n) = c \frac{d}{dx}(x^n) = cnx^{n-1}$. The derivative of a constant is $0$, i.e., $\frac{d}{dx}(c) = 0$.

Apply the sum/difference rule to the given function:

$f'(x) = \frac{d}{dx}(6x^{100} - x^{55} + x)$

$f'(x) = \frac{d}{dx}(6x^{100}) - \frac{d}{dx}(x^{55}) + \frac{d}{dx}(x)$

Now, apply the power rule to each term:

For the first term, $6x^{100}$: $c=6$, $n=100$.

$\frac{d}{dx}(6x^{100}) = 6 \times 100 x^{100-1} = 600x^{99}$

For the second term, $x^{55}$: This is $1 \times x^{55}$. $c=1$, $n=55$.

$\frac{d}{dx}(x^{55}) = 1 \times 55 x^{55-1} = 55x^{54}$

For the third term, $x$: This is $1 \times x^1$. $c=1$, $n=1$.

$\frac{d}{dx}(x) = 1 \times 1 x^{1-1} = 1 \times x^0 = 1 \times 1 = 1$

Combine the derivatives of the individual terms:

$f'(x) = 600x^{99} - 55x^{54} + 1$

Therefore, the derivative of the function $f(x) = 6x^{100} – x^{55} + x$ is $600x^{99} - 55x^{54} + 1$.

The derivative is $f'(x) = 600x^{99} - 55x^{54} + 1$.

Given:

The function $f(x) = 1 + x + x^2 + x^3 + \dots + x^{50}$.

To Find:

The derivative of $f(x)$ at $x = 1$, denoted as $f'(1)$.

Solution:

The function $f(x)$ is a polynomial, which is a sum of terms of the form $x^n$. We can find the derivative of $f(x)$ using the sum rule and the power rule for differentiation.

The derivative of a sum of functions is the sum of their derivatives:

$f'(x) = \frac{d}{dx}(1 + x + x^2 + x^3 + \dots + x^{50})$

$f'(x) = \frac{d}{dx}(1) + \frac{d}{dx}(x) + \frac{d}{dx}(x^2) + \frac{d}{dx}(x^3) + \dots + \frac{d}{dx}(x^{50})$

Using the power rule $\frac{d}{dx}(x^n) = nx^{n-1}$ and the rule for the derivative of a constant $\frac{d}{dx}(c) = 0$:

• $\frac{d}{dx}(1) = \frac{d}{dx}(x^0) = 0x^{-1} = 0$
• $\frac{d}{dx}(x) = \frac{d}{dx}(x^1) = 1x^{1-1} = x^0 = 1$
• $\frac{d}{dx}(x^2) = 2x^{2-1} = 2x^1 = 2x$
• $\frac{d}{dx}(x^3) = 3x^{3-1} = 3x^2$
• ...
• $\frac{d}{dx}(x^{50}) = 50x^{50-1} = 50x^{49}$

Summing these derivatives, we get the general derivative $f'(x)$:

$f'(x) = 0 + 1 + 2x + 3x^2 + \dots + 50x^{49}$

$f'(x) = 1 + 2x + 3x^2 + \dots + 50x^{49}$

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 $(1)^n = 1$ for any positive integer $n$, the expression 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 sum of the first $N$ positive integers is given by the formula $\frac{N(N+1)}{2}$. Here, $N = 50$.

$f'(1) = \frac{50(50+1)}{2}$

$f'(1) = \frac{50 \times 51}{2}$

$f'(1) = \frac{2550}{2}$

$f'(1) = 1275$

Therefore, the derivative of the function $f(x) = 1 + x + x^2 + \dots + x^{50}$ at $x = 1$ is $1275$.

The derivative at $x=1$ is $f'(1) = 1275$.

Given:

The function $f(x) = \frac{x + 1}{x}$, where $x \neq 0$.

To Find:

The derivative of $f(x)$, denoted as $f'(x)$ or $\frac{d}{dx}(f(x))$.

Solution:

We can find the derivative of $f(x)$ in two ways: using the quotient rule or by first simplifying the function.

Method 1: Using the Quotient Rule

The quotient rule 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 case, let $u(x) = x + 1$ and $v(x) = x$.

Find the derivatives of $u(x)$ and $v(x)$:

$u'(x) = \frac{d}{dx}(x + 1) = \frac{d}{dx}(x) + \frac{d}{dx}(1) = 1 + 0 = 1$

$v'(x) = \frac{d}{dx}(x) = 1$

Now, apply the quotient rule formula:

$f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2}$

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

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

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

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

Method 2: Simplifying the function first

We can rewrite the function $f(x)$ by dividing each term in the numerator by $x$:

$f(x) = \frac{x + 1}{x} = \frac{x}{x} + \frac{1}{x}$

$f(x) = 1 + \frac{1}{x}$

Now, we can find the derivative using the sum rule and the power rule.

$f'(x) = \frac{d}{dx}(1 + \frac{1}{x})$

$f'(x) = \frac{d}{dx}(1) + \frac{d}{dx}(\frac{1}{x})$

The derivative of a constant (1) is $0$.

The term $\frac{1}{x}$ can be written as $x^{-1}$. Using the power rule $\frac{d}{dx}(x^n) = nx^{n-1}$ with $n=-1$:

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

Combining the derivatives of the terms:

$f'(x) = 0 + (-\frac{1}{x^2})$

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

Both methods yield the same result.

Therefore, the derivative of the function $f(x) = \frac{x + 1}{x}$ is $-\frac{1}{x^2}$.

The derivative is $f'(x) = -\frac{1}{x^2}$.

Given:

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

To Compute:

The derivative of $f(x)$, denoted as $f'(x)$ or $\frac{d}{dx}(\sin x)$.

Solution:

We will compute the derivative of $f(x) = \sin x$ using the definition of the derivative (first principle):

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

First, find $f(x+h)$ by substituting $(x+h)$ into the function $f(x) = \sin x$:

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

The difference $f(x+h) - f(x)$ is:

$f(x+h) - f(x) = \sin(x+h) - \sin x$

We use the trigonometric identity for the difference of sines: $\sin A - \sin B = 2 \cos \left(\frac{A+B}{2}\right) \sin \left(\frac{A-B}{2}\right)$.

Let $A = x+h$ and $B = x$. Then $\frac{A+B}{2} = \frac{(x+h) + x}{2} = \frac{2x+h}{2} = x + \frac{h}{2}$ and $\frac{A-B}{2} = \frac{(x+h) - x}{2} = \frac{h}{2}$.

So, $\sin(x+h) - \sin x = 2 \cos \left(x + \frac{h}{2}\right) \sin \left(\frac{h}{2}\right)$.

Substitute this into the limit definition of the derivative:

$f'(x) = \lim\limits_{h \to 0} \frac{2 \cos \left(x + \frac{h}{2}\right) \sin \left(\frac{h}{2}\right)}{h}$

Rearrange the terms to separate the cosine part and the part involving $\sin(\frac{h}{2})$ and $h$:

$f'(x) = \lim\limits_{h \to 0} \left( \cos \left(x + \frac{h}{2}\right) \times \frac{2 \sin \left(\frac{h}{2}\right)}{h} \right)$

We can rewrite the second part as $\frac{\sin(\frac{h}{2})}{\frac{h}{2}}$.

$f'(x) = \lim\limits_{h \to 0} \left( \cos \left(x + \frac{h}{2}\right) \times \frac{\sin \left(\frac{h}{2}\right)}{\frac{h}{2}} \right)$

Using the product rule for limits, we can split this into two limits:

$f'(x) = \left( \lim\limits_{h \to 0} \cos \left(x + \frac{h}{2}\right) \right) \times \left( \lim\limits_{h \to 0} \frac{\sin \left(\frac{h}{2}\right)}{\frac{h}{2}} \right)$

Evaluate the first limit: As $h \to 0$, $x + \frac{h}{2} \to x$. Since the cosine function is continuous, $\lim\limits_{h \to 0} \cos \left(x + \frac{h}{2}\right) = \cos(x)$.

Evaluate the second limit: Let $k = \frac{h}{2}$. As $h \to 0$, $k \to 0$. The limit becomes $\lim\limits_{k \to 0} \frac{\sin k}{k}$. This is a standard trigonometric limit which equals 1.

So, $\lim\limits_{h \to 0} \frac{\sin \left(\frac{h}{2}\right)}{\frac{h}{2}} = 1$.

Combine the results of the two limits:

$f'(x) = \cos(x) \times 1$

$f'(x) = \cos x$

Therefore, the derivative of the function $f(x) = \sin x$ is $\cos x$.

The derivative is $f'(x) = \cos x$.

Given:

The function $f(x) = \tan x$.

To Compute:

The derivative of $f(x)$, denoted as $f'(x)$ or $\frac{d}{dx}(\tan x)$.

Solution:

We can compute the derivative of $f(x) = \tan x$ by rewriting $\tan x$ in terms of $\sin x$ and $\cos x$ and then using the quotient rule.

$f(x) = \tan x = \frac{\sin x}{\cos x}$

The quotient rule 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 case, let $u(x) = \sin x$ and $v(x) = \cos x$.

We need to find the derivatives of $u(x)$ and $v(x)$:

$u'(x) = \frac{d}{dx}(\sin x) = \cos x$

$v'(x) = \frac{d}{dx}(\cos x) = -\sin x$

Now, substitute $u(x)$, $v(x)$, $u'(x)$, and $v'(x)$ into the quotient rule formula:

$f'(x) = \frac{(\cos x)(\cos x) - (\sin x)(-\sin x)}{(\cos x)^2}$

$f'(x) = \frac{\cos^2 x - (-\sin^2 x)}{\cos^2 x}$

$f'(x) = \frac{\cos^2 x + \sin^2 x}{\cos^2 x}$

Using the fundamental trigonometric identity $\sin^2 x + \cos^2 x = 1$ in the numerator:

$f'(x) = \frac{1}{\cos^2 x}$

Recall that $\sec x = \frac{1}{\cos x}$. Therefore, $\frac{1}{\cos^2 x} = \left(\frac{1}{\cos x}\right)^2 = \sec^2 x$.

$f'(x) = \sec^2 x$

Therefore, the derivative of the function $f(x) = \tan x$ is $\sec^2 x$.

The derivative is $f'(x) = \sec^2 x$.

Given:

The function $f(x) = \sin^2 x$.

To Compute:

The derivative of $f(x)$, denoted as $f'(x)$ or $\frac{d}{dx}(f(x))$.

Solution:

We can rewrite the function as $f(x) = (\sin x)^2$. To find the derivative of this function, we need to use the chain rule.

The chain rule is used when differentiating a composite function, $f(x) = g(h(x))$. It states that the derivative is $f'(x) = g'(h(x)) \times h'(x)$.

In this function, $f(x) = (\sin x)^2$, we can identify the outer function $g(u) = u^2$ and the inner function $h(x) = \sin x$.

First, find the derivative of the outer function $g(u)$ with respect to $u$:

$g'(u) = \frac{d}{du}(u^2)$

Using the power rule, $\frac{d}{du}(u^n) = nu^{n-1}$:

$g'(u) = 2u^{2-1} = 2u$

Next, find the derivative of the inner function $h(x)$ with respect to $x$:

$h'(x) = \frac{d}{dx}(\sin x)$

The derivative of $\sin x$ is $\cos x$:

$h'(x) = \cos x$

Now, apply the chain rule: $f'(x) = g'(h(x)) \times h'(x)$.

Substitute $h(x) = \sin x$ into $g'(u) = 2u$ to get $g'(h(x)) = 2 \sin x$.

$f'(x) = (2 \sin x) \times (\cos x)$

$f'(x) = 2 \sin x \cos x$

This result can also be written using the trigonometric identity $\sin(2x) = 2 \sin x \cos x$.

So, $f'(x) = \sin(2x)$.

Therefore, the derivative of the function $f(x) = \sin^2 x$ is $2 \sin x \cos x$ or $\sin(2x)$.

The derivative is $f'(x) = 2 \sin x \cos x$ or $f'(x) = \sin(2x)$.

Given:

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

To Find:

The derivative of $f(x)$ at $x = 10$, denoted as $f'(10)$.

Solution:

We can find the derivative of $f(x)$ at $x = 10$ using the limit definition of the derivative at a point:

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

In this case, $c = 10$ and $f(x) = x^2 - 2$.

First, we find $f(c)$, which is $f(10)$:

$f(10) = (10)^2 - 2 = 100 - 2 = 98$

Next, we find $f(c+h)$, which is $f(10+h)$:

$f(10+h) = (10+h)^2 - 2$

Expand $(10+h)^2$ using the formula $(a+b)^2 = a^2 + 2ab + b^2$:

$f(10+h) = (10^2 + 2(10)h + h^2) - 2$

$f(10+h) = 100 + 20h + h^2 - 2$

$f(10+h) = 98 + 20h + h^2$

Now, substitute these into the limit definition for $f'(10)$:

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

$f'(10) = \lim\limits_{h \to 0} \frac{(98 + 20h + h^2) - 98}{h}$

Simplify the numerator:

$f'(10) = \lim\limits_{h \to 0} \frac{98 + 20h + h^2 - 98}{h}$

$f'(10) = \lim\limits_{h \to 0} \frac{20h + h^2}{h}$

Factor out $h$ from the numerator:

$f'(10) = \lim\limits_{h \to 0} \frac{h(20 + h)}{h}$

Since we are evaluating the limit as $h \to 0$, $h$ is approaching $0$ but is not equal to $0$. Therefore, we can cancel the $h$ terms:

$f'(10) = \lim\limits_{h \to 0} (20 + h)$

Substitute $h = 0$ to evaluate the limit:

$f'(10) = 20 + 0 = 20$

Alternate Solution (Using Power Rule):

We can first find the general derivative of $f(x) = x^2 - 2$ using the power rule and the difference rule.

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

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

Using the power rule $\frac{d}{dx}(x^n) = nx^{n-1}$, $\frac{d}{dx}(x^2) = 2x^{2-1} = 2x$.

The derivative of a constant ($2$) is $0$.

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

$f'(x) = 2x$

Now, evaluate the general derivative at $x = 10$:

$f'(10) = 2(10) = 20$

This matches the result obtained from the limit definition.

Therefore, the derivative of the function $f(x) = x^2 - 2$ at $x = 10$ is $20$.

The derivative at $x=10$ is $f'(10) = 20$.

Given:

The function $f(x) = x$.

To Find:

The derivative of $f(x)$ at $x = 1$, denoted as $f'(1)$.

Solution:

We can find the derivative of a function $f(x)$ at a specific point $x=c$ using the limit definition of the derivative at a point:

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

In this case, $c = 1$ and $f(x) = x$.

First, we find $f(c)$, which is $f(1)$:

$f(1) = 1$

Next, we find $f(c+h)$, which is $f(1+h)$:

$f(1+h) = 1+h$

Now, substitute these into the limit definition for $f'(1)$:

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

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

Simplify the numerator:

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

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

Since we are evaluating the limit as $h \to 0$, $h$ is approaching $0$ but is not equal to $0$. Therefore, we can cancel the $h$ terms:

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

The limit of a constant as $h$ approaches any value is the constant itself:

$f'(1) = 1$

Alternate Solution (Using Power Rule):

We can first find the general derivative of $f(x) = x$ using the power rule. The function is $f(x) = x^1$. Using the power rule $\frac{d}{dx}(x^n) = nx^{n-1}$ with $n=1$:

$f'(x) = \frac{d}{dx}(x^1) = 1 \times x^{1-1} = 1 \times x^0 = 1 \times 1 = 1$

The general derivative is $f'(x) = 1$. Now, evaluate this at $x=1$:

$f'(1) = 1$

This matches the result obtained from the limit definition.

Therefore, the derivative of the function $f(x) = x$ at $x = 1$ is $1$.

The derivative at $x=1$ is $f'(1) = 1$.

Given:

The function $f(x) = 99x$.

To Find:

The derivative of $f(x)$ at $x = 100$, denoted as $f'(100)$.

Solution:

We can find the derivative of a function $f(x)$ at a specific point $x=c$ using the limit definition of the derivative at a point:

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

In this case, $c = 100$ and $f(x) = 99x$.

First, we find $f(c)$, which is $f(100)$:

$f(100) = 99 \times 100 = 9900$

Next, we find $f(c+h)$, which is $f(100+h)$:

$f(100+h) = 99(100+h)$

$f(100+h) = 9900 + 99h$

Now, substitute these into the limit definition for $f'(100)$:

$f'(100) = \lim\limits_{h \to 0} \frac{f(100+h) - f(100)}{h}$

$f'(100) = \lim\limits_{h \to 0} \frac{(9900 + 99h) - 9900}{h}$

Simplify the numerator:

$f'(100) = \lim\limits_{h \to 0} \frac{9900 + 99h - 9900}{h}$

$f'(100) = \lim\limits_{h \to 0} \frac{99h}{h}$

Since we are evaluating the limit as $h \to 0$, $h$ is approaching $0$ but is not equal to $0$. Therefore, we can cancel the $h$ terms:

$f'(100) = \lim\limits_{h \to 0} 99$

The limit of a constant as $h$ approaches any value is the constant itself:

$f'(100) = 99$

Alternate Solution (Using Power Rule):

We can first find the general derivative of $f(x) = 99x$ using the power rule. The function is $f(x) = 99x^1$. Using the power rule $\frac{d}{dx}(cx^n) = cnx^{n-1}$ with $c=99$ and $n=1$:

$f'(x) = \frac{d}{dx}(99x^1) = 99 \times 1 \times x^{1-1} = 99 \times x^0 = 99 \times 1 = 99$

The general derivative is $f'(x) = 99$. Now, evaluate this at $x=100$:

$f'(100) = 99$

This matches the result obtained from the limit definition.

Therefore, the derivative of the function $f(x) = 99x$ at $x = 100$ is $99$.

The derivative at $x=100$ is $f'(100) = 99$.

We will find the derivative of each function using the first principle, which is given by the limit definition:

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

(i) $f(x) = x^3 - 27$

First, find $f(x+h)$:

$f(x+h) = (x+h)^3 - 27$

Expand $(x+h)^3 = x^3 + 3x^2h + 3xh^2 + h^3$:

$f(x+h) = x^3 + 3x^2h + 3xh^2 + h^3 - 27$

Next, find the difference $f(x+h) - f(x)$:

$f(x+h) - f(x) = (x^3 + 3x^2h + 3xh^2 + h^3 - 27) - (x^3 - 27)$

$f(x+h) - f(x) = x^3 + 3x^2h + 3xh^2 + h^3 - 27 - x^3 + 27$

$f(x+h) - f(x) = 3x^2h + 3xh^2 + h^3$

Now, divide by $h$:

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

Factor out $h$ from the numerator:

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

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

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

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

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

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

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

$\mathbf{f'(x) = 3x^2}$

(ii) $f(x) = (x - 1)(x - 2)$

First, simplify the function:

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

Next, find $f(x+h)$:

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

$f(x+h) = (x^2 + 2xh + h^2) - (3x + 3h) + 2$

$f(x+h) = x^2 + 2xh + h^2 - 3x - 3h + 2$

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

$f(x+h) - f(x) = (x^2 + 2xh + h^2 - 3x - 3h + 2) - (x^2 - 3x + 2)$

$f(x+h) - f(x) = x^2 + 2xh + h^2 - 3x - 3h + 2 - x^2 + 3x - 2$

$f(x+h) - f(x) = 2xh + h^2 - 3h$

Now, divide by $h$:

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

Factor out $h$ from the numerator:

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

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

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

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

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

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

$\mathbf{f'(x) = 2x - 3}$

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

First, find $f(x+h)$:

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

Next, find the difference $f(x+h) - f(x)$:

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

Combine the fractions using a common denominator $x^2(x+h)^2$:

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

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

Expand $(x+h)^2 = x^2 + 2xh + h^2$ in the numerator:

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

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

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

Factor out $h$ from the numerator:

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

Now, divide by $h$:

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

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

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

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

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

Substitute $h = 0$:

$f'(x) = \frac{-2x - 0}{x^2(x+0)^2} = \frac{-2x}{x^2(x^2)} = \frac{-2x}{x^4}$

Simplify the fraction:

$\mathbf{f'(x) = -\frac{2}{x^3}}$

(iv) $f(x) = \frac{ x \;+\; 1}{x \;-\; 1}$

First, find $f(x+h)$:

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

Next, find the difference $f(x+h) - f(x)$:

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

Combine the fractions using the common denominator $(x+h-1)(x-1)$:

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

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 expansions back into the numerator difference:

$f(x+h) - f(x) = (x^2 + hx - h - 1) - (x^2 + xh + h - 1)$

$f(x+h) - f(x) = x^2 + hx - h - 1 - x^2 - xh - h + 1$

Cancel out terms:

$f(x+h) - f(x) = (x^2 - x^2) + (hx - xh) + (-h - h) + (-1 + 1)$

$f(x+h) - f(x) = 0 + 0 - 2h + 0$

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

Now, divide by $h$:

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

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

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

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

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

Substitute $h = 0$:

$f'(x) = \frac{-2}{(x+0-1)(x-1)} = \frac{-2}{(x-1)(x-1)} = \frac{-2}{(x-1)^2}$

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

Given:

The function $f(x) = \frac{x^{100}}{100} + \frac{x^{99}}{99} + \dots + \frac{x^{2}}{2} + x + 1$.

To Prove:

$f'(1) = 100 f'(0)$.

Solution:

First, we need to find the general derivative of the function $f(x)$. The function is a sum of terms. We can use the sum rule and the power rule for differentiation.

The sum rule states that the derivative of a sum of functions is the sum of their derivatives: $\frac{d}{dx}[g(x) + h(x)] = \frac{d}{dx}[g(x)] + \frac{d}{dx}[h(x)]$.

The power rule states that the derivative of $x^n$ is $nx^{n-1}$, i.e., $\frac{d}{dx}(x^n) = nx^{n-1}$. Also, for a constant $c$, $\frac{d}{dx}(cx^n) = c \frac{d}{dx}(x^n) = cnx^{n-1}$. The derivative of a constant is $0$, i.e., $\frac{d}{dx}(c) = 0$.

Apply the derivative rules to each term of $f(x) = \frac{x^{100}}{100} + \frac{x^{99}}{99} + \dots + \frac{x^2}{2} + x + 1$.

$f'(x) = \frac{d}{dx}\left(\frac{x^{100}}{100}\right) + \frac{d}{dx}\left(\frac{x^{99}}{99}\right) + \dots + \frac{d}{dx}\left(\frac{x^{2}}{2}\right) + \frac{d}{dx}(x) + \frac{d}{dx}(1)$

• For the term $\frac{x^n}{n} = \frac{1}{n}x^n$, the derivative is $\frac{1}{n} \cdot nx^{n-1} = x^{n-1}$.
• $\frac{d}{dx}\left(\frac{x^{100}}{100}\right) = x^{100-1} = x^{99}$
• $\frac{d}{dx}\left(\frac{x^{99}}{99}\right) = x^{99-1} = x^{98}$
• ...
• $\frac{d}{dx}\left(\frac{x^{2}}{2}\right) = x^{2-1} = x^{1} = x$
• $\frac{d}{dx}(x) = \frac{d}{dx}(x^1) = 1 \cdot x^{1-1} = x^0 = 1$
• $\frac{d}{dx}(1) = 0$

Summing these derivatives, the general derivative $f'(x)$ is:

$f'(x) = x^{99} + x^{98} + \dots + x + 1 + 0$

$f'(x) = 1 + x + x^2 + \dots + x^{98} + x^{99}$

Next, we evaluate the derivative at $x=1$. Substitute $x=1$ into the expression for $f'(x)$:

$f'(1) = 1 + (1) + (1)^2 + \dots + (1)^{98} + (1)^{99}$

Since $1^n = 1$ for any positive integer $n$, each term in the sum is 1.

$f'(1) = 1 + 1 + 1 + \dots + 1 + 1$

To find the number of terms, note that the powers of $x$ are from $0$ (in the term 1, which is $x^0$) up to $99$. So there are $99 - 0 + 1 = 100$ terms.

$f'(1) = \underbrace{1 + 1 + \dots + 1}_{100 \text{ terms}}$

$f'(1) = 100 \times 1$

$f'(1) = 100$

Now, we evaluate the derivative at $x=0$. Substitute $x=0$ into the expression for $f'(x)$:

$f'(0) = 1 + (0) + (0)^2 + \dots + (0)^{98} + (0)^{99}$

Except for the first term (1), all other terms containing $x$ raised to a positive power will become $0$ when $x=0$.

$f'(0) = 1 + 0 + 0 + \dots + 0 + 0$

$f'(0) = 1$

Finally, we need to prove that $f'(1) = 100 f'(0)$.

Substitute the values we found for $f'(1)$ and $f'(0)$ into this equation:

LHS = $f'(1) = 100$

RHS = $100 f'(0) = 100 \times 1 = 100$

Since LHS = RHS ($100 = 100$), the relationship is proven.

Thus, we have shown that $f'(1) = 100 f'(0)$.

Given:

The function $f(x) = x^n + ax^{n-1} + a^2x^{n-2} + \dots + a^{n-1}x + a^n$, where $a$ is a fixed real number.

To Find:

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

Solution:

The function $f(x)$ is a polynomial, which is a sum of terms of the form $cx^k$, where $c$ is a constant (in this case, powers of $a$) and $k$ is a non-negative integer (the power of $x$). We can find the derivative of $f(x)$ using the sum rule, the constant multiple rule, and the power rule.

The sum rule allows us to differentiate each term separately:

$f'(x) = \frac{d}{dx}(x^n) + \frac{d}{dx}(ax^{n-1}) + \frac{d}{dx}(a^2x^{n-2}) + \dots + \frac{d}{dx}(a^{n-1}x) + \frac{d}{dx}(a^n)$

For each term of the form $a^k x^{n-k}$ (where $a^k$ is the constant coefficient and $n-k$ is the power of $x$), we use the power rule $\frac{d}{dx}(cx^m) = cmx^{m-1}$.

• For the first term, $x^n = 1 \cdot x^n$: $c=1$, $m=n$. Derivative is $1 \cdot nx^{n-1} = nx^{n-1}$.
• For the second term, $ax^{n-1}$: $c=a$, $m=n-1$. Derivative is $a \cdot (n-1)x^{(n-1)-1} = a(n-1)x^{n-2}$.
• For the third term, $a^2x^{n-2}$: $c=a^2$, $m=n-2$. Derivative is $a^2 \cdot (n-2)x^{(n-2)-1} = a^2(n-2)x^{n-3}$.
• ...
• The second to last term is $a^{n-1}x = a^{n-1}x^1$: $c=a^{n-1}$, $m=1$. Derivative is $a^{n-1} \cdot 1x^{1-1} = a^{n-1}x^0 = a^{n-1}$.
• The last term is $a^n$: This is a constant, since $a$ is a fixed real number and $n$ is the power. The derivative of a constant is $0$. $\frac{d}{dx}(a^n) = 0$.

Combine the derivatives of all the terms:

$f'(x) = nx^{n-1} + a(n-1)x^{n-2} + a^2(n-2)x^{n-3} + \dots + a^{n-1}(1)x^{1-1} + 0$

$f'(x) = nx^{n-1} + a(n-1)x^{n-2} + a^2(n-2)x^{n-3} + \dots + a^{n-1}$

Therefore, the derivative of the function $f(x) = x^n + ax^{n-1} + a^2x^{n-2} + \dots + a^{n-1}x + a^n$ is $nx^{n-1} + a(n-1)x^{n-2} + a^2(n-2)x^{n-3} + \dots + a^{n-1}$.

The derivative is $f'(x) = nx^{n-1} + a(n-1)x^{n-2} + a^2(n-2)x^{n-3} + \dots + a^{n-1}$.

We will find the derivative of each function using appropriate differentiation rules (product rule, chain rule, quotient rule, power rule, sum/difference rule).

(i) $f(x) = (x - a)(x - b)$

Method 1: Using the Product Rule

The product rule states that if $f(x) = u(x)v(x)$, then $f'(x) = u'(x)v(x) + u(x)v'(x)$.

Let $u(x) = x - a$ and $v(x) = x - b$.

Find the derivatives of $u(x)$ and $v(x)$:

$u'(x) = \frac{d}{dx}(x - a) = \frac{d}{dx}(x) - \frac{d}{dx}(a) = 1 - 0 = 1$

$v'(x) = \frac{d}{dx}(x - b) = \frac{d}{dx}(x) - \frac{d}{dx}(b) = 1 - 0 = 1$

Apply the product rule:

$f'(x) = u'(x)v(x) + u(x)v'(x)$

$f'(x) = (1)(x - b) + (x - a)(1)$

$f'(x) = x - b + x - a$

$f'(x) = 2x - a - b$

Method 2: Expanding the function first

Expand the function $f(x)$: $f(x) = (x - a)(x - b) = x^2 - bx - ax + ab = x^2 - (a+b)x + ab$.

Now, differentiate the polynomial using the sum/difference rule and power rule:

$f'(x) = \frac{d}{dx}(x^2 - (a+b)x + ab)$

$f'(x) = \frac{d}{dx}(x^2) - \frac{d}{dx}((a+b)x) + \frac{d}{dx}(ab)$

$f'(x) = 2x - (a+b)\frac{d}{dx}(x) + 0 \quad$ (since $a$ and $b$ are constants, $a+b$ and $ab$ are constants)

$f'(x) = 2x - (a+b)(1)$

$f'(x) = 2x - a - b$

Both methods give the same result.

The derivative is $\mathbf{f'(x) = 2x - a - b}$.

(ii) $f(x) = (ax^2 + b)^2$

Method 1: Using the Chain Rule

This is a composite function of the form $g(h(x))$, where the outer function is squaring, $g(u) = u^2$, and the inner function is a polynomial, $h(x) = ax^2 + b$. The chain rule states $f'(x) = g'(h(x)) \times h'(x)$.

Find the derivative of the outer function $g(u)$ with respect to $u$:

$g'(u) = \frac{d}{du}(u^2) = 2u^{2-1} = 2u$

Find the derivative of the inner function $h(x)$ with respect to $x$:

$h'(x) = \frac{d}{dx}(ax^2 + b) = \frac{d}{dx}(ax^2) + \frac{d}{dx}(b) = a \cdot 2x^{2-1} + 0 = 2ax$

Apply the chain rule $f'(x) = g'(h(x)) \times h'(x)$: Substitute $h(x) = ax^2 + b$ into $g'(u) = 2u$ to get $g'(h(x)) = 2(ax^2 + b)$.

$f'(x) = (2(ax^2 + b)) \times (2ax)$

$f'(x) = 4ax(ax^2 + b)$

We can expand this result:

$f'(x) = 4a^2x^3 + 4abx$

Method 2: Expanding the function first

Expand the function $f(x)$: $f(x) = (ax^2 + b)^2 = (ax^2)^2 + 2(ax^2)(b) + b^2 = a^2x^4 + 2abx^2 + b^2$.

Now, differentiate the polynomial using the sum rule and power rule:

$f'(x) = \frac{d}{dx}(a^2x^4 + 2abx^2 + b^2)$

$f'(x) = \frac{d}{dx}(a^2x^4) + \frac{d}{dx}(2abx^2) + \frac{d}{dx}(b^2)$

$f'(x) = a^2 \cdot 4x^{4-1} + 2ab \cdot 2x^{2-1} + 0 \quad$ (since $a, b$ are constants, $a^2, 2ab, b^2$ are constants)

$f'(x) = 4a^2x^3 + 4abx^1$

$f'(x) = 4a^2x^3 + 4abx$

Both methods give the same result.

The derivative is $\mathbf{f'(x) = 4ax(ax^2 + b) = 4a^2x^3 + 4abx}$.

(iii) $f(x) = \frac{x \;-\; a}{x \;-\;b}$

We use 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 $u(x) = x - a$ and $v(x) = x - b$.

Find the derivatives of $u(x)$ and $v(x)$:

$u'(x) = \frac{d}{dx}(x - a) = \frac{d}{dx}(x) - \frac{d}{dx}(a) = 1 - 0 = 1$

$v'(x) = \frac{d}{dx}(x - b) = \frac{d}{dx}(x) - \frac{d}{dx}(b) = 1 - 0 = 1$

Apply the quotient rule formula:

$f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2}$

$f'(x) = \frac{(1)(x - b) - (x - a)(1)}{(x - b)^2}$

$f'(x) = \frac{(x - b) - (x - a)}{(x - b)^2}$

$f'(x) = \frac{x - b - x + a}{(x - b)^2}$

$f'(x) = \frac{-b + a}{(x - b)^2}$

$f'(x) = \frac{a - b}{(x - b)^2}$

The derivative is $\mathbf{f'(x) = \frac{a - b}{(x - b)^2}}$.

Given:

The function $f(x) = \frac{x^n - a^n}{x - a}$, where $a$ is a constant and $x \neq a$.

To Find:

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

Solution:

We can simplify the function $f(x)$ first by factoring the numerator using the difference of powers formula: $x^n - a^n = (x-a)(x^{n-1} + ax^{n-2} + a^2x^{n-3} + \dots + a^{n-1})$.

For $x \neq a$, we can write:

$f(x) = \frac{(x-a)(x^{n-1} + ax^{n-2} + a^2x^{n-3} + \dots + a^{n-1})}{x - a}$

Cancel the $(x-a)$ terms (since $x \neq a$):

$f(x) = x^{n-1} + ax^{n-2} + a^2x^{n-3} + \dots + a^{n-1}$

This is a polynomial in $x$, where $a$ is a constant. We can find the derivative using the sum rule, constant multiple rule, and power rule.

$f'(x) = \frac{d}{dx}(x^{n-1} + ax^{n-2} + a^2x^{n-3} + \dots + a^{n-1})$

$f'(x) = \frac{d}{dx}(x^{n-1}) + \frac{d}{dx}(ax^{n-2}) + \frac{d}{dx}(a^2x^{n-3}) + \dots + \frac{d}{dx}(a^{n-1})$

Apply the power rule $\frac{d}{dx}(cx^m) = cmx^{m-1}$ and $\frac{d}{dx}(c) = 0$:

• $\frac{d}{dx}(x^{n-1}) = (n-1)x^{(n-1)-1} = (n-1)x^{n-2}$
• $\frac{d}{dx}(ax^{n-2}) = a \cdot (n-2)x^{(n-2)-1} = a(n-2)x^{n-3}$
• $\frac{d}{dx}(a^2x^{n-3}) = a^2 \cdot (n-3)x^{(n-3)-1} = a^2(n-3)x^{n-4}$
• ...
• The second to last term in $f(x)$ is $a^{n-2}x^1$: $\frac{d}{dx}(a^{n-2}x) = a^{n-2} \cdot 1 x^{1-1} = a^{n-2}$.
• The last term in $f(x)$ is $a^{n-1}$: This is a constant term. $\frac{d}{dx}(a^{n-1}) = 0$.

Combining the derivatives of all the terms:

$f'(x) = (n-1)x^{n-2} + a(n-2)x^{n-3} + a^2(n-3)x^{n-4} + \dots + a^{n-2} + 0$

$f'(x) = (n-1)x^{n-2} + a(n-2)x^{n-3} + a^2(n-3)x^{n-4} + \dots + a^{n-2}$

Therefore, the derivative of the function $f(x) = \frac{x^n - a^n}{x - a}$ is $(n-1)x^{n-2} + a(n-2)x^{n-3} + a^2(n-3)x^{n-4} + \dots + a^{n-2}$.

The derivative is $f'(x) = (n-1)x^{n-2} + a(n-2)x^{n-3} + a^2(n-3)x^{n-4} + \dots + a^{n-2}$.

We will find the derivative of each function using the rules of differentiation (sum rule, difference rule, constant multiple rule, power rule, product rule, quotient rule).

(i) $f(x) = 2x - \frac{3}{4}$

Given: $f(x) = 2x - \frac{3}{4}$.

To Find: $f'(x)$.

Solution: Use the difference rule, constant multiple rule, and power rule.

$f'(x) = \frac{d}{dx}\left(2x - \frac{3}{4}\right)$

$f'(x) = \frac{d}{dx}(2x) - \frac{d}{dx}\left(\frac{3}{4}\right)$

Using the constant multiple rule and power rule $\frac{d}{dx}(cx^n) = cnx^{n-1}$, $\frac{d}{dx}(2x) = 2 \cdot 1 x^{1-1} = 2x^0 = 2 \cdot 1 = 2$.

The derivative of a constant is $0$, so $\frac{d}{dx}\left(\frac{3}{4}\right) = 0$.

$f'(x) = 2 - 0$

$\mathbf{f'(x) = 2}$

(ii) $f(x) = (5x^3 + 3x - 1) (x - 1)$

Given: $f(x) = (5x^3 + 3x - 1) (x - 1)$.

To Find: $f'(x)$.

Solution: Expand the function first, then use the sum/difference rule and power rule.

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

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

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

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

Now, differentiate term by term:

$f'(x) = \frac{d}{dx}(5x^4) - \frac{d}{dx}(5x^3) + \frac{d}{dx}(3x^2) - \frac{d}{dx}(4x) + \frac{d}{dx}(1)$

Using the power rule and constant multiple rule:

$f'(x) = 5 \cdot 4x^{4-1} - 5 \cdot 3x^{3-1} + 3 \cdot 2x^{2-1} - 4 \cdot 1x^{1-1} + 0$

$f'(x) = 20x^3 - 15x^2 + 6x^1 - 4x^0 + 0$

$f'(x) = 20x^3 - 15x^2 + 6x - 4$

The derivative is $\mathbf{f'(x) = 20x^3 - 15x^2 + 6x - 4}$.

(iii) $f(x) = x^{-3} (5 + 3x)$

Given: $f(x) = x^{-3} (5 + 3x)$.

To Find: $f'(x)$.

Solution: Expand the function first, then use the sum rule, constant multiple rule, and power rule.

$f(x) = x^{-3}(5) + x^{-3}(3x)$

$f(x) = 5x^{-3} + 3x^{-3+1}$

$f(x) = 5x^{-3} + 3x^{-2}$

Now, differentiate term by term:

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

Using the power rule and constant multiple rule:

$f'(x) = 5 \cdot (-3)x^{-3-1} + 3 \cdot (-2)x^{-2-1}$

$f'(x) = -15x^{-4} - 6x^{-3}$

The derivative is $\mathbf{f'(x) = -15x^{-4} - 6x^{-3}}$.

(iv) $f(x) = x^{5} (3 - 6x^{-9})$

Given: $f(x) = x^{5} (3 - 6x^{-9})$.

To Find: $f'(x)$.

Solution: Expand the function first, then use the difference rule, constant multiple rule, and power rule.

$f(x) = x^5(3) - x^5(6x^{-9})$

$f(x) = 3x^5 - 6x^{5-9}$

$f(x) = 3x^5 - 6x^{-4}$

Now, differentiate term by term:

$f'(x) = \frac{d}{dx}(3x^5) - \frac{d}{dx}(6x^{-4})$

Using the power rule and constant multiple rule:

$f'(x) = 3 \cdot 5x^{5-1} - 6 \cdot (-4)x^{-4-1}$

$f'(x) = 15x^4 - (-24)x^{-5}$

$f'(x) = 15x^4 + 24x^{-5}$

The derivative is $\mathbf{f'(x) = 15x^4 + 24x^{-5}}$.

(v) $f(x) = x^{-4} (3 - 4x^{-5})$

Given: $f(x) = x^{-4} (3 - 4x^{-5})$.

To Find: $f'(x)$.

Solution: Expand the function first, then use the difference rule, constant multiple rule, and power rule.

$f(x) = x^{-4}(3) - x^{-4}(4x^{-5})$

$f(x) = 3x^{-4} - 4x^{-4-5}$

$f(x) = 3x^{-4} - 4x^{-9}$

Now, differentiate term by term:

$f'(x) = \frac{d}{dx}(3x^{-4}) - \frac{d}{dx}(4x^{-9})$

Using the power rule and constant multiple rule:

$f'(x) = 3 \cdot (-4)x^{-4-1} - 4 \cdot (-9)x^{-9-1}$

$f'(x) = -12x^{-5} - (-36)x^{-10}$

$f'(x) = -12x^{-5} + 36x^{-10}$

The derivative is $\mathbf{f'(x) = -12x^{-5} + 36x^{-10}}$.

(vi) $f(x) = \frac{2}{x \;+\; 1} - \frac{x^{2}}{3x\;-\;1}$

Given: $f(x) = \frac{2}{x + 1} - \frac{x^{2}}{3x - 1}$.

To Find: $f'(x)$.

Solution: Use the difference rule and the quotient rule for each term.

$f'(x) = \frac{d}{dx}\left(\frac{2}{x + 1}\right) - \frac{d}{dx}\left(\frac{x^{2}}{3x - 1}\right)$

For the first term, $\frac{d}{dx}\left(\frac{2}{x + 1}\right)$: Use the quotient rule $\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2}$ with $u=2$ and $v=x+1$.

$u' = \frac{d}{dx}(2) = 0$

$v' = \frac{d}{dx}(x + 1) = 1$

$\frac{d}{dx}\left(\frac{2}{x + 1}\right) = \frac{(0)(x+1) - (2)(1)}{(x+1)^2} = \frac{0 - 2}{(x+1)^2} = \frac{-2}{(x+1)^2}$

For the second term, $\frac{d}{dx}\left(\frac{x^{2}}{3x - 1}\right)$: Use the quotient rule with $u=x^2$ and $v=3x-1$.

$u' = \frac{d}{dx}(x^2) = 2x$

$v' = \frac{d}{dx}(3x - 1) = 3$

$\frac{d}{dx}\left(\frac{x^{2}}{3x - 1}\right) = \frac{(2x)(3x-1) - (x^2)(3)}{(3x-1)^2} = \frac{6x^2 - 2x - 3x^2}{(3x-1)^2} = \frac{3x^2 - 2x}{(3x-1)^2}$

Combine the derivatives:

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

The derivative is $\mathbf{f'(x) = -\frac{2}{(x+1)^2} - \frac{3x^2 - 2x}{(3x-1)^2}}$.

Given:

The function $f(x) = \cos x$.

To Find:

The derivative of $f(x)$ from the first principle.

Solution:

We use the definition of the derivative (first principle):

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

First, find $f(x+h)$ by substituting $(x+h)$ into the function $f(x) = \cos x$:

$f(x+h) = \cos(x+h)$

The difference $f(x+h) - f(x)$ is:

$f(x+h) - f(x) = \cos(x+h) - \cos x$

We use the trigonometric identity for the difference of cosines: $\cos A - \cos B = -2 \sin \left(\frac{A+B}{2}\right) \sin \left(\frac{A-B}{2}\right)$.

Let $A = x+h$ and $B = x$. Then $\frac{A+B}{2} = \frac{(x+h) + x}{2} = \frac{2x+h}{2} = x + \frac{h}{2}$ and $\frac{A-B}{2} = \frac{(x+h) - x}{2} = \frac{h}{2}$.

So, $\cos(x+h) - \cos x = -2 \sin \left(x + \frac{h}{2}\right) \sin \left(\frac{h}{2}\right)$.

Substitute this into the limit definition of the derivative:

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

Rearrange the terms to separate the sine parts:

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

We can rewrite the second part as $\frac{\sin(\frac{h}{2})}{\frac{h}{2}}$.

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

Using the product rule for limits, we can split this into two limits:

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

Evaluate the first limit: As $h \to 0$, $x + \frac{h}{2} \to x$. Since the sine function is continuous, $\lim\limits_{h \to 0} - \sin \left(x + \frac{h}{2}\right) = - \sin(x)$.

Evaluate the second limit: Let $k = \frac{h}{2}$. As $h \to 0$, $k \to 0$. The limit becomes $\lim\limits_{k \to 0} \frac{\sin k}{k}$. This is a standard trigonometric limit which equals 1.

So, $\lim\limits_{h \to 0} \frac{\sin \left(\frac{h}{2}\right)}{\frac{h}{2}} = 1$.

Combine the results of the two limits:

$f'(x) = (-\sin(x)) \times 1$

$f'(x) = -\sin x$

Therefore, the derivative of the function $f(x) = \cos x$ from the first principle is $-\sin x$.

The derivative is $f'(x) = -\sin x$.

We will find the derivative of each function using the rules of differentiation (product rule, sum/difference rule, constant multiple rule) and known derivatives of trigonometric functions.

Recall the standard derivatives:

• $\frac{d}{dx}(\sin x) = \cos x$
• $\frac{d}{dx}(\cos x) = -\sin x$
• $\frac{d}{dx}(\tan x) = \sec^2 x$
• $\frac{d}{dx}(\cot x) = -\text{cosec}^2 x$
• $\frac{d}{dx}(\sec x) = \sec x \tan x$
• $\frac{d}{dx}(\text{cosec} x) = -\text{cosec} x \cot x$
• $\frac{d}{dx}(c) = 0$ (for a constant $c$)

(i) $f(x) = \sin x \cos x$

Given: $f(x) = \sin x \cos x$.

To Find: $f'(x)$.

Solution: Use the product rule: $\frac{d}{dx}(uv) = u'v + uv'$. Let $u = \sin x$ and $v = \cos x$.

$u' = \frac{d}{dx}(\sin x) = \cos x$

$v' = \frac{d}{dx}(\cos x) = -\sin x$

$f'(x) = (\cos x)(\cos x) + (\sin x)(-\sin x)$

$f'(x) = \cos^2 x - \sin^2 x$

This can also be written using the double angle identity for cosine: $\cos(2x) = \cos^2 x - \sin^2 x$.

So, $f'(x) = \cos(2x)$.

Alternatively, use the double angle identity for sine first: $f(x) = \frac{1}{2}(2 \sin x \cos x) = \frac{1}{2}\sin(2x)$.

Using the chain rule, $\frac{d}{dx}(\frac{1}{2}\sin(2x)) = \frac{1}{2} \frac{d}{dx}(\sin(2x)) = \frac{1}{2} \cos(2x) \cdot \frac{d}{dx}(2x) = \frac{1}{2} \cos(2x) \cdot 2 = \cos(2x)$.

The derivative is $\mathbf{f'(x) = \cos^2 x - \sin^2 x = \cos(2x)}$.

(ii) $f(x) = \sec x$

Given: $f(x) = \sec x$.

To Find: $f'(x)$.

Solution: Use the standard derivative of $\sec x$.

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

The derivative is $\mathbf{f'(x) = \sec x \tan x}$.

(iii) $f(x) = 5\sec x + 4\cos x$

Given: $f(x) = 5\sec x + 4\cos x$.

To Find: $f'(x)$.

Solution: Use the sum rule and constant multiple rule.

$f'(x) = \frac{d}{dx}(5\sec x + 4\cos x) = \frac{d}{dx}(5\sec x) + \frac{d}{dx}(4\cos x)$

$f'(x) = 5 \frac{d}{dx}(\sec x) + 4 \frac{d}{dx}(\cos x)$

$f'(x) = 5 (\sec x \tan x) + 4 (-\sin x)$

$f'(x) = 5 \sec x \tan x - 4 \sin x$

The derivative is $\mathbf{f'(x) = 5 \sec x \tan x - 4 \sin x}$.

(iv) $f(x) = \text{cosec } x$

Given: $f(x) = \text{cosec } x$.

To Find: $f'(x)$.

Solution: Use the standard derivative of $\text{cosec } x$.

$f'(x) = \frac{d}{dx}(\text{cosec } x) = -\text{cosec } x \cot x$

The derivative is $\mathbf{f'(x) = -\text{cosec } x \cot x}$.

(v) $f(x) = 3\cot x + 5\text{cosec } x$

Given: $f(x) = 3\cot x + 5\text{cosec } x$.

To Find: $f'(x)$.

Solution: Use the sum rule and constant multiple rule.

$f'(x) = \frac{d}{dx}(3\cot x + 5\text{cosec } x) = \frac{d}{dx}(3\cot x) + \frac{d}{dx}(5\text{cosec } x)$

$f'(x) = 3 \frac{d}{dx}(\cot x) + 5 \frac{d}{dx}(\text{cosec } x)$

$f'(x) = 3 (-\text{cosec}^2 x) + 5 (-\text{cosec } x \cot x)$

$f'(x) = -3 \text{cosec}^2 x - 5 \text{cosec } x \cot x$

The derivative is $\mathbf{f'(x) = -3 \text{cosec}^2 x - 5 \text{cosec } x \cot x}$.

(vi) $f(x) = 5\sin x - 6\cos x + 7$

Given: $f(x) = 5\sin x - 6\cos x + 7$.

To Find: $f'(x)$.

Solution: Use the sum/difference rule, constant multiple rule, and derivative of a constant.

$f'(x) = \frac{d}{dx}(5\sin x - 6\cos x + 7) = \frac{d}{dx}(5\sin x) - \frac{d}{dx}(6\cos x) + \frac{d}{dx}(7)$

$f'(x) = 5 \frac{d}{dx}(\sin x) - 6 \frac{d}{dx}(\cos x) + 0$

$f'(x) = 5 (\cos x) - 6 (-\sin x)$

$f'(x) = 5 \cos x + 6 \sin x$

The derivative is $\mathbf{f'(x) = 5 \cos x + 6 \sin x}$.

(vii) $f(x) = 2\tan x - 7\sec x$

Given: $f(x) = 2\tan x - 7\sec x$.

To Find: $f'(x)$.

Solution: Use the difference rule and constant multiple rule.

$f'(x) = \frac{d}{dx}(2\tan x - 7\sec x) = \frac{d}{dx}(2\tan x) - \frac{d}{dx}(7\sec x)$

$f'(x) = 2 \frac{d}{dx}(\tan x) - 7 \frac{d}{dx}(\sec x)$

$f'(x) = 2 (\sec^2 x) - 7 (\sec x \tan x)$

$f'(x) = 2 \sec^2 x - 7 \sec x \tan x$

The derivative is $\mathbf{f'(x) = 2 \sec^2 x - 7 \sec x \tan x}$.

We will find the derivative of each function using the first principle, which is given by the limit definition:

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

(i) $f(x) = \frac{2x \;+\; 3}{x \;-\; 2}$

Given: $f(x) = \frac{2x + 3}{x - 2}$, where $x \neq 2$.

To Find: $f'(x)$ using the first principle.

Solution:

Find $f(x+h)$:

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

Find the difference $f(x+h) - f(x)$:

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

Combine the fractions using a common denominator $(x + h - 2)(x - 2)$:

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

Expand the numerator:

Numerator $= (2x^2 - 4x + 2xh - 4h + 3x - 6) - (2x^2 + 2xh - 4x + 3x + 3h - 6)$

Numerator $= (2x^2 - x + 2xh - 4h - 6) - (2x^2 - x + 2xh + 3h - 6)$

Numerator $= 2x^2 - x + 2xh - 4h - 6 - 2x^2 + x - 2xh - 3h + 6$

Combine like terms in the numerator:

Numerator $= (2x^2 - 2x^2) + (-x + x) + (2xh - 2xh) + (-4h - 3h) + (-6 + 6)$

Numerator $= 0 + 0 + 0 - 7h + 0 = -7h$

So, $f(x+h) - f(x) = \frac{-7h}{(x + h - 2)(x - 2)}$

Now, divide by $h$:

$\frac{f(x+h) - f(x)}{h} = \frac{\frac{-7h}{(x + h - 2)(x - 2)}}{h} = \frac{-7h}{h(x + h - 2)(x - 2)}$

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

$\frac{f(x+h) - f(x)}{h} = \frac{-7}{(x + h - 2)(x - 2)}$

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

$f'(x) = \lim\limits_{h \to 0} \frac{-7}{(x + h - 2)(x - 2)}$

Substitute $h = 0$ into the expression:

$f'(x) = \frac{-7}{(x + 0 - 2)(x - 2)} = \frac{-7}{(x - 2)(x - 2)}$

$\mathbf{f'(x) = \frac{-7}{(x - 2)^2}}$

(ii) $f(x) = x + \frac{1}{x}$

Given: $f(x) = x + \frac{1}{x}$, where $x \neq 0$.

To Find: $f'(x)$ using the first principle.

Solution:

Find $f(x+h)$:

$f(x+h) = (x+h) + \frac{1}{x+h}$

Find the difference $f(x+h) - f(x)$:

$f(x+h) - f(x) = \left((x+h) + \frac{1}{x+h}\right) - \left(x + \frac{1}{x}\right)$

Rearrange the terms:

$f(x+h) - f(x) = (x+h - x) + \left(\frac{1}{x+h} - \frac{1}{x}\right)$

Simplify the first part and combine the fractions in the second part:

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

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

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

Now, divide by $h$:

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

Divide each term in the numerator by $h$:

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

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

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

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

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

$f'(x) = \lim\limits_{h \to 0} \left(1 - \frac{1}{x(x+h)}\right)$

Using the properties of limits, we can evaluate the limit of each term:

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

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

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

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

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

We will find the derivative of each function using the first principle, which is given by the limit definition:

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

(i) $f(x) = \sin x + \cos x$

Given: $f(x) = \sin x + \cos x$.

To Find: $f'(x)$ using the first principle.

Solution:

Find $f(x+h)$:

$f(x+h) = \sin(x+h) + \cos(x+h)$

Find the difference $f(x+h) - f(x)$:

$f(x+h) - f(x) = (\sin(x+h) + \cos(x+h)) - (\sin x + \cos x)$

$f(x+h) - f(x) = (\sin(x+h) - \sin x) + (\cos(x+h) - \cos x)$

Use the sum-to-product trigonometric identities:

$\sin A - \sin B = 2 \cos \left(\frac{A+B}{2}\right) \sin \left(\frac{A-B}{2}\right)$

$\cos A - \cos B = -2 \sin \left(\frac{A+B}{2}\right) \sin \left(\frac{A-B}{2}\right)$

Let $A = x+h$ and $B = x$. Then $\frac{A+B}{2} = x + \frac{h}{2}$ and $\frac{A-B}{2} = \frac{h}{2}$.

So, the difference becomes:

$f(x+h) - f(x) = 2 \cos\left(x + \frac{h}{2}\right) \sin\left(\frac{h}{2}\right) - 2 \sin\left(x + \frac{h}{2}\right) \sin\left(\frac{h}{2}\right)$

Factor out $2 \sin\left(\frac{h}{2}\right)$:

$f(x+h) - f(x) = 2 \sin\left(\frac{h}{2}\right) \left[\cos\left(x + \frac{h}{2}\right) - \sin\left(x + \frac{h}{2}\right)\right]$

Now, divide by $h$:

$\frac{f(x+h) - f(x)}{h} = \frac{2 \sin\left(\frac{h}{2}\right) \left[\cos\left(x + \frac{h}{2}\right) - \sin\left(x + \frac{h}{2}\right)\right]}{h}$

Rearrange the terms:

$\frac{f(x+h) - f(x)}{h} = \frac{\sin\left(\frac{h}{2}\right)}{\frac{h}{2}} \times \left[\cos\left(x + \frac{h}{2}\right) - \sin\left(x + \frac{h}{2}\right)\right]$

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

$f'(x) = \lim\limits_{h \to 0} \left(\frac{\sin\left(\frac{h}{2}\right)}{\frac{h}{2}}\right) \times \lim\limits_{h \to 0} \left[\cos\left(x + \frac{h}{2}\right) - \sin\left(x + \frac{h}{2}\right)\right]$

We know that $\lim\limits_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$. Let $\theta = \frac{h}{2}$. As $h \to 0$, $\theta \to 0$. So, $\lim\limits_{h \to 0} \frac{\sin\left(\frac{h}{2}\right)}{\frac{h}{2}} = 1$.

Also, as $h \to 0$, $\cos\left(x + \frac{h}{2}\right) \to \cos(x)$ and $\sin\left(x + \frac{h}{2}\right) \to \sin(x)$ due to the continuity of cosine and sine functions.

So, $\lim\limits_{h \to 0} \left[\cos\left(x + \frac{h}{2}\right) - \sin\left(x + \frac{h}{2}\right)\right] = \cos x - \sin x$.

Substitute these limit values:

$f'(x) = 1 \times (\cos x - \sin x)$

$\mathbf{f'(x) = \cos x - \sin x}$

(ii) $f(x) = x \sin x$

Given: $f(x) = x \sin x$.

To Find: $f'(x)$ using the first principle.

Solution:

Find $f(x+h)$:

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

Find the difference $f(x+h) - f(x)$:

$f(x+h) - f(x) = (x+h)\sin(x+h) - x \sin x$

Expand and rearrange terms:

$f(x+h) - f(x) = x\sin(x+h) + h\sin(x+h) - x \sin x$

$f(x+h) - f(x) = x(\sin(x+h) - \sin x) + h\sin(x+h)$

Now, divide by $h$:

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

Split the fraction:

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

For $h \neq 0$, cancel $h$ in the second term:

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

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

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

Using the properties of limits, we can take the limit of each term:

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

For the first term, $x$ is a constant with respect to $h$, so we can factor it out of the limit:

$\lim\limits_{h \to 0} \left(x \frac{\sin(x+h) - \sin x}{h}\right) = x \lim\limits_{h \to 0} \frac{\sin(x+h) - \sin x}{h}$

The limit $\lim\limits_{h \to 0} \frac{\sin(x+h) - \sin x}{h}$ is the definition of the derivative of $\sin x$ with respect to $x$, which is $\cos x$.

For the second term, as $h \to 0$, $\sin(x+h) \to \sin(x)$ due to the continuity of the sine function.

Substitute these limit values:

$f'(x) = x (\cos x) + \sin x$

$\mathbf{f'(x) = x \cos x + \sin x}$

We will compute the derivative of each function using appropriate differentiation rules.

(i) $f(x) = \sin 2x$

Given: $f(x) = \sin 2x$.

To Compute: $f'(x)$.

Solution: Use the chain rule. The function is a composite function $f(x) = g(h(x))$, where the outer function is sine, $g(u) = \sin u$, and the inner function is $h(x) = 2x$. The chain rule states $f'(x) = g'(h(x)) \times h'(x)$.

Find the derivative of the outer function $g(u)$ with respect to $u$:

$g'(u) = \frac{d}{du}(\sin u) = \cos u$

Find the derivative of the inner function $h(x)$ with respect to $x$:

$h'(x) = \frac{d}{dx}(2x) = 2 \cdot 1 x^{1-1} = 2x^0 = 2$

Apply the chain rule $f'(x) = g'(h(x)) \times h'(x)$: Substitute $h(x) = 2x$ into $g'(u) = \cos u$ to get $g'(h(x)) = \cos(2x)$.

$f'(x) = (\cos(2x)) \times (2)$

$\mathbf{f'(x) = 2 \cos(2x)}$

(ii) $g(x) = \cot x$

Given: $g(x) = \cot x$.

To Compute: $g'(x)$.

Solution: We can compute the derivative by writing $\cot x = \frac{\cos x}{\sin x}$ and using the quotient rule.

The quotient rule states that if $f(x) = \frac{u(x)}{v(x)}$, then $f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2}$.

Here, let $u(x) = \cos x$ and $v(x) = \sin x$.

Find the derivatives of $u(x)$ and $v(x)$:

$u'(x) = \frac{d}{dx}(\cos x) = -\sin x$

$v'(x) = \frac{d}{dx}(\sin x) = \cos x$

Apply the quotient rule formula:

$g'(x) = \frac{(-\sin x)(\sin x) - (\cos x)(\cos x)}{(\sin x)^2}$

$g'(x) = \frac{-\sin^2 x - \cos^2 x}{\sin^2 x}$

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, $\frac{1}{\sin^2 x} = \left(\frac{1}{\sin x}\right)^2 = \text{cosec}^2 x$.

$g'(x) = -\text{cosec}^2 x$

This is a standard derivative of $\cot x$.

The derivative is $\mathbf{g'(x) = -\text{cosec}^2 x}$.

We will find the derivative of each function using the quotient rule.

The quotient rule states that if $f(x) = \frac{u(x)}{v(x)}$, then $f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2}$.

(i) $f(x) = \frac{x^{5}\;-\;cos\; x}{\sin\; x}$

Given: $f(x) = \frac{x^{5} - \cos x}{\sin x}$, where $\sin x \neq 0$.

To Find: $f'(x)$.

Solution: Use the quotient rule. Let $u(x) = x^5 - \cos x$ and $v(x) = \sin x$.

Find the derivatives of $u(x)$ and $v(x)$:

$u'(x) = \frac{d}{dx}(x^5 - \cos x) = \frac{d}{dx}(x^5) - \frac{d}{dx}(\cos x) = 5x^{5-1} - (-\sin x) = 5x^4 + \sin x$

$v'(x) = \frac{d}{dx}(\sin x) = \cos x$

Apply the quotient rule formula:

$f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2}$

$f'(x) = \frac{(5x^4 + \sin x)(\sin x) - (x^5 - \cos x)(\cos x)}{(\sin x)^2}$

Expand the numerator:

Numerator $= (5x^4 \sin x + \sin^2 x) - (x^5 \cos x - \cos^2 x)$

Numerator $= 5x^4 \sin x + \sin^2 x - x^5 \cos x + \cos^2 x$

Group the $\sin^2 x + \cos^2 x$ terms:

Numerator $= 5x^4 \sin x - x^5 \cos x + (\sin^2 x + \cos^2 x)$

Using the identity $\sin^2 x + \cos^2 x = 1$:

Numerator $= 5x^4 \sin x - x^5 \cos x + 1$

Substitute the simplified numerator back into the fraction:

$f'(x) = \frac{5x^4 \sin x - x^5 \cos x + 1}{\sin^2 x}$

The derivative is $\mathbf{f'(x) = \frac{5x^4 \sin x - x^5 \cos x + 1}{\sin^2 x}}$.

(ii) $g(x) = \frac{x \;+\; cos\; x }{tan\; x}$

Given: $g(x) = \frac{x + \cos x}{\tan x}$, where $\tan x \neq 0$.

To Find: $g'(x)$.

Solution: Use the quotient rule. Let $u(x) = x + \cos x$ and $v(x) = \tan x$.

Find the derivatives of $u(x)$ and $v(x)$:

$u'(x) = \frac{d}{dx}(x + \cos x) = \frac{d}{dx}(x) + \frac{d}{dx}(\cos x) = 1 + (-\sin x) = 1 - \sin x$

$v'(x) = \frac{d}{dx}(\tan x) = \sec^2 x$

Apply the quotient rule formula:

$g'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2}$

$g'(x) = \frac{(1 - \sin x)(\tan x) - (x + \cos x)(\sec^2 x)}{(\tan x)^2}$

Substitute $\tan x = \frac{\sin x}{\cos x}$ and $\sec^2 x = \frac{1}{\cos^2 x}$:

Numerator $= (1 - \sin x)\left(\frac{\sin x}{\cos x}\right) - (x + \cos x)\left(\frac{1}{\cos^2 x}\right)$

Numerator $= \frac{\sin x}{\cos x} - \frac{\sin^2 x}{\cos x} - \frac{x}{\cos^2 x} - \frac{\cos x}{\cos^2 x}$

Numerator $= \frac{\sin x \cos x}{\cos^2 x} - \frac{\sin^2 x \cos x}{\cos^2 x} - \frac{x}{\cos^2 x} - \frac{1}{\cos x}$

Numerator $= \frac{\sin x \cos x - \sin^2 x \cos x - x - \cos x}{\cos^2 x}$

Denominator $= (\tan x)^2 = \left(\frac{\sin x}{\cos x}\right)^2 = \frac{\sin^2 x}{\cos^2 x}$

Now, divide the numerator by the denominator:

$g'(x) = \frac{\frac{\sin x \cos x - \sin^2 x \cos x - x - \cos x}{\cos^2 x}}{\frac{\sin^2 x}{\cos^2 x}}$

$g'(x) = \frac{\sin x \cos x - \sin^2 x \cos x - x - \cos x}{\cos^2 x} \times \frac{\cos^2 x}{\sin^2 x}$

$g'(x) = \frac{\sin x \cos x - \sin^2 x \cos x - x - \cos x}{\sin^2 x}$

This can be written in various forms. One form is by separating the terms:

$g'(x) = \frac{\sin x \cos x}{\sin^2 x} - \frac{\sin^2 x \cos x}{\sin^2 x} - \frac{x}{\sin^2 x} - \frac{\cos x}{\sin^2 x}$

$g'(x) = \frac{\cos x}{\sin x} - \cos x - x \cdot \frac{1}{\sin^2 x} - \frac{\cos x}{\sin x} \cdot \frac{1}{\sin x}$

$g'(x) = \cot x - \cos x - x \text{cosec}^2 x - \cot x \text{cosec } x$

The derivative is $\mathbf{g'(x) = \cot x - \cos x - x \text{cosec}^2 x - \cot x \text{cosec } x}$.

We will find the derivative of each function using the first principle, which is given by the limit definition:

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

(i) $f(x) = -x$

Given: $f(x) = -x$.

To Find: $f'(x)$ using the first principle.

Solution:

Find $f(x+h)$:

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

Find the difference $f(x+h) - f(x)$:

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

Now, divide by $h$:

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

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

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

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

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

The limit of a constant is the constant itself:

$\mathbf{f'(x) = -1}$

(ii) $f(x) = (-x)^{-1}$

Given: $f(x) = (-x)^{-1} = -\frac{1}{x}$, where $x \neq 0$.

To Find: $f'(x)$ using the first principle.

Solution:

Find $f(x+h)$:

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

Find the difference $f(x+h) - f(x)$:

$f(x+h) - f(x) = -\frac{1}{x+h} - \left(-\frac{1}{x}\right) = \frac{1}{x} - \frac{1}{x+h}$

Combine the fractions using a common denominator $x(x+h)$:

$f(x+h) - f(x) = \frac{1 \cdot (x+h)}{x(x+h)} - \frac{1 \cdot x}{x(x+h)} = \frac{x+h - x}{x(x+h)}$

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

Now, divide by $h$:

$\frac{f(x+h) - f(x)}{h} = \frac{\frac{h}{x(x+h)}}{h} = \frac{h}{x(x+h)} \times \frac{1}{h}$

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

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

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

$f'(x) = \lim\limits_{h \to 0} \frac{1}{x(x+h)}$

Substitute $h = 0$:

$f'(x) = \frac{1}{x(x+0)} = \frac{1}{x(x)} = \frac{1}{x^2}$

$\mathbf{f'(x) = \frac{1}{x^2}}$

(iii) $f(x) = \sin (x + 1)$

Given: $f(x) = \sin (x + 1)$.

To Find: $f'(x)$ using the first principle.

Solution:

Find $f(x+h)$:

$f(x+h) = \sin ((x+h) + 1) = \sin(x + 1 + h)$

Find the difference $f(x+h) - f(x)$:

$f(x+h) - f(x) = \sin(x + 1 + h) - \sin(x + 1)$

Use the sum-to-product identity: $\sin A - \sin B = 2 \cos \left(\frac{A+B}{2}\right) \sin \left(\frac{A-B}{2}\right)$.

Let $A = x + 1 + h$ and $B = x + 1$.

$\frac{A+B}{2} = \frac{(x+1+h) + (x+1)}{2} = \frac{2(x+1) + h}{2} = x+1 + \frac{h}{2}$

$\frac{A-B}{2} = \frac{(x+1+h) - (x+1)}{2} = \frac{h}{2}$

So, $f(x+h) - f(x) = 2 \cos\left(x+1 + \frac{h}{2}\right) \sin\left(\frac{h}{2}\right)$.

Now, divide by $h$:

$\frac{f(x+h) - f(x)}{h} = \frac{2 \cos\left(x+1 + \frac{h}{2}\right) \sin\left(\frac{h}{2}\right)}{h}$

Rearrange the terms:

$\frac{f(x+h) - f(x)}{h} = \cos\left(x+1 + \frac{h}{2}\right) \times \frac{2 \sin\left(\frac{h}{2}\right)}{h}$

Rewrite the second part as $\frac{\sin(\frac{h}{2})}{\frac{h}{2}}$:

$\frac{f(x+h) - f(x)}{h} = \cos\left(x+1 + \frac{h}{2}\right) \times \frac{\sin\left(\frac{h}{2}\right)}{\frac{h}{2}}$

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

$f'(x) = \lim\limits_{h \to 0} \left( \cos\left(x+1 + \frac{h}{2}\right) \times \frac{\sin\left(\frac{h}{2}\right)}{\frac{h}{2}} \right)$

Using the product rule for limits:

$f'(x) = \left( \lim\limits_{h \to 0} \cos\left(x+1 + \frac{h}{2}\right) \right) \times \left( \lim\limits_{h \to 0} \frac{\sin\left(\frac{h}{2}\right)}{\frac{h}{2}} \right)$

As $h \to 0$, $x+1 + \frac{h}{2} \to x+1$. Since cosine is continuous, $\lim\limits_{h \to 0} \cos\left(x+1 + \frac{h}{2}\right) = \cos(x+1)$.

The standard limit $\lim\limits_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$. Here $\theta = \frac{h}{2}$. As $h \to 0$, $\theta \to 0$. So, $\lim\limits_{h \to 0} \frac{\sin\left(\frac{h}{2}\right)}{\frac{h}{2}} = 1$.

Substitute the limit values:

$f'(x) = \cos(x+1) \times 1$

$\mathbf{f'(x) = \cos(x+1)}$

(iv) $f(x) = \cos \left(x - \frac{\pi}{8}\right)$

Given: $f(x) = \cos \left(x - \frac{\pi}{8}\right)$.

To Find: $f'(x)$ using the first principle.

Solution:

Find $f(x+h)$:

$f(x+h) = \cos \left((x+h) - \frac{\pi}{8}\right) = \cos \left(x - \frac{\pi}{8} + h\right)$

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 sum-to-product identity: $\cos A - \cos B = -2 \sin \left(\frac{A+B}{2}\right) \sin \left(\frac{A-B}{2}\right)$.

Let $A = x - \frac{\pi}{8} + h$ and $B = x - \frac{\pi}{8}$.

$\frac{A+B}{2} = \frac{\left(x - \frac{\pi}{8} + h\right) + \left(x - \frac{\pi}{8}\right)}{2} = \frac{2\left(x - \frac{\pi}{8}\right) + h}{2} = x - \frac{\pi}{8} + \frac{h}{2}$

$\frac{A-B}{2} = \frac{\left(x - \frac{\pi}{8} + h\right) - \left(x - \frac{\pi}{8}\right)}{2} = \frac{h}{2}$

So, $f(x+h) - f(x) = -2 \sin\left(x - \frac{\pi}{8} + \frac{h}{2}\right) \sin\left(\frac{h}{2}\right)$.

Now, divide by $h$:

$\frac{f(x+h) - f(x)}{h} = \frac{-2 \sin\left(x - \frac{\pi}{8} + \frac{h}{2}\right) \sin\left(\frac{h}{2}\right)}{h}$

Rearrange the terms:

$\frac{f(x+h) - f(x)}{h} = -\sin\left(x - \frac{\pi}{8} + \frac{h}{2}\right) \times \frac{2 \sin\left(\frac{h}{2}\right)}{h}$

Rewrite the second part as $\frac{\sin(\frac{h}{2})}{\frac{h}{2}}$:

$\frac{f(x+h) - f(x)}{h} = -\sin\left(x - \frac{\pi}{8} + \frac{h}{2}\right) \times \frac{\sin\left(\frac{h}{2}\right)}{\frac{h}{2}}$

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

$f'(x) = \lim\limits_{h \to 0} \left( -\sin\left(x - \frac{\pi}{8} + \frac{h}{2}\right) \times \frac{\sin\left(\frac{h}{2}\right)}{\frac{h}{2}} \right)$

Using the product rule for limits:

$f'(x) = \left( \lim\limits_{h \to 0} -\sin\left(x - \frac{\pi}{8} + \frac{h}{2}\right) \right) \times \left( \lim\limits_{h \to 0} \frac{\sin\left(\frac{h}{2}\right)}{\frac{h}{2}} \right)$

As $h \to 0$, $x - \frac{\pi}{8} + \frac{h}{2} \to x - \frac{\pi}{8}$. Since sine is continuous, $\lim\limits_{h \to 0} -\sin\left(x - \frac{\pi}{8} + \frac{h}{2}\right) = -\sin\left(x - \frac{\pi}{8}\right)$.

The standard limit $\lim\limits_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$. Here $\theta = \frac{h}{2}$. As $h \to 0$, $\theta \to 0$. So, $\lim\limits_{h \to 0} \frac{\sin\left(\frac{h}{2}\right)}{\frac{h}{2}} = 1$.

Substitute the limit values:

$f'(x) = -\sin\left(x - \frac{\pi}{8}\right) \times 1$

$\mathbf{f'(x) = -\sin\left(x - \frac{\pi}{8}\right)}$

Given:

The function $f(x) = x + a$, where $a$ is a fixed non-zero constant.

To Find:

The derivative of $f(x)$ from the first principle.

Solution:

We use the definition of the derivative from the first principle:

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

First, find $f(x+h)$ by substituting $(x+h)$ for $x$ in the function $f(x) = x + a$:

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

Next, find the difference $f(x+h) - f(x)$:

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

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

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

Now, substitute this difference into the limit definition and divide by $h$:

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

Since we are evaluating the limit as $h \to 0$, $h$ is approaching $0$ but is not equal to $0$. Therefore, we can cancel the $h$ terms:

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

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

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

The limit of a constant as $h$ approaches any value is the constant itself:

$f'(x) = 1$

Therefore, the derivative of the function $f(x) = x + a$ from the first principle is $1$.

The derivative is $f'(x) = 1$.

Given:

The function $f(x) = (px + q) \left( \frac{r}{x} + s\right)$, where $p, q, r, s$ are fixed non-zero constants.

To Find:

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

Solution:

We can find the derivative by first expanding the function and then differentiating term by term using the sum rule, constant multiple rule, and power rule.

Expand the given expression:

$f(x) = (px + q) \left( \frac{r}{x} + s\right)$

$f(x) = px \left(\frac{r}{x}\right) + px(s) + q\left(\frac{r}{x}\right) + q(s)$

$f(x) = \frac{prx}{x} + psx + \frac{qr}{x} + qs$

Assuming $x \neq 0$, we can simplify $\frac{prx}{x} = pr$. Also, rewrite $\frac{qr}{x}$ as $qrx^{-1}$.

$f(x) = pr + psx + qrx^{-1} + qs$

Now, differentiate each term with respect to $x$:

$f'(x) = \frac{d}{dx}(pr + psx + qrx^{-1} + qs)$

Using the sum rule, constant multiple rule, and power rule:

$f'(x) = \frac{d}{dx}(pr) + \frac{d}{dx}(psx) + \frac{d}{dx}(qrx^{-1}) + \frac{d}{dx}(qs)$

• $\frac{d}{dx}(pr)$: Since $p$ and $r$ are constants, $pr$ is a constant. The derivative of a constant is $0$. $\frac{d}{dx}(pr) = 0$.
• $\frac{d}{dx}(psx)$: Since $p$ and $s$ are constants, $ps$ is a constant coefficient. Using the power rule $\frac{d}{dx}(cx^n) = cnx^{n-1}$ with $c=ps$ and $n=1$: $\frac{d}{dx}(psx) = ps \cdot 1 \cdot x^{1-1} = psx^0 = ps \cdot 1 = ps$.
• $\frac{d}{dx}(qrx^{-1})$: Since $q$ and $r$ are constants, $qr$ is a constant coefficient. Using the power rule $\frac{d}{dx}(cx^n) = cnx^{n-1}$ with $c=qr$ and $n=-1$: $\frac{d}{dx}(qrx^{-1}) = qr \cdot (-1)x^{-1-1} = -qrx^{-2} = -\frac{qr}{x^2}$.
• $\frac{d}{dx}(qs)$: Since $q$ and $s$ are constants, $qs$ is a constant. The derivative of a constant is $0$. $\frac{d}{dx}(qs) = 0$.

Combine the derivatives of the individual terms:

$f'(x) = 0 + ps + \left(-\frac{qr}{x^2}\right) + 0$

$f'(x) = ps - \frac{qr}{x^2}$

Therefore, the derivative of the function $f(x) = (px + q) \left( \frac{r}{x} + s\right)$ is $ps - \frac{qr}{x^2}$.

The derivative is $f'(x) = ps - \frac{qr}{x^2}$.

Given:

The function $f(x) = (ax + b) (cx + d)^2$, where $a, b, c, d$ are fixed non-zero constants.

To Find:

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

Solution:

We can find the derivative using the product rule in combination with the chain rule and power rule.

Let $u(x) = ax + b$ and $v(x) = (cx + d)^2$.

The product rule states that if $f(x) = u(x)v(x)$, then $f'(x) = u'(x)v(x) + u(x)v'(x)$.

Find the derivative of $u(x)$:

$u'(x) = \frac{d}{dx}(ax + b) = \frac{d}{dx}(ax) + \frac{d}{dx}(b) = a \cdot 1 x^{1-1} + 0 = a$

Find the derivative of $v(x) = (cx + d)^2$. Use the chain rule and power rule. The outer function is squaring, $g(y) = y^2$, and the inner function is $h(x) = cx + d$. The chain rule states $v'(x) = g'(h(x)) \times h'(x)$.

$g'(y) = \frac{d}{dy}(y^2) = 2y^{2-1} = 2y$

$h'(x) = \frac{d}{dx}(cx + d) = \frac{d}{dx}(cx) + \frac{d}{dx}(d) = c \cdot 1 x^{1-1} + 0 = c$

So, $v'(x) = g'(h(x)) \times h'(x) = (2(cx + d)) \times (c) = 2c(cx + d)$.

Now, apply the product rule formula $f'(x) = u'(x)v(x) + u(x)v'(x)$:

$f'(x) = (a) \cdot (cx + d)^2 + (ax + b) \cdot (2c(cx + d))$

Factor out the common term $(cx+d)$:

$f'(x) = (cx + d) [a(cx + d) + (ax + b)(2c)]$

Expand the terms inside the square brackets:

$f'(x) = (cx + d) [acx + ad + 2acx + 2bc]$

Combine like terms inside the square brackets:

$f'(x) = (cx + d) [(acx + 2acx) + (ad + 2bc)]$

$f'(x) = (cx + d) [3acx + ad + 2bc]$

Alternatively, we could expand the function first, although this is usually more work for higher powers.

$f(x) = (ax + b)(c^2x^2 + 2cdx + d^2)$

$f(x) = ax(c^2x^2 + 2cdx + d^2) + b(c^2x^2 + 2cdx + d^2)$

$f(x) = ac^2x^3 + 2acd x^2 + ad^2 x + bc^2x^2 + 2bcd x + bd^2$

$f(x) = ac^2x^3 + (2acd + bc^2)x^2 + (ad^2 + 2bcd)x + bd^2$

Now, differentiate term by term using the power rule:

$f'(x) = \frac{d}{dx}(ac^2x^3) + \frac{d}{dx}((2acd + bc^2)x^2) + \frac{d}{dx}((ad^2 + 2bcd)x) + \frac{d}{dx}(bd^2)$

$f'(x) = ac^2(3x^2) + (2acd + bc^2)(2x) + (ad^2 + 2bcd)(1) + 0$

$f'(x) = 3ac^2x^2 + (4acd + 2bc^2)x + ad^2 + 2bcd$

To check if this matches the factored form $(cx + d)(3acx + ad + 2bc)$, expand the factored form:

$(cx + d)(3acx + ad + 2bc) = cx(3acx + ad + 2bc) + d(3acx + ad + 2bc)$

$= 3ac^2x^2 + acdx + 2bc^2x + 3acdx + ad^2 + 2bcd$

$= 3ac^2x^2 + (acd + 3acd)x + 2bc^2x + ad^2 + 2bcd$

$= 3ac^2x^2 + 4acdx + 2bc^2x + ad^2 + 2bcd$

$= 3ac^2x^2 + (4acd + 2bc^2)x + ad^2 + 2bcd$

The results match.

The derivative is $f'(x) = (cx + d) (3acx + ad + 2bc)$ or $f'(x) = 3ac^2x^2 + (4acd + 2bc^2)x + ad^2 + 2bcd$.

Given:

The function $f(x) = \frac{ax + b}{cx + d}$, where $a, b, c, d$ are fixed non-zero constants and $cx + d \neq 0$.

To Find:

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

Solution:

We can find the derivative using the quotient rule.

The quotient rule 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 case, let $u(x) = ax + b$ and $v(x) = cx + d$.

Find the derivatives of $u(x)$ and $v(x)$:

$u'(x) = \frac{d}{dx}(ax + b) = \frac{d}{dx}(ax) + \frac{d}{dx}(b) = a \cdot 1 x^{1-1} + 0 = a$

$v'(x) = \frac{d}{dx}(cx + d) = \frac{d}{dx}(cx) + \frac{d}{dx}(d) = c \cdot 1 x^{1-1} + 0 = c$

Now, apply the quotient rule formula:

$f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2}$

$f'(x) = \frac{(a)(cx + d) - (ax + b)(c)}{(cx + d)^2}$

Expand the numerator:

$f'(x) = \frac{acx + ad - (acx + bc)}{(cx + d)^2}$

$f'(x) = \frac{acx + ad - acx - bc}{(cx + d)^2}$

Combine like terms in the numerator:

$f'(x) = \frac{(acx - acx) + (ad - bc)}{(cx + d)^2}$

$f'(x) = \frac{ad - bc}{(cx + d)^2}$

Therefore, the derivative of the function $f(x) = \frac{ax + b}{cx + d}$ is $\frac{ad - bc}{(cx + d)^2}$.

The derivative is $f'(x) = \frac{ad - bc}{(cx + d)^2}$.

Given:

The function $f(x) = \frac{1 + \frac{1}{x}}{1 - \frac{1}{x}}$.

To Find:

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

Solution:

First, we can simplify the function $f(x)$ by finding a common denominator in the numerator and the denominator of the main fraction.

Numerator: $1 + \frac{1}{x} = \frac{x}{x} + \frac{1}{x} = \frac{x+1}{x}$

Denominator: $1 - \frac{1}{x} = \frac{x}{x} - \frac{1}{x} = \frac{x-1}{x}$

So the function becomes:

$f(x) = \frac{\frac{x+1}{x}}{\frac{x-1}{x}}$

Assuming $x \neq 0$ and $x-1 \neq 0$ (so $x \neq 1$), we can simplify by multiplying the numerator by the reciprocal of the denominator:

$f(x) = \frac{x+1}{x} \times \frac{x}{x-1} = \frac{x+1}{x-1}$

Now we need to find the derivative of the simplified function $f(x) = \frac{x+1}{x-1}$. We will use the quotient rule.

The quotient rule 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 case, let $u(x) = x + 1$ and $v(x) = x - 1$.

Find the derivatives of $u(x)$ and $v(x)$:

$u'(x) = \frac{d}{dx}(x + 1) = \frac{d}{dx}(x) + \frac{d}{dx}(1) = 1 + 0 = 1$

$v'(x) = \frac{d}{dx}(x - 1) = \frac{d}{dx}(x) - \frac{d}{dx}(1) = 1 - 0 = 1$

Now, apply the quotient rule formula:

$f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2}$

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

Expand the numerator:

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

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

Combine like terms in the numerator:

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

$f'(x) = \frac{0 - 2}{(x - 1)^2}$

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

Therefore, the derivative of the function $f(x) = \frac{1 + \frac{1}{x}}{1 - \frac{1}{x}}$ is $-\frac{2}{(x - 1)^2}$.

The derivative is $f'(x) = -\frac{2}{(x - 1)^2}$.

Given:

The function $f(x) = \frac{1}{ax^{2} + bx + c}$, where $a, b, c$ are fixed non-zero constants and $ax^2 + bx + c \neq 0$.

To Find:

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

Solution:

We can find the derivative of $f(x)$ using the quotient rule.

The quotient rule states that if $f(x) = \frac{u(x)}{v(x)}$, then the derivative is $f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2}$.

In this case, let $u(x) = 1$ and $v(x) = ax^2 + bx + c$.

Find the derivatives of $u(x)$ and $v(x)$:

$u'(x) = \frac{d}{dx}(1)$

The derivative of a constant is $0$.

$u'(x) = 0$

$v'(x) = \frac{d}{dx}(ax^2 + bx + c)$

Using the sum rule, constant multiple rule, and power rule:

$v'(x) = \frac{d}{dx}(ax^2) + \frac{d}{dx}(bx) + \frac{d}{dx}(c)$

$v'(x) = a \cdot 2x^{2-1} + b \cdot 1x^{1-1} + 0$

$v'(x) = 2ax + b$

Now, apply the quotient rule formula:

$f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2}$

$f'(x) = \frac{(0)(ax^2 + bx + c) - (1)(2ax + b)}{(ax^2 + bx + c)^2}$

Simplify the numerator:

$f'(x) = \frac{0 - (2ax + b)}{(ax^2 + bx + c)^2}$

$f'(x) = \frac{-(2ax + b)}{(ax^2 + bx + c)^2}$

$\mathbf{f'(x) = -\frac{2ax + b}{(ax^2 + bx + c)^2}}$

Alternatively, we can rewrite the function as $f(x) = (ax^2 + bx + c)^{-1}$ and use the chain rule.

Let $g(u) = u^{-1}$ and $h(x) = ax^2 + bx + c$. Then $f(x) = g(h(x))$.

$g'(u) = -1u^{-2} = -\frac{1}{u^2}$

$h'(x) = 2ax + b$

By the chain rule, $f'(x) = g'(h(x)) \times h'(x)$:

$f'(x) = -\frac{1}{(ax^2 + bx + c)^2} \times (2ax + b)$

$f'(x) = -\frac{2ax + b}{(ax^2 + bx + c)^2}$

Both methods give the same result.

The derivative is $f'(x) = -\frac{2ax + b}{(ax^2 + bx + c)^2}$.

Given:

The function $f(x) = \frac{ax + b}{px^{2} + qx + r}$, where $a, b, p, q, r$ are fixed non-zero constants and $px^2 + qx + r \neq 0$.

To Find:

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

Solution:

We will find the derivative using the quotient rule.

The quotient rule 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 case, let $u(x) = ax + b$ and $v(x) = px^2 + qx + r$.

Find the derivatives of $u(x)$ and $v(x)$:

$u'(x) = \frac{d}{dx}(ax + b) = \frac{d}{dx}(ax) + \frac{d}{dx}(b) = a \cdot 1 x^{1-1} + 0 = a$

$v'(x) = \frac{d}{dx}(px^2 + qx + r) = \frac{d}{dx}(px^2) + \frac{d}{dx}(qx) + \frac{d}{dx}(r)$

$v'(x) = p \cdot 2x^{2-1} + q \cdot 1x^{1-1} + 0 = 2px + q$

Now, apply the quotient rule formula:

$f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2}$

$f'(x) = \frac{(a)(px^2 + qx + r) - (ax + b)(2px + q)}{(px^2 + qx + r)^2}$

Expand the numerator:

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 fraction:

$f'(x) = \frac{-apx^2 - 2bpx + ar - bq}{(px^2 + qx + r)^2}$

The derivative is $\mathbf{f'(x) = \frac{-apx^2 - 2bpx + ar - bq}{(px^2 + qx + r)^2}}$.

Given:

The function $f(x) = \frac{px^{2} + qx + r}{ax + b}$, where $a, b, p, q, r$ are fixed non-zero constants and $ax + b \neq 0$.

To Find:

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

Solution:

We will find the derivative using the quotient rule.

The quotient rule 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 case, let $u(x) = px^2 + qx + r$ and $v(x) = ax + b$.

Find the derivative of the numerator, $u'(x)$:

$u'(x) = \frac{d}{dx}(px^2 + qx + r) = \frac{d}{dx}(px^2) + \frac{d}{dx}(qx) + \frac{d}{dx}(r)$

$u'(x) = p \cdot 2x^{2-1} + q \cdot 1x^{1-1} + 0 = 2px + q$

Find the derivative of the denominator, $v'(x)$:

$v'(x) = \frac{d}{dx}(ax + b) = \frac{d}{dx}(ax) + \frac{d}{dx}(b)$

$v'(x) = a \cdot 1x^{1-1} + 0 = a$

Now, apply the quotient rule formula:

$f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2}$

$f'(x) = \frac{(2px + q)(ax + b) - (px^2 + qx + r)(a)}{(ax + b)^2}$

Expand the terms in the numerator:

Numerator $= (2px)(ax) + (2px)(b) + (q)(ax) + (q)(b) - [(px^2)(a) + (qx)(a) + (r)(a)]$

Numerator $= (2apx^2 + 2pbx + aqx + bq) - (apx^2 + aqx + ar)$

Numerator $= 2apx^2 + 2pbx + aqx + bq - apx^2 - aqx - ar$

Combine like terms in the numerator:

Numerator $= (2apx^2 - apx^2) + (2pbx + aqx - aqx) + (bq - ar)$

Numerator $= apx^2 + 2pbx + bq - ar$

Substitute the simplified numerator back into the fraction:

$f'(x) = \frac{apx^2 + 2pbx + bq - ar}{(ax + b)^2}$

Therefore, the derivative of the function $f(x) = \frac{px^{2} + qx + r}{ax + b}$ is $\frac{apx^2 + 2pbx + bq - ar}{(ax + b)^2}$.

The derivative is $f'(x) = \frac{apx^2 + 2pbx + bq - ar}{(ax + b)^2}$.

Given:

The function $f(x) = \frac{a}{x^{4}} - \frac{b}{x^{2}} + \cos x$, where $a, b$ are fixed non-zero constants.

To Find:

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

Solution:

We can rewrite the terms involving $x$ using negative exponents: $f(x) = ax^{-4} - bx^{-2} + \cos x$.

Now, we can find the derivative using the difference rule, sum rule, constant multiple rule, power rule, and the derivative of cosine.

$f'(x) = \frac{d}{dx}(ax^{-4} - bx^{-2} + \cos x)$

$f'(x) = \frac{d}{dx}(ax^{-4}) - \frac{d}{dx}(bx^{-2}) + \frac{d}{dx}(\cos x)$

Using the constant multiple rule and power rule $\frac{d}{dx}(cx^n) = cnx^{n-1}$:

$\frac{d}{dx}(ax^{-4}) = a \cdot (-4)x^{-4-1} = -4ax^{-5}$

$\frac{d}{dx}(bx^{-2}) = b \cdot (-2)x^{-2-1} = -2bx^{-3}$

The derivative of $\cos x$ is $-\sin x$:

$\frac{d}{dx}(\cos x) = -\sin x$

Combine the derivatives of the individual terms:

$f'(x) = (-4ax^{-5}) - (-2bx^{-3}) + (-\sin x)$

$f'(x) = -4ax^{-5} + 2bx^{-3} - \sin x$

We can write the terms with negative exponents back in fraction form:

$f'(x) = -\frac{4a}{x^5} + \frac{2b}{x^3} - \sin x$

Therefore, the derivative of the function $f(x) = \frac{a}{x^{4}} - \frac{b}{x^{2}} + \cos x$ is $-\frac{4a}{x^5} + \frac{2b}{x^3} - \sin x$.

The derivative is $f'(x) = -\frac{4a}{x^5} + \frac{2b}{x^3} - \sin x$.

Given:

The function $f(x) = 4\sqrt{x} - 2$.

To Find:

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

Solution:

We can rewrite the square root term using a fractional exponent: $\sqrt{x} = x^{\frac{1}{2}}$.

So, the function becomes $f(x) = 4x^{\frac{1}{2}} - 2$.

Now, we can find the derivative using the difference rule, constant multiple rule, and power rule.

$f'(x) = \frac{d}{dx}(4x^{\frac{1}{2}} - 2)$

$f'(x) = \frac{d}{dx}(4x^{\frac{1}{2}}) - \frac{d}{dx}(2)$

Using the constant multiple rule and power rule $\frac{d}{dx}(cx^n) = cnx^{n-1}$:

$\frac{d}{dx}(4x^{\frac{1}{2}}) = 4 \cdot \frac{1}{2} x^{\frac{1}{2}-1} = 2x^{\frac{1}{2} - \frac{2}{2}} = 2x^{-\frac{1}{2}}$

The derivative of a constant (2) is $0$:

$\frac{d}{dx}(2) = 0$

Combine the derivatives of the individual terms:

$f'(x) = 2x^{-\frac{1}{2}} - 0$

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

We can write the term with the negative exponent back in radical form: $x^{-\frac{1}{2}} = \frac{1}{x^{\frac{1}{2}}} = \frac{1}{\sqrt{x}}$.

$f'(x) = 2 \cdot \frac{1}{\sqrt{x}} = \frac{2}{\sqrt{x}}$

Therefore, the derivative of the function $f(x) = 4\sqrt{x} - 2$ is $\frac{2}{\sqrt{x}}$.

The derivative is $f'(x) = \frac{2}{\sqrt{x}}$.

Given:

The function $f(x) = (ax + b)^n$, where $a$ and $b$ are fixed non-zero constants and $n$ is an integer.

To Find:

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

Solution:

We will find the derivative of $f(x)$ using the chain rule.

The chain rule is used to differentiate composite functions. If $f(x) = g(h(x))$, then the derivative is $f'(x) = g'(h(x)) \times h'(x)$.

In this case, the function $f(x) = (ax + b)^n$ can be viewed as a composite function:

• Outer function $g(u) = u^n$
• Inner function $h(x) = ax + b$

First, find the derivative of the outer function $g(u)$ with respect to $u$. Using the power rule $\frac{d}{du}(u^n) = nu^{n-1}$:

$g'(u) = \frac{d}{du}(u^n) = nu^{n-1}$

Next, find the derivative of the inner function $h(x)$ with respect to $x$. Using the sum rule, constant multiple rule, and power rule:

$h'(x) = \frac{d}{dx}(ax + b) = \frac{d}{dx}(ax) + \frac{d}{dx}(b)$

$h'(x) = a \cdot 1x^{1-1} + 0 = a$

Now, apply the chain rule formula $f'(x) = g'(h(x)) \times h'(x)$. Substitute $h(x) = ax + b$ into the expression for $g'(u)$:

$g'(h(x)) = n(ax + b)^{n-1}$

Multiply this by $h'(x) = a$:

$f'(x) = n(ax + b)^{n-1} \times a$

$f'(x) = an(ax + b)^{n-1}$

Therefore, the derivative of the function $f(x) = (ax + b)^n$ is $an(ax + b)^{n-1}$.

The derivative is $f'(x) = an(ax + b)^{n-1}$.

Given:

The function $f(x) = (ax + b)^n (cx + d)^m$, where $a, b, c, d$ are fixed non-zero constants and $m, n$ are integers.

To Find:

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

Solution:

We will use the product rule to find the derivative of this function, as it is a product of two functions.

The product rule states that if $f(x) = u(x)v(x)$, then $f'(x) = u'(x)v(x) + u(x)v'(x)$.

Let $u(x) = (ax + b)^n$ and $v(x) = (cx + d)^m$.

We need to find the derivatives $u'(x)$ and $v'(x)$ using the chain rule (as in Question 12).

Find the derivative of $u(x) = (ax + b)^n$. Using the chain rule with outer function $g(y) = y^n$ and inner function $h(x) = ax + b$, we get $u'(x) = n(ax + b)^{n-1} \cdot \frac{d}{dx}(ax + b) = n(ax + b)^{n-1} \cdot a = an(ax + b)^{n-1}$.

Find the derivative of $v(x) = (cx + d)^m$. Using the chain rule with outer function $p(z) = z^m$ and inner function $q(x) = cx + d$, we get $v'(x) = m(cx + d)^{m-1} \cdot \frac{d}{dx}(cx + d) = m(cx + d)^{m-1} \cdot c = cm(cx + d)^{m-1}$.

Now, apply the product rule formula $f'(x) = u'(x)v(x) + u(x)v'(x)$:

$f'(x) = (an(ax + b)^{n-1}) \cdot ((cx + d)^m) + ((ax + b)^n) \cdot (cm(cx + d)^{m-1})$

$f'(x) = an(ax + b)^{n-1} (cx + d)^m + cm(ax + b)^n (cx + d)^{m-1}$

We can factor out common terms from both parts of the sum. The common terms are $(ax+b)^{n-1}$ and $(cx+d)^{m-1}$.

$f'(x) = (ax + b)^{n-1} (cx + d)^{m-1} [an (cx + d)^{m - (m-1)} + cm (ax + b)^{n - (n-1)}]$

$f'(x) = (ax + b)^{n-1} (cx + d)^{m-1} [an (cx + d)^{1} + cm (ax + b)^{1}]$

$f'(x) = (ax + b)^{n-1} (cx + d)^{m-1} [an(cx + d) + cm(ax + b)]$

Expand the terms inside the square brackets:

$f'(x) = (ax + b)^{n-1} (cx + d)^{m-1} [ancx + and + acmx + bcm]$

Combine the terms with $x$ inside the square brackets:

$f'(x) = (ax + b)^{n-1} (cx + d)^{m-1} [(anc + acm)x + (and + bcm)]$

$f'(x) = (ax + b)^{n-1} (cx + d)^{m-1} [ac(n + m)x + (and + bcm)]$

Therefore, the derivative of the function $f(x) = (ax + b)^n (cx + d)^m$ is $(ax + b)^{n-1} (cx + d)^{m-1} [ac(n + m)x + adn + bcm]$.

The derivative is $f'(x) = (ax + b)^{n-1} (cx + d)^{m-1} [ac(n + m)x + adn + bcm]$.

Given:

The function $f(x) = \sin(x + a)$, where $a$ is a fixed non-zero constant.

To Find:

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

Solution:

We will find the derivative of $f(x)$ using the chain rule.

The chain rule is used to differentiate composite functions. If $f(x) = g(h(x))$, then the derivative is $f'(x) = g'(h(x)) \times h'(x)$.

In this case, the function $f(x) = \sin(x + a)$ can be viewed as a composite function:

• Outer function $g(u) = \sin u$
• Inner function $h(x) = x + a$

First, find the derivative of the outer function $g(u)$ with respect to $u$. The derivative of $\sin u$ is $\cos u$:

$g'(u) = \frac{d}{du}(\sin u) = \cos u$

Next, find the derivative of the inner function $h(x)$ with respect to $x$. Using the sum rule and the derivative of a constant:

$h'(x) = \frac{d}{dx}(x + a) = \frac{d}{dx}(x) + \frac{d}{dx}(a)$

Using the power rule $\frac{d}{dx}(x^1) = 1x^0 = 1$ and the derivative of a constant $\frac{d}{dx}(a) = 0$:

$h'(x) = 1 + 0 = 1$

Now, apply the chain rule formula $f'(x) = g'(h(x)) \times h'(x)$. Substitute $h(x) = x + a$ into the expression for $g'(u)$:

$g'(h(x)) = \cos(x + a)$

Multiply this by $h'(x) = 1$:

$f'(x) = \cos(x + a) \times 1$

$f'(x) = \cos(x + a)$

Therefore, the derivative of the function $f(x) = \sin(x + a)$ is $\cos(x + a)$.

The derivative is $f'(x) = \cos(x + a)$.

Given:

The function $f(x) = \text{cosec } x \cot x$.

To Find:

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

Solution:

We will use the product rule to find the derivative of this function.

The product rule states that if $f(x) = u(x)v(x)$, then $f'(x) = u'(x)v(x) + u(x)v'(x)$.

Let $u(x) = \text{cosec } x$ and $v(x) = \cot x$.

We need to find the derivatives $u'(x)$ and $v'(x)$. Recall the standard derivatives:

• $\frac{d}{dx}(\text{cosec } x) = -\text{cosec } x \cot x$
• $\frac{d}{dx}(\cot x) = -\text{cosec}^2 x$

So, $u'(x) = -\text{cosec } x \cot x$ and $v'(x) = -\text{cosec}^2 x$.

Now, apply the product rule formula $f'(x) = u'(x)v(x) + u(x)v'(x)$:

$f'(x) = (-\text{cosec } x \cot x) (\cot x) + (\text{cosec } x) (-\text{cosec}^2 x)$

$f'(x) = -\text{cosec } x \cot^2 x - \text{cosec}^3 x$

We can factor out $-\text{cosec } x$ from both terms:

$f'(x) = -\text{cosec } x (\cot^2 x + \text{cosec}^2 x)$

Using the trigonometric identity $\cot^2 x + 1 = \text{cosec}^2 x$, we have $\cot^2 x = \text{cosec}^2 x - 1$.

Substitute this into the expression inside the parentheses:

$f'(x) = -\text{cosec } x ((\text{cosec}^2 x - 1) + \text{cosec}^2 x)$

$f'(x) = -\text{cosec } x (2 \text{cosec}^2 x - 1)$

Another way to express the result is by keeping the original form:

$f'(x) = -\text{cosec } x \cot^2 x - \text{cosec}^3 x$

Therefore, the derivative of the function $f(x) = \text{cosec } x \cot x$ is $-\text{cosec } x \cot^2 x - \text{cosec}^3 x$.

The derivative is $f'(x) = -\text{cosec } x (\cot^2 x + \text{cosec}^2 x)$ or $f'(x) = -\text{cosec } x (2 \text{cosec}^2 x - 1)$ or $f'(x) = -\text{cosec } x \cot^2 x - \text{cosec}^3 x$.

Given:

The function $f(x) = \frac{\cos x}{1 + \sin x}$, where $1 + \sin x \neq 0$.

To Find:

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

Solution:

We will use the quotient rule to find the derivative of this function.

The quotient rule 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 case, let $u(x) = \cos x$ and $v(x) = 1 + \sin x$.

Find the derivatives of $u(x)$ and $v(x)$:

$u'(x) = \frac{d}{dx}(\cos x) = -\sin x$

$v'(x) = \frac{d}{dx}(1 + \sin x) = \frac{d}{dx}(1) + \frac{d}{dx}(\sin x) = 0 + \cos x = \cos x$

Now, apply the quotient rule formula:

$f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2}$

$f'(x) = \frac{(-\sin x)(1 + \sin x) - (\cos x)(\cos x)}{(1 + \sin x)^2}$

Expand the numerator:

Numerator $= -\sin x(1) - \sin x(\sin x) - \cos x(\cos x)$

Numerator $= -\sin x - \sin^2 x - \cos^2 x$

Factor out $-1$ from the terms $\sin^2 x + \cos^2 x$:

Numerator $= -\sin x - (\sin^2 x + \cos^2 x)$

Using the fundamental trigonometric identity $\sin^2 x + \cos^2 x = 1$:

Numerator $= -\sin x - (1)$

Numerator $= -(\sin x + 1)$

Substitute the simplified numerator back into the fraction:

$f'(x) = \frac{-(\sin x + 1)}{(1 + \sin x)^2}$

Since $1 + \sin x = \sin x + 1$, we can simplify the fraction by cancelling one factor of $(\sin x + 1)$ (assuming $\sin x + 1 \neq 0$):

$f'(x) = \frac{-(\sin x + 1)}{(\sin x + 1)(\sin x + 1)} = -\frac{1}{\sin x + 1}$

Recall that $\text{cosec } x = \frac{1}{\sin x}$. The term $\frac{1}{\sin x + 1}$ doesn't simplify nicely into a single standard trigonometric function.

So, the simplified form is $-\frac{1}{1 + \sin x}$.

Therefore, the derivative of the function $f(x) = \frac{\cos x}{1 + \sin x}$ is $-\frac{1}{1 + \sin x}$.

The derivative is $f'(x) = -\frac{1}{1 + \sin x}$.

Given:

The function $f(x) = \frac{\sin x + \cos x}{\sin x - \cos x}$, where $\sin x - \cos x \neq 0$.

To Find:

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

Solution:

We will use the quotient rule to find the derivative of this function.

The quotient rule 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 case, let $u(x) = \sin x + \cos x$ and $v(x) = \sin x - \cos x$.

Find the derivative of the numerator, $u'(x)$:

$u'(x) = \frac{d}{dx}(\sin x + \cos x) = \frac{d}{dx}(\sin x) + \frac{d}{dx}(\cos x)$

$u'(x) = \cos x + (-\sin x) = \cos x - \sin x$

Find the derivative of the denominator, $v'(x)$:

$v'(x) = \frac{d}{dx}(\sin x - \cos x) = \frac{d}{dx}(\sin x) - \frac{d}{dx}(\cos x)$

$v'(x) = \cos x - (-\sin x) = \cos x + \sin x$

Now, apply the quotient rule formula:

$f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2}$

$f'(x) = \frac{(\cos x - \sin x)(\sin x - \cos x) - (\sin x + \cos x)(\cos x + \sin x)}{(\sin x - \cos x)^2}$

Expand the terms in the numerator:

Numerator $= (\cos x)(\sin x) + (\cos x)(-\cos x) + (-\sin x)(\sin x) + (-\sin x)(-\cos x) - [(\sin x)(\cos x) + (\sin x)(\sin x) + (\cos x)(\cos x) + (\cos x)(\sin x)]$

Numerator $= (\sin x \cos x - \cos^2 x - \sin^2 x + \sin x \cos x) - (\sin x \cos x + \sin^2 x + \cos^2 x + \sin x \cos x)$

Numerator $= (2 \sin x \cos x - (\cos^2 x + \sin^2 x)) - ((\sin^2 x + \cos^2 x) + 2 \sin x \cos x)$

Using the identity $\sin^2 x + \cos^2 x = 1$:

Numerator $= (2 \sin x \cos x - 1) - (1 + 2 \sin x \cos x)$

Numerator $= 2 \sin x \cos x - 1 - 1 - 2 \sin x \cos x$

Numerator $= (2 \sin x \cos x - 2 \sin x \cos x) + (-1 - 1)$

Numerator $= 0 - 2 = -2$

Substitute the simplified numerator back into the fraction:

$f'(x) = \frac{-2}{(\sin x - \cos x)^2}$

The denominator can also be written as $(\cos x - \sin x)^2$ since $(a-b)^2 = (b-a)^2$.

The derivative is $\mathbf{f'(x) = \frac{-2}{(\sin x - \cos x)^2}}$.

Given:

The function $f(x) = \frac{\sec x - 1}{\sec x + 1}$, where $\sec x + 1 \neq 0$.

To Find:

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

Solution:

We can simplify the function first before differentiating. Recall that $\sec x = \frac{1}{\cos x}$.

$f(x) = \frac{\frac{1}{\cos x} - 1}{\frac{1}{\cos x} + 1}$

Find a common denominator in the numerator and denominator:

$f(x) = \frac{\frac{1 - \cos x}{\cos x}}{\frac{1 + \cos x}{\cos x}}$

Assuming $\cos x \neq 0$, we can multiply the numerator by the reciprocal of the denominator:

$f(x) = \frac{1 - \cos x}{\cos x} \times \frac{\cos x}{1 + \cos x} = \frac{1 - \cos x}{1 + \cos x}$

Now we need to find the derivative of the simplified function $f(x) = \frac{1 - \cos x}{1 + \cos x}$ using the quotient rule.

The quotient rule 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 $u(x) = 1 - \cos x$ and $v(x) = 1 + \cos x$.

Find the derivative of the numerator, $u'(x)$:

$u'(x) = \frac{d}{dx}(1 - \cos x) = \frac{d}{dx}(1) - \frac{d}{dx}(\cos x) = 0 - (-\sin x) = \sin x$

Find the derivative of the denominator, $v'(x)$:

$v'(x) = \frac{d}{dx}(1 + \cos x) = \frac{d}{dx}(1) + \frac{d}{dx}(\cos x) = 0 + (-\sin x) = -\sin x$

Now, apply the quotient rule formula:

$f'(x) = \frac{(\sin x)(1 + \cos x) - (1 - \cos x)(-\sin x)}{(1 + \cos x)^2}$

Expand the terms in 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 like terms in the numerator:

Numerator $= (\sin x + \sin x) + (\sin x \cos x - \sin x \cos x)$

Numerator $= 2 \sin x + 0 = 2 \sin x$

Substitute the simplified numerator back into the fraction:

$f'(x) = \frac{2 \sin x}{(1 + \cos x)^2}$

Therefore, the derivative of the function $f(x) = \frac{\sec x - 1}{\sec x + 1}$ is $\frac{2 \sin x}{(1 + \cos x)^2}$.

The derivative is $f'(x) = \frac{2 \sin x}{(1 + \cos x)^2}$.

Given:

The function $f(x) = \sin^n x$, where $n$ is an integer.

To Find:

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

Solution:

We can rewrite the function as $f(x) = (\sin x)^n$. To find the derivative of this function, we need to use the chain rule.

The chain rule states that for a composite function $f(x) = g(h(x))$, the derivative is $f'(x) = g'(h(x)) \times h'(x)$.

In this function, $f(x) = (\sin x)^n$, we can identify the outer function $g(u) = u^n$ and the inner function $h(x) = \sin x$.

First, find the derivative of the outer function $g(u)$ with respect to $u$. Using the power rule $\frac{d}{du}(u^n) = nu^{n-1}$:

$g'(u) = \frac{d}{du}(u^n) = nu^{n-1}$

Next, find the derivative of the inner function $h(x)$ with respect to $x$:

$h'(x) = \frac{d}{dx}(\sin x)$

The derivative of $\sin x$ is $\cos x$:

$h'(x) = \cos x$

Now, apply the chain rule: $f'(x) = g'(h(x)) \times h'(x)$.

Substitute $h(x) = \sin x$ into $g'(u) = nu^{n-1}$ to get $g'(h(x)) = n(\sin x)^{n-1} = n\sin^{n-1} x$.

Multiply this by $h'(x) = \cos x$:

$f'(x) = (n\sin^{n-1} x) \times (\cos x)$

$f'(x) = n\sin^{n-1} x \cos x$

Therefore, the derivative of the function $f(x) = \sin^n x$ is $n\sin^{n-1} x \cos x$.

The derivative is $f'(x) = n\sin^{n-1} x \cos x$.

Given:

The function $f(x) = \frac{a + b \sin x}{c + d \sin x}$, where $a, b, c, d$ are fixed non-zero constants and $c + d \sin x \neq 0$.

To Find:

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

Solution:

We will use the quotient rule to find the derivative of this function.

The quotient rule 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 case, let $u(x) = a + b \sin x$ and $v(x) = c + d \sin x$.

Find the derivative of the numerator, $u'(x)$:

$u'(x) = \frac{d}{dx}(a + b \sin x) = \frac{d}{dx}(a) + \frac{d}{dx}(b \sin x)$

$u'(x) = 0 + b \frac{d}{dx}(\sin x) = b \cos x$

Find the derivative of the denominator, $v'(x)$:

$v'(x) = \frac{d}{dx}(c + d \sin x) = \frac{d}{dx}(c) + \frac{d}{dx}(d \sin x)$

$v'(x) = 0 + d \frac{d}{dx}(\sin x) = d \cos x$

Now, apply the quotient rule formula:

$f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2}$

$f'(x) = \frac{(b \cos x)(c + d \sin x) - (a + b \sin x)(d \cos x)}{(c + d \sin x)^2}$

Expand the terms in the numerator:

Numerator $= (b \cos x)(c) + (b \cos x)(d \sin x) - [(a)(d \cos x) + (b \sin x)(d \cos x)]$

Numerator $= bc \cos x + bd \sin x \cos x - [ad \cos x + bd \sin x \cos x]$

Numerator $= bc \cos x + bd \sin x \cos x - ad \cos x - bd \sin x \cos x$

Combine like terms in the numerator:

Numerator $= (bd \sin x \cos x - bd \sin x \cos x) + bc \cos x - ad \cos x$

Numerator $= 0 + (bc - ad) \cos x$

Numerator $= (bc - ad) \cos x$

Substitute the simplified numerator back into the fraction:

$f'(x) = \frac{(bc - ad) \cos x}{(c + d \sin x)^2}$

Therefore, the derivative of the function $f(x) = \frac{a + b \sin x}{c + d \sin x}$ is $\frac{(bc - ad) \cos x}{(c + d \sin x)^2}$.

The derivative is $f'(x) = \frac{(bc - ad) \cos x}{(c + d \sin x)^2}$.

Given:

The function $f(x) = \frac{\sin (x + a)}{\cos x}$, where $a$ is a fixed constant and $\cos x \neq 0$.

To Find:

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

Solution:

We will use the quotient rule to find the derivative of this function.

The quotient rule 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 case, let $u(x) = \sin(x + a)$ and $v(x) = \cos x$.

Find the derivative of the numerator, $u'(x)$. Use the chain rule for $u(x) = \sin(x+a)$. Let the outer function be $g(y) = \sin y$ and the inner function be $h(x) = x+a$.

$g'(y) = \frac{d}{dy}(\sin y) = \cos y$

$h'(x) = \frac{d}{dx}(x + a) = \frac{d}{dx}(x) + \frac{d}{dx}(a) = 1 + 0 = 1$

$u'(x) = g'(h(x)) \times h'(x) = \cos(x+a) \times 1 = \cos(x+a)$

Find the derivative of the denominator, $v'(x)$:

$v'(x) = \frac{d}{dx}(\cos x) = -\sin x$

Now, apply the quotient rule formula:

$f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2}$

$f'(x) = \frac{(\cos(x+a))(\cos x) - (\sin(x+a))(-\sin x)}{(\cos x)^2}$

Expand the numerator:

Numerator $= \cos(x+a)\cos x - (-\sin(x+a)\sin x)$

Numerator $= \cos(x+a)\cos x + \sin(x+a)\sin x$

Use the trigonometric identity for the cosine of a difference: $\cos(A - B) = \cos A \cos B + \sin A \sin B$. Let $A = x+a$ and $B = x$.

Numerator $= \cos((x+a) - x) = \cos(a)$

Substitute the simplified numerator back into the fraction:

$f'(x) = \frac{\cos a}{(\cos x)^2}$

$f'(x) = \frac{\cos a}{\cos^2 x}$

This can also be written as $f'(x) = \cos a \sec^2 x$, since $\frac{1}{\cos^2 x} = \sec^2 x$.

Therefore, the derivative of the function $f(x) = \frac{\sin (x + a)}{\cos x}$ is $\frac{\cos a}{\cos^2 x}$.

The derivative is $f'(x) = \frac{\cos a}{\cos^2 x} = \cos a \sec^2 x$.

Given:

The function $f(x) = x^4 (5 \sin x - 3 \cos x)$.

To Find:

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

Solution:

We will use the product rule to find the derivative of this function.

The product rule states that if $f(x) = u(x)v(x)$, then $f'(x) = u'(x)v(x) + u(x)v'(x)$.

In this case, let $u(x) = x^4$ and $v(x) = 5 \sin x - 3 \cos x$.

Find the derivative of $u(x)$ using the power rule:

$u'(x) = \frac{d}{dx}(x^4) = 4x^{4-1} = 4x^3$

Find the derivative of $v(x)$ using the difference rule, constant multiple rule, and derivatives of sine and cosine:

$v'(x) = \frac{d}{dx}(5 \sin x - 3 \cos x) = \frac{d}{dx}(5 \sin x) - \frac{d}{dx}(3 \cos x)$

$v'(x) = 5 \frac{d}{dx}(\sin x) - 3 \frac{d}{dx}(\cos x)$

$v'(x) = 5 (\cos x) - 3 (-\sin x)$

$v'(x) = 5 \cos x + 3 \sin x$

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:

$f'(x) = 4x^3 (5 \sin x) - 4x^3 (3 \cos x) + x^4 (5 \cos x) + x^4 (3 \sin x)$

$f'(x) = 20x^3 \sin x - 12x^3 \cos x + 5x^4 \cos x + 3x^4 \sin x$

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 common terms:

$f'(x) = (20x^3 + 3x^4) \sin x + (-12x^3 + 5x^4) \cos x$

$f'(x) = x^3 (20 + 3x) \sin x + x^3 (-12 + 5x) \cos x$

We can also factor out $x^3$ from the entire expression:

$f'(x) = x^3 [(20 + 3x) \sin x + (5x - 12) \cos x]$

Therefore, the derivative of the function $f(x) = x^4 (5 \sin x - 3 \cos x)$ is $x^3 [ (3x + 20) \sin x + (5x - 12) \cos x ]$.

The derivative is $f'(x) = x^3 [ (3x + 20) \sin x + (5x - 12) \cos x ]$.

Given:

The function $f(x) = (x^2 + 1) \cos x$.

To Find:

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

Solution:

We will use the product rule to find the derivative of this function, as it is a product of two functions.

The product rule states that if $f(x) = u(x)v(x)$, then the derivative is $f'(x) = u'(x)v(x) + u(x)v'(x)$.

In this case, let $u(x) = x^2 + 1$ and $v(x) = \cos x$.

First, find the derivative of $u(x)$: $u'(x) = \frac{d}{dx}(x^2 + 1)$.

Using the sum rule and the power rule:

$u'(x) = \frac{d}{dx}(x^2) + \frac{d}{dx}(1) = 2x^{2-1} + 0 = 2x$

Next, find the derivative of $v(x)$: $v'(x) = \frac{d}{dx}(\cos x)$.

The standard derivative of $\cos x$ is $-\sin x$:

$v'(x) = -\sin x$

Now, apply the product rule formula $f'(x) = u'(x)v(x) + u(x)v'(x)$:

$f'(x) = (2x)(\cos x) + (x^2 + 1)(-\sin x)$

Expand the second term:

$f'(x) = 2x \cos x - (x^2 \sin x + 1 \cdot \sin x)$

$f'(x) = 2x \cos x - x^2 \sin x - \sin x$

We can also factor out $\sin x$ from the last two terms:

$f'(x) = 2x \cos x - (x^2 + 1) \sin x$

Therefore, the derivative of the function $f(x) = (x^2 + 1) \cos x$ is $2x \cos x - (x^2 + 1) \sin x$.

The derivative is $f'(x) = 2x \cos x - (x^2 + 1) \sin x$.

Given:

The function $f(x) = (ax^2 + \sin x ) (p + q \cos x)$, where $a, p, q$ are fixed non-zero constants.

To Find:

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

Solution:

We will use the product rule to find the derivative of this function.

The product rule states that if $f(x) = u(x)v(x)$, then the derivative is $f'(x) = u'(x)v(x) + u(x)v'(x)$.

In this case, let $u(x) = ax^2 + \sin x$ and $v(x) = p + q \cos x$.

First, find the derivative of $u(x)$: $u'(x) = \frac{d}{dx}(ax^2 + \sin x)$.

Using the sum rule, constant multiple rule, and derivative of sine:

$u'(x) = \frac{d}{dx}(ax^2) + \frac{d}{dx}(\sin x) = a \cdot 2x^{2-1} + \cos x = 2ax + \cos x$

Next, find the derivative of $v(x)$: $v'(x) = \frac{d}{dx}(p + q \cos x)$.

Using the sum rule, derivative of a constant, constant multiple rule, and derivative of cosine:

$v'(x) = \frac{d}{dx}(p) + \frac{d}{dx}(q \cos x) = 0 + q \frac{d}{dx}(\cos x) = q (-\sin x) = -q \sin x$

Now, apply the product rule formula $f'(x) = u'(x)v(x) + u(x)v'(x)$:

$f'(x) = (2ax + \cos x)(p + q \cos x) + (ax^2 + \sin x)(-q \sin x)$

Expand the terms:

$f'(x) = (2ax)(p) + (2ax)(q \cos x) + (\cos x)(p) + (\cos x)(q \cos x) + (ax^2)(-q \sin x) + (\sin x)(-q \sin x)$

$f'(x) = 2apx + 2axq \cos x + p \cos x + q \cos^2 x - ax^2 q \sin x - q \sin^2 x$

Rearrange the terms and group those involving $q \cos^2 x$ and $-q \sin^2 x$:

$f'(x) = 2apx + 2aqx \cos x + p \cos x - ax^2 q \sin x + q (\cos^2 x - \sin^2 x)$

Using the identity $\cos(2x) = \cos^2 x - \sin^2 x$:

$f'(x) = 2apx + 2aqx \cos x + p \cos x - ax^2 q \sin x + q \cos(2x)$

We can also group terms with common factors:

$f'(x) = (2ap)x + (2aqx + p) \cos x - (aqx^2) \sin x + q \cos(2x)$

Therefore, the derivative of the function $f(x) = (ax^2 + \sin x ) (p + q \cos x)$ is $2apx + (2aqx + p) \cos x - aqx^2 \sin x + q \cos(2x)$.

The derivative is $f'(x) = 2apx + (2aqx + p) \cos x - aqx^2 \sin x + q \cos(2x)$.

Given:

The function $f(x) = (x + \cos x) (x - \tan x)$.

To Find:

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

Solution:

We will use the product rule to find the derivative of this function. The product rule states that if $f(x) = u(x)v(x)$, then the derivative is $f'(x) = u'(x)v(x) + u(x)v'(x)$.

In this case, let $u(x) = x + \cos x$ and $v(x) = x - \tan x$.

First, find the derivative of $u(x)$: $u'(x) = \frac{d}{dx}(x + \cos x)$.

Using the sum rule and the derivatives of $x$ and $\cos x$:

$u'(x) = \frac{d}{dx}(x) + \frac{d}{dx}(\cos x) = 1 + (-\sin x) = 1 - \sin x$

Next, find the derivative of $v(x)$: $v'(x) = \frac{d}{dx}(x - \tan x)$.

Using the difference rule and the derivatives of $x$ and $\tan x$:

$v'(x) = \frac{d}{dx}(x) - \frac{d}{dx}(\tan x) = 1 - \sec^2 x$

Recall the identity $1 + \tan^2 x = \sec^2 x$, so $1 - \sec^2 x = 1 - (1 + \tan^2 x) = -\tan^2 x$.

So, $v'(x) = -\tan^2 x$.

Now, apply the product rule formula $f'(x) = u'(x)v(x) + u(x)v'(x)$:

$f'(x) = (1 - \sin x)(x - \tan x) + (x + \cos x)(-\tan^2 x)$

Expand the terms:

$f'(x) = (1)(x) - (1)(\tan x) - (\sin x)(x) - (\sin x)(-\tan x) + (x)(-\tan^2 x) + (\cos x)(-\tan^2 x)$

$f'(x) = x - \tan x - x \sin x + \sin x \tan x - x \tan^2 x - \cos x \tan^2 x$

Substitute $\tan x = \frac{\sin x}{\cos x}$ and $\tan^2 x = \frac{\sin^2 x}{\cos^2 x}$ into the terms involving $\tan x$ and $\tan^2 x$:

$\sin x \tan x = \sin x \left(\frac{\sin x}{\cos x}\right) = \frac{\sin^2 x}{\cos x}$

$\cos x \tan^2 x = \cos x \left(\frac{\sin^2 x}{\cos^2 x}\right) = \frac{\sin^2 x}{\cos x}$

Substitute these back into the expression for $f'(x)$:

$f'(x) = x - \tan x - x \sin x + \frac{\sin^2 x}{\cos x} - x \tan^2 x - \frac{\sin^2 x}{\cos x}$

Observe that the terms $+\frac{\sin^2 x}{\cos x}$ and $-\frac{\sin^2 x}{\cos x}$ cancel each other out.

$f'(x) = x - \tan x - x \sin x - x \tan^2 x$

Therefore, the derivative of the function $f(x) = (x + \cos x) (x - \tan x)$ is $x - \tan x - x \sin x - x \tan^2 x$.

The derivative is $f'(x) = x - \tan x - x \sin x - x \tan^2 x$.

Given:

The function $f(x) = \frac{4x + 5 \sin x}{3x + 7 \cos x}$, where $3x + 7 \cos x \neq 0$.

To Find:

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

Solution:

We will use the quotient rule to find the derivative of this function.

The quotient rule 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 case, let $u(x) = 4x + 5 \sin x$ and $v(x) = 3x + 7 \cos x$.

Find the derivative of the numerator, $u'(x)$:

$u'(x) = \frac{d}{dx}(4x + 5 \sin x) = \frac{d}{dx}(4x) + \frac{d}{dx}(5 \sin x)$

$u'(x) = 4 \cdot 1 + 5 \cos x = 4 + 5 \cos x$

Find the derivative of the denominator, $v'(x)$:

$v'(x) = \frac{d}{dx}(3x + 7 \cos x) = \frac{d}{dx}(3x) + \frac{d}{dx}(7 \cos x)$

$v'(x) = 3 \cdot 1 + 7 (-\sin x) = 3 - 7 \sin x$

Now, apply the quotient rule formula:

$f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2}$

$f'(x) = \frac{(4 + 5 \cos x)(3x + 7 \cos x) - (4x + 5 \sin x)(3 - 7 \sin x)}{(3x + 7 \cos x)^2}$

Expand the terms in the numerator:

$(4 + 5 \cos x)(3x + 7 \cos x) = 4(3x) + 4(7 \cos x) + (5 \cos x)(3x) + (5 \cos x)(7 \cos x)$

$= 12x + 28 \cos x + 15x \cos x + 35 \cos^2 x$

$(4x + 5 \sin x)(3 - 7 \sin x) = 4x(3) + 4x(-7 \sin x) + (5 \sin x)(3) + (5 \sin x)(-7 \sin x)$

$= 12x - 28x \sin x + 15 \sin x - 35 \sin^2 x$

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 in the numerator:

Numerator $= (12x - 12x) + 28 \cos x + 15x \cos x + 28x \sin x - 15 \sin x + (35 \cos^2 x + 35 \sin^2 x)$

Numerator $= 0 + 28 \cos x + 15x \cos x + 28x \sin x - 15 \sin x + 35 (\cos^2 x + \sin^2 x)$

Using the identity $\cos^2 x + \sin^2 x = 1$:

Numerator $= 28 \cos x + 15x \cos x + 28x \sin x - 15 \sin x + 35(1)$

Numerator $= 35 + 28 \cos x + 15x \cos x + 28x \sin x - 15 \sin x$

Substitute the simplified numerator back into the fraction:

$f'(x) = \frac{35 + 28 \cos x + 15x \cos x + 28x \sin x - 15 \sin x}{(3x + 7 \cos x)^2}$

Therefore, the derivative of the function $f(x) = \frac{4x + 5 \sin x}{3x + 7 \cos x}$ is $\frac{35 + 28 \cos x + 15x \cos x + 28x \sin x - 15 \sin x}{(3x + 7 \cos x)^2}$.

The derivative is $f'(x) = \frac{35 + (28 + 15x) \cos x + (28x - 15) \sin x}{(3x + 7 \cos x)^2}$.

Given:

The function $f(x) = \frac{x^{2} \cos\left( \frac{\pi}{4} \right)}{\sin x}$, where $\sin x \neq 0$.

To Find:

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

Solution:

First, evaluate the constant term $\cos\left(\frac{\pi}{4}\right)$. We know that $\cos\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}$.

So, the function can be rewritten as $f(x) = \frac{x^2 \cdot \frac{1}{\sqrt{2}}}{\sin x} = \frac{1}{\sqrt{2}} \frac{x^2}{\sin x}$.

We can find the derivative using the constant multiple rule and the quotient rule.

$f'(x) = \frac{d}{dx}\left(\frac{1}{\sqrt{2}} \frac{x^2}{\sin x}\right)$

Using the constant multiple rule:

$f'(x) = \frac{1}{\sqrt{2}} \frac{d}{dx}\left(\frac{x^2}{\sin x}\right)$

Now, apply the quotient rule to $\frac{x^2}{\sin x}$. Let $u(x) = x^2$ and $v(x) = \sin x$.

Find the derivative of the numerator, $u'(x)$:

$u'(x) = \frac{d}{dx}(x^2) = 2x$

Find the derivative of the denominator, $v'(x)$:

$v'(x) = \frac{d}{dx}(\sin x) = \cos x$

Apply the quotient rule formula $\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2}$:

$\frac{d}{dx}\left(\frac{x^2}{\sin x}\right) = \frac{(2x)(\sin x) - (x^2)(\cos x)}{(\sin x)^2} = \frac{2x \sin x - x^2 \cos x}{\sin^2 x}$

Now, substitute this back into the expression for $f'(x)$:

$f'(x) = \frac{1}{\sqrt{2}} \left(\frac{2x \sin x - x^2 \cos x}{\sin^2 x}\right)$

$\mathbf{f'(x) = \frac{2x \sin x - x^2 \cos x}{\sqrt{2} \sin^2 x}}$

Therefore, the derivative of the function $f(x) = \frac{x^{2} \cos\left( \frac{\pi}{4} \right)}{\sin x}$ is $\frac{2x \sin x - x^2 \cos x}{\sqrt{2} \sin^2 x}$.

The derivative is $f'(x) = \frac{2x \sin x - x^2 \cos x}{\sqrt{2} \sin^2 x}$.

Given:

The function $f(x) = \frac{x}{1 + \tan x}$, where $1 + \tan x \neq 0$.

To Find:

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

Solution:

We will use the quotient rule to find the derivative of this function.

The quotient rule 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 case, let $u(x) = x$ and $v(x) = 1 + \tan x$.

Find the derivative of the numerator, $u'(x)$:

$u'(x) = \frac{d}{dx}(x) = 1$

Find the derivative of the denominator, $v'(x)$:

$v'(x) = \frac{d}{dx}(1 + \tan x) = \frac{d}{dx}(1) + \frac{d}{dx}(\tan x)$

$v'(x) = 0 + \sec^2 x = \sec^2 x$

Now, apply the quotient rule formula:

$f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2}$

$f'(x) = \frac{(1)(1 + \tan x) - (x)(\sec^2 x)}{(1 + \tan x)^2}$

Simplify the numerator:

$f'(x) = \frac{1 + \tan x - x \sec^2 x}{(1 + \tan x)^2}$

Therefore, the derivative of the function $f(x) = \frac{x}{1 + \tan x}$ is $\frac{1 + \tan x - x \sec^2 x}{(1 + \tan x)^2}$.

The derivative is $f'(x) = \frac{1 + \tan x - x \sec^2 x}{(1 + \tan x)^2}$.

Given:

The function $f(x) = (x + \sec x) (x - \tan x)$.

To Find:

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

Solution:

We will use the product rule to find the derivative of this function.

The product rule states that if $f(x) = u(x)v(x)$, then the derivative is $f'(x) = u'(x)v(x) + u(x)v'(x)$.

In this case, let $u(x) = x + \sec x$ and $v(x) = x - \tan x$.

First, find the derivative of $u(x)$: $u'(x) = \frac{d}{dx}(x + \sec x)$.

Using the sum rule and the derivatives of $x$ and $\sec x$:

$u'(x) = \frac{d}{dx}(x) + \frac{d}{dx}(\sec x) = 1 + \sec x \tan x$

Next, find the derivative of $v(x)$: $v'(x) = \frac{d}{dx}(x - \tan x)$.

Using the difference rule and the derivatives of $x$ and $\tan x$:

$v'(x) = \frac{d}{dx}(x) - \frac{d}{dx}(\tan x) = 1 - \sec^2 x$

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:

$f'(x) = 1(x - \tan x) + \sec x \tan x (x - \tan x) + x(1 - \sec^2 x) + \sec x (1 - \sec^2 x)$

$f'(x) = x - \tan x + x \sec x \tan x - \sec x \tan^2 x + x - x \sec^2 x + \sec x - \sec^3 x$

Combine the terms with $x$:

$f'(x) = (x + x) - \tan x + x \sec x \tan x - \sec x \tan^2 x - x \sec^2 x + \sec x - \sec^3 x$

$f'(x) = 2x - \tan x + x \sec x \tan x - \sec x \tan^2 x - x \sec^2 x + \sec x - \sec^3 x$

This expression can be rearranged, but often leaving it in terms of the individual functions is sufficient unless further simplification is requested or possible using identities.

Therefore, the derivative of the function $f(x) = (x + \sec x) (x - \tan x)$ is $2x - \tan x + x \sec x \tan x - \sec x \tan^2 x - x \sec^2 x + \sec x - \sec^3 x$.

The derivative is $f'(x) = 2x - \tan x + x \sec x \tan x - \sec x \tan^2 x - x \sec^2 x + \sec x - \sec^3 x$.

Given:

The function $f(x) = \frac{x}{\sin^{n}x}$, where $n$ is an integer and $\sin x \neq 0$.

To Find:

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

Solution:

We will use the quotient rule to find the derivative of this function.

The quotient rule 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 case, let $u(x) = x$ and $v(x) = \sin^{n} x = (\sin x)^n$.

Find the derivative of the numerator, $u'(x)$:

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

Using the power rule $\frac{d}{dx}(x^1) = 1x^{1-1} = x^0 = 1$:

$u'(x) = 1$

Find the derivative of the denominator, $v'(x)$. This requires the chain rule.

$v(x) = (\sin x)^n$. Let the outer function be $g(y) = y^n$ and the inner function be $h(x) = \sin x$.

The derivative of the outer function is $g'(y) = \frac{d}{dy}(y^n) = ny^{n-1}$.

The derivative of the inner function is $h'(x) = \frac{d}{dx}(\sin x) = \cos x$.

By the chain rule, $v'(x) = g'(h(x)) \times h'(x) = n(\sin x)^{n-1} \times \cos x = n \sin^{n-1} x \cos x$.

Now, apply the quotient rule formula $f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2}$:

$f'(x) = \frac{(1)(\sin^n x) - (x)(n \sin^{n-1} x \cos x)}{(\sin^n x)^2}$

Simplify the expression:

$f'(x) = \frac{\sin^n x - nx \sin^{n-1} x \cos x}{\sin^{2n} x}$

Factor out $\sin^{n-1} x$ from the numerator:

$f'(x) = \frac{\sin^{n-1} x (\sin^{n - (n-1)} x - nx \cos x)}{\sin^{2n} x}$

$f'(x) = \frac{\sin^{n-1} x (\sin x - nx \cos x)}{\sin^{2n} x}$

Cancel $\sin^{n-1} x$ from the numerator and denominator (assuming $\sin^{n-1} x \neq 0$):

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

So, the derivative is:

$f'(x) = \frac{\sin x - nx \cos x}{\sin^{n+1} x}$

Therefore, the derivative of the function $f(x) = \frac{x}{\sin^{n}x}$ is $\frac{\sin x - nx \cos x}{\sin^{n+1} x}$.

The derivative is $f'(x) = \frac{\sin x - nx \cos x}{\sin^{n+1} x}$.