Differentiation: Basic Rules and Standard Formulas

Derivative of a Constant and Power Rule

Building upon the limit definition of the derivative (the "first principle"), we can derive general rules for differentiating common types of functions. The most basic rules apply to constant functions and power functions.

1. Derivative of a Constant Function

Consider a function $f(x) = c$, where $c$ is a real number constant. The graph of this function is a horizontal line $y=c$. Intuitively, a horizontal line has a slope of 0 everywhere, so its rate of change is always 0.

Let's derive this using the definition of the derivative from first principles:

Since $f(x) = c$ for any value of $x$, we have $f(x+h) = c$ and $f(x) = c$.

Substitute these values into the limit expression:

Simplify the numerator:

When evaluating the limit as $h \to 0$, we consider values of $h$ that are very close to 0 but not equal to 0 ($h \neq 0$). For any non-zero value of $h$, $\frac{0}{h} = 0$.

The limit of a constant is the constant itself:

Rule: The derivative of a constant function is always zero.

$\frac{d}{dx}(c) = 0$

Geometric Interpretation: This confirms the geometric intuition that a horizontal line has a slope of zero everywhere.

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2. The Power Rule

The Power Rule is one of the most fundamental rules for differentiation. It provides a straightforward formula for finding the derivative of a power function of the form $f(x) = x^n$.

The rule states that for any real number $n$, the derivative of $x^n$ with respect to $x$ is:

$\frac{d}{dx}(x^n) = nx^{n-1}$

This rule is remarkably versatile and applies when $n$ is a positive integer, zero, a negative integer, or a rational/irrational number.

Derivation for Positive Integer $n$:

We can derive the Power Rule for a positive integer $n$ using the alternative limit definition of the derivative at a point $x$ and the standard algebraic limit $\lim\limits_{z \to a} \frac{z^n - a^n}{z - a} = na^{n-1}$.

Let $f(x) = x^n$. The derivative $f'(x)$ is given by:

Substitute $f(z) = z^n$ and $f(x) = x^n$ into the limit expression:

This limit is exactly in the standard algebraic form $\lim\limits_{y \to a} \frac{y^n - a^n}{y - a}$ where $y=z$ and $a=x$. Using the result of this standard limit, which we previously established as $na^{n-1}$:

Therefore, $f'(x) = nx^{n-1}$.

The Power Rule can also be derived for positive integers using the first principle definition $f'(x) = \lim\limits_{h \to 0} \frac{(x+h)^n - x^n}{h}$ by expanding $(x+h)^n$ using the binomial theorem, subtracting $x^n$, dividing by $h$, and then taking the limit as $h \to 0$. This derivation is longer but confirms the result.

The proof for negative integers and rational exponents can be done using the quotient rule and chain rule (which are derived using first principles), or through implicit differentiation or the definition involving $\frac{x^n-a^n}{x-a}$ and careful consideration of domains.

Illustrative Examples using the Power Rule:

• Let $f(x) = x^5$. Here $n=5$.

  $\frac{d}{dx}(x^5) = 5 x^{5-1} = 5x^4$.

• Let $f(x) = x$. Here $n=1$.

  $\frac{d}{dx}(x) = \frac{d}{dx}(x^1) = 1 x^{1-1} = 1 x^0$. Since $x^0 = 1$ (for $x \neq 0$), $\frac{d}{dx}(x) = 1 \cdot 1 = 1$. (For $x=0$, the derivative is also 1 from the linear function $y=x$).

• Let $f(x) = \sqrt{x}$. Rewrite using fractional exponents: $f(x) = x^{1/2}$. Here $n=1/2$.

  $\frac{d}{dx}(\sqrt{x}) = \frac{d}{dx}(x^{1/2}) = \frac{1}{2} x^{(1/2)-1} = \frac{1}{2} x^{-1/2}$. Rewrite using positive exponents and radicals: $\frac{1}{2} x^{-1/2} = \frac{1}{2 \sqrt{x}}$ (for $x>0$).

• Let $f(x) = \frac{1}{x^3}$. Rewrite using negative exponents: $f(x) = x^{-3}$. Here $n=-3$.

  $\frac{d}{dx}\left(\frac{1}{x^3}\right) = \frac{d}{dx}(x^{-3}) = -3 x^{-3-1} = -3 x^{-4}$. Rewrite using positive exponents: $-3 x^{-4} = -\frac{3}{x^4}$ (for $x \neq 0$).

The Power Rule is a cornerstone of differentiation techniques, allowing us to differentiate any polynomial or expression that can be written as $x$ raised to a power.

Algebra of Derivatives (Sum, Difference, Product, Quotient Rules)

Just as we have rules for finding the limit of sums, differences, products, and quotients of functions (the Algebra of Limits), we also have corresponding rules for finding the derivatives of functions combined through these operations. These rules are derived from the first principle definition of the derivative and the algebra of limits.

Let $u(x)$ and $v(x)$ be two functions that are differentiable at a point $x$.

1. Constant Multiple Rule

The derivative of a constant multiplied by a function is equal to the constant multiplied by the derivative of the function.

If $u(x)$ is differentiable at $x$ and $c$ is any real constant, then $c \cdot u(x)$ is differentiable at $x$, and:

$\frac{d}{dx}[c \cdot u(x)] = c \cdot \frac{d}{dx}[u(x)] = c \cdot u'(x)$

Proof Idea: Using the definition: $\lim\limits_{h \to 0} \frac{c \cdot u(x+h) - c \cdot u(x)}{h} = \lim\limits_{h \to 0} c \cdot \frac{u(x+h) - u(x)}{h} = c \cdot \lim\limits_{h \to 0} \frac{u(x+h) - u(x)}{h} = c \cdot u'(x)$.

Example: To find the derivative of $f(x) = 5x^3$. Using the Constant Multiple Rule and the Power Rule: $\frac{d}{dx}(5x^3) = 5 \cdot \frac{d}{dx}(x^3) = 5 \cdot (3x^{3-1}) = 5 \cdot (3x^2) = 15x^2$.

2. Sum Rule

The derivative of the sum of two differentiable functions is the sum of their individual derivatives.

If $u(x)$ and $v(x)$ are differentiable at $x$, then $u(x) + v(x)$ is differentiable at $x$, and:

$\frac{d}{dx}[u(x) + v(x)] = \frac{d}{dx}[u(x)] + \frac{d}{dx}[v(x)] = u'(x) + v'(x)$

Proof Idea: Using the definition: $\lim\limits_{h \to 0} \frac{[u(x+h) + v(x+h)] - [u(x) + v(x)]}{h} = \lim\limits_{h \to 0} \frac{[u(x+h) - u(x)] + [v(x+h) - v(x)]}{h} = \lim\limits_{h \to 0} \frac{u(x+h) - u(x)}{h} + \lim\limits_{h \to 0} \frac{v(x+h) - v(x)}{h} = u'(x) + v'(x)$.

3. Difference Rule

The derivative of the difference of two differentiable functions is the difference of their individual derivatives.

If $u(x)$ and $v(x)$ are differentiable at $x$, then $u(x) - v(x)$ is differentiable at $x$, and:

$\frac{d}{dx}[u(x) - v(x)] = \frac{d}{dx}[u(x)] - \frac{d}{dx}[v(x)] = u'(x) - v'(x)$

Proof Idea: Similar to the Sum Rule, or by using the Sum Rule and Constant Multiple Rule with $c=-1$: $\frac{d}{dx}[u(x) + (-1)v(x)] = \frac{d}{dx}u(x) + \frac{d}{dx}(-1)v(x) = u'(x) + (-1)v'(x) = u'(x) - v'(x)$.

The Sum and Difference rules, along with the Constant Multiple and Power Rules, allow us to differentiate any polynomial function term by term.

4. Product Rule

The derivative of the product of two differentiable functions $u(x)$ and $v(x)$ is given by a specific formula:

$\frac{d}{dx}[u(x) \cdot v(x)] = u(x) \cdot v'(x) + v(x) \cdot u'(x)$

This is read as: "the first function times the derivative of the second, plus the second function times the derivative of the first."

Proof Idea: Using the definition, we consider $\lim\limits_{h \to 0} \frac{u(x+h)v(x+h) - u(x)v(x)}{h}$. This limit is manipulated by adding and subtracting $u(x)v(x+h)$ in the numerator: $\lim\limits_{h \to 0} \frac{u(x+h)v(x+h) - u(x)v(x+h) + u(x)v(x+h) - u(x)v(x)}{h} = \lim\limits_{h \to 0} \left[ v(x+h) \frac{u(x+h) - u(x)}{h} + u(x) \frac{v(x+h) - v(x)}{h} \right]$. Since $v$ is differentiable, it is continuous, so $\lim\limits_{h \to 0} v(x+h) = v(x)$. Using limit properties, this becomes $v(x) u'(x) + u(x) v'(x)$.

Mnemonic: A common mnemonic is "uv' + vu'" or "first d-second plus second d-first".

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Answer:

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5. Quotient Rule

The derivative of the quotient of two differentiable functions $u(x)$ and $v(x)$ is given by:

$\frac{d}{dx}\left[\frac{u(x)}{v(x)}\right] = \frac{v(x) \cdot u'(x) - u(x) \cdot v'(x)}{[v(x)]^2}$

This rule is valid at any point $x$ where $v(x) \neq 0$ (as the function $\frac{u(x)}{v(x)}$ is undefined or has an infinite discontinuity where $v(x)=0$).

Proof Idea: This rule can be derived using the limit definition or by combining the product rule and chain rule (after showing $\frac{d}{dx}\left(\frac{1}{v(x)}\right) = -\frac{v'(x)}{[v(x)]^2}$). Using the limit definition involves a similar algebraic manipulation as the product rule.

Mnemonic: A common mnemonic is "Low dee High minus High dee Low, over Low squared". Let $u=$ High and $v=$ Low: $\frac{\text{Low} \cdot (\text{derivative of High}) - \text{High} \cdot (\text{derivative of Low})}{(\text{Low})^2}$.

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Derivatives of Polynomial Functions

A polynomial function is a function that can be written in the form $P(x) = c_n x^n + c_{n-1} x^{n-1} + \dots + c_1 x + c_0$, where $n$ is a non-negative integer, and $c_0, c_1, \dots, c_n$ are real number coefficients ($c_n \neq 0$ for degree $n$). Polynomials are fundamental building blocks in calculus because they are continuous and differentiable everywhere.

To find the derivative of a polynomial function, we can apply the properties of the Algebra of Derivatives (Sum/Difference Rule and Constant Multiple Rule) and the Power Rule term by term.

Let $P(x) = c_n x^n + c_{n-1} x^{n-1} + \dots + c_1 x + c_0$. We want to find $P'(x) = \frac{d}{dx}P(x)$.

Using the Sum/Difference Rule, we can differentiate each term separately:

$P'(x) = \frac{d}{dx}(c_n x^n) + \frac{d}{dx}(c_{n-1} x^{n-1}) + \dots + \frac{d}{dx}(c_1 x^1) + \frac{d}{dx}(c_0)$

Using the Constant Multiple Rule for each term with a coefficient and the Constant Rule for the last term:

$P'(x) = c_n \frac{d}{dx}(x^n) + c_{n-1} \frac{d}{dx}(x^{n-1}) + \dots + c_1 \frac{d}{dx}(x) + \frac{d}{dx}(c_0)$

Now, apply the Power Rule $\frac{d}{dx}(x^k) = kx^{k-1}$ to each power of $x$, and the Constant Rule $\frac{d}{dx}(c_0)=0$:

$P'(x) = c_n (nx^{n-1}) + c_{n-1} ((n-1)x^{(n-1)-1}) + \dots + c_1 (1x^{1-1}) + 0$

$P'(x) = nc_n x^{n-1} + (n-1)c_{n-1} x^{n-2} + \dots + c_1 x^0 + 0$

Since $x^0=1$ (for $x \neq 0$, and the derivative exists at $x=0$ for polynomials), this simplifies to:

$P'(x) = nc_n x^{n-1} + (n-1)c_{n-1} x^{n-2} + \dots + c_1$

The derivative of a polynomial of degree $n$ is a polynomial of degree $n-1$.

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Standard Derivatives of Trigonometric Functions

The derivatives of the six basic trigonometric functions ($\sin x$, $\cos x$, $\tan x$, $\cot x$, $\sec x$, $\text{cosec } x$) are fundamental results that must be known for efficient differentiation. These formulas are derived using the definition of the derivative (first principle), often combined with trigonometric identities and the standard trigonometric limits $\lim\limits_{h \to 0} \frac{\sin h}{h} = 1$ and $\lim\limits_{h \to 0} \frac{1 - \cos h}{h} = 0$.

Here, $x$ is assumed to be in radians.

Derivatives of $\sin x$ and $\cos x$

These are the two most basic trigonometric derivatives, from which the others can be derived.

• $\frac{d}{dx}(\sin x) = \cos x$

Derivation of $\frac{d}{dx}(\sin x)$ using First Principle:

Let $f(x) = \sin x$. Using the definition $f'(x) = \lim\limits_{h \to 0} \frac{f(x+h) - f(x)}{h}$:

Use the trigonometric sum-to-product identity: $\sin A - \sin B = 2 \cos\left(\frac{A+B}{2}\right) \sin\left(\frac{A-B}{2}\right)$.

Let $A = x+h$ and $B = x$. Then $A-B = (x+h) - x = h$ and $A+B = (x+h) + x = 2x+h$.

Rewrite the expression by grouping terms to utilize the standard limit $\lim\limits_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$. Let $\theta = h/2$. As $h \to 0$, $\theta \to 0$.

We need the denominator for the sine term to match its argument. Multiply and divide by 2:

Now, take the limit as $h \to 0$. Use the product rule for limits and the fact that $\cos(x + h/2)$ is continuous (so we can substitute $h=0$):

And using the standard limit $\lim\limits_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$ with $\theta = h/2$:

Combine these limits:

• $\frac{d}{dx}(\cos x) = -\sin x$

Derivation of $\frac{d}{dx}(\cos x)$ using First Principle:

Let $f(x) = \cos x$. Using the definition $f'(x) = \lim\limits_{h \to 0} \frac{f(x+h) - f(x)}{h}$:

Use the trigonometric sum-to-product identity: $\cos A - \cos B = -2 \sin\left(\frac{A+B}{2}\right) \sin\left(\frac{A-B}{2}\right)$.

Let $A = x+h$ and $B = x$. Then $A-B = h$ and $A+B = 2x+h$.

Rewrite the expression:

Multiply and divide by 2:

Now, take the limit as $h \to 0$. Use the product rule for limits and the continuity of $\sin(x + h/2)$:

And using the standard limit $\lim\limits_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$ with $\theta = h/2$:

Combine these limits:

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Derivatives of Other Trigonometric Functions

The derivatives of the remaining four trigonometric functions can be derived using their definitions in terms of $\sin x$ and $\cos x$ and applying the Quotient Rule for differentiation.

• $\frac{d}{dx}(\tan x) = \sec^2 x$

  Derivation using Quotient Rule:

  Recall that $\tan x = \frac{\sin x}{\cos x}$. Let $u(x) = \sin x$ and $v(x) = \cos x$.

  Then $u'(x) = \cos x$ and $v'(x) = -\sin x$.

  Apply the Quotient Rule $\frac{d}{dx}\left[\frac{u(x)}{v(x)}\right] = \frac{v(x) u'(x) - u(x) v'(x)}{[v(x)]^2}$:

  $\frac{d}{dx}(\tan x) = \frac{(\cos x)(\cos x) - (\sin x)(-\sin x)}{(\cos x)^2}$

  [Using Quotient Rule]

  Simplify the numerator and denominator:

  $= \frac{\cos^2 x + \sin^2 x}{\cos^2 x}$

  Using the Pythagorean identity $\sin^2 x + \cos^2 x = 1$:

  $= \frac{1}{\cos^2 x}$

  [Using $\sin^2 x + \cos^2 x = 1$]

  Using the definition $\sec x = \frac{1}{\cos x}$:

  $= \left(\frac{1}{\cos x}\right)^2 = \sec^2 x$

  [Using $\sec x = 1/\cos x$]

• $\frac{d}{dx}(\cot x) = -\text{cosec}^2 x$

  Derivation using Quotient Rule:

  Recall that $\cot x = \frac{\cos x}{\sin x}$. Let $u(x) = \cos x$ and $v(x) = \sin x$.

  Then $u'(x) = -\sin x$ and $v'(x) = \cos x$.

  Apply the Quotient Rule:

  $\frac{d}{dx}(\cot x) = \frac{(\sin x)(-\sin x) - (\cos x)(\cos x)}{(\sin x)^2}$

  [Using Quotient Rule]

  Simplify the numerator:

  $= \frac{-\sin^2 x - \cos^2 x}{\sin^2 x} = \frac{-(\sin^2 x + \cos^2 x)}{\sin^2 x}$

  Using the Pythagorean identity $\sin^2 x + \cos^2 x = 1$:

  $= \frac{-1}{\sin^2 x}$

  [Using $\sin^2 x + \cos^2 x = 1$]

  Using the definition $\text{cosec } x = \frac{1}{\sin x}$:

  $= -\left(\frac{1}{\sin x}\right)^2 = -\text{cosec}^2 x$

  [Using $\text{cosec } x = 1/\sin x$]

• $\frac{d}{dx}(\sec x) = \sec x \tan x$

  Derivation using Quotient Rule:

  Recall that $\sec x = \frac{1}{\cos x}$. Let $u(x) = 1$ and $v(x) = \cos x$.

  Then $u'(x) = \frac{d}{dx}(1) = 0$ (derivative of a constant) and $v'(x) = -\sin x$.

  Apply the Quotient Rule:

  $\frac{d}{dx}(\sec x) = \frac{(\cos x)(0) - (1)(-\sin x)}{(\cos x)^2}$

  [Using Quotient Rule]

  Simplify the numerator and denominator:

  $= \frac{0 - (-\sin x)}{\cos^2 x} = \frac{\sin x}{\cos^2 x}$

  Rewrite the fraction as a product to get the desired form:

  $= \frac{1}{\cos x} \cdot \frac{\sin x}{\cos x}$

  Using the definitions $\sec x = \frac{1}{\cos x}$ and $\tan x = \frac{\sin x}{\cos x}$:

  $= \sec x \tan x$

  [Using definitions]

• $\frac{d}{dx}(\text{cosec } x) = -\text{cosec } x \cot x$

  Derivation using Quotient Rule:

  Recall that $\text{cosec } x = \frac{1}{\sin x}$. Let $u(x) = 1$ and $v(x) = \sin x$.

  Then $u'(x) = 0$ and $v'(x) = \cos x$.

  Apply the Quotient Rule:

  $\frac{d}{dx}(\text{cosec } x) = \frac{(\sin x)(0) - (1)(\cos x)}{(\sin x)^2}$

  [Using Quotient Rule]

  Simplify the numerator and denominator:

  $= \frac{0 - \cos x}{\sin^2 x} = \frac{-\cos x}{\sin^2 x}$

  Rewrite the fraction as a product:

  $= -\frac{1}{\sin x} \cdot \frac{\cos x}{\sin x}$

  Using the definitions $\text{cosec } x = \frac{1}{\sin x}$ and $\cot x = \frac{\cos x}{\sin x}$:

  $= -\text{cosec } x \cot x$

  [Using definitions]

These six formulas are essential for differentiating expressions involving trigonometric functions.

Derivatives of Exponential Functions

Exponential functions are crucial in many areas of science and mathematics due to their properties related to growth and decay. Their derivatives follow simple and elegant rules, especially for the natural exponential function with base $e$.

Derivative of the Natural Exponential Function ($e^x$)

The natural exponential function $f(x) = e^x$ is unique because its derivative is the function itself. This property makes $e^x$ and related functions fundamental in calculus.

$\frac{d}{dx}(e^x) = e^x$

Derivation using First Principle:

Let $f(x) = e^x$. Using the definition $f'(x) = \lim\limits_{h \to 0} \frac{f(x+h) - f(x)}{h}$:

Use the property of exponents $e^{x+h} = e^x e^h$:

Factor out $e^x$ from the numerator. Since $e^x$ does not depend on $h$, it can be moved outside the limit:

Recall the standard limit $\lim\limits_{h \to 0} \frac{e^h - 1}{h} = 1$.

So, the derivative is:

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Derivative of the General Exponential Function ($a^x$)

For a general exponential function $f(x) = a^x$, where $a$ is a positive constant and $a \neq 1$.

$\frac{d}{dx}(a^x) = a^x \ln a$

Derivation using First Principle:

Let $f(x) = a^x$. Using the definition $f'(x) = \lim\limits_{h \to 0} \frac{f(x+h) - f(x)}{h}$:

Use the property of exponents $a^{x+h} = a^x a^h$:

Factor out $a^x$ from the numerator. Since $a^x$ does not depend on $h$, it can be moved outside the limit:

Recall the standard limit $\lim\limits_{h \to 0} \frac{a^h - 1}{h} = \ln a$.

So, the derivative is:

Alternate Derivation using $a^x = e^{x \ln a}$ and Chain Rule:

Although the Chain Rule is typically introduced later, we can use the identity $a^x = e^{\ln(a^x)} = e^{x \ln a}$. Let $y = a^x = e^{x \ln a}$.

Let $u = x \ln a$. Then $y = e^u$. The derivative of $u$ with respect to $x$ is $\frac{du}{dx} = \frac{d}{dx}(x \ln a) = \ln a \cdot \frac{d}{dx}(x) = \ln a \cdot 1 = \ln a$ (since $\ln a$ is a constant).

The derivative of $y$ with respect to $u$ is $\frac{dy}{du} = \frac{d}{du}(e^u) = e^u$.

Using the Chain Rule $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$:

Substitute back $u = x \ln a$ and recall $e^{x \ln a} = a^x$:

This confirms the result using the Chain Rule.

Derivatives of Logarithmic Functions

Logarithmic functions are the inverses of exponential functions, and their derivatives also have standard formulas that are frequently used.

Derivative of the Natural Logarithmic Function ($\ln x$)

For the natural logarithmic function $f(x) = \ln x$, which is defined for $x > 0$, its derivative is:

$\frac{d}{dx}(\ln x) = \frac{1}{x}$

Derivation using First Principle:

Let $f(x) = \ln x$ for $x > 0$. Using the definition $f'(x) = \lim\limits_{h \to 0} \frac{f(x+h) - f(x)}{h}$. Since the domain is $x>0$, $x+h$ must also be positive. If $x>0$, we must consider $h$ such that $x+h>0$. As $h \to 0$, this interval around $x$ for $h$ will be small enough to keep $x+h$ positive if $x>0$.

Use the property of logarithms $\ln A - \ln B = \ln(A/B)$:

Rewrite $\frac{x+h}{x}$ as $1 + \frac{h}{x}$ and use the property $k \ln A = \ln(A^k)$:

We need to relate this to the standard limit definition of $e$: $\lim\limits_{t \to 0} (1 + t)^{1/t} = e$. Let $t = h/x$. As $h \to 0$, $t = h/x \to 0/x = 0$ (since $x$ is treated as a constant in this limit). Also, from $t = h/x$, we get $h = xt$, so $1/h = 1/(xt)$.

Substitute $t$ into the expression:

Use the property $(A^m)^k = A^{mk}$ to rewrite the exponent $\frac{1}{xt}$ as $\frac{1}{x} \cdot \frac{1}{t}$:

Using the property $\ln(A^k) = k \ln A$ again, or moving the constant $1/x$ outside the limit (since it doesn't depend on $t$):

Since the natural logarithm function $\ln u$ is continuous, we can move the limit inside the logarithm:

Evaluate the standard limit inside the brackets, which is the definition of $e$:

Recall that $\ln e = 1$:

So, the derivative is:

Alternate Derivation using Inverse Functions and Chain Rule:

Let $y = \ln x$. The inverse function is $x = e^y$. We know $\frac{dx}{dy} = \frac{d}{dy}(e^y) = e^y$.

Using the inverse function theorem $\frac{dy}{dx} = \frac{1}{dx/dy}$:

Substitute $e^y = x$ back:

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Derivative of the General Logarithmic Function ($\log_a x$)

For a general logarithmic function $f(x) = \log_a x$, where $a$ is a positive constant and $a \neq 1$, defined for $x > 0$.

$\frac{d}{dx}(\log_a x) = \frac{1}{x \ln a}$

Derivation using Change of Base Formula:

We can use the change of base formula for logarithms to express $\log_a x$ in terms of the natural logarithm:

Since $a$ is a constant, $\ln a$ is also a constant (and $\ln a \neq 0$ because $a \neq 1$).

So, the function can be written as $\frac{1}{\ln a} \cdot \ln x$.

Now, differentiate this expression using the Constant Multiple Rule and the derivative of $\ln x$:

Substitute the derivative of $\ln x$, which is $\frac{1}{x}$:

So, the derivative is:

Note that for the special case where $a=e$, $\ln e = 1$, and the formula becomes $\frac{d}{dx}(\log_e x) = \frac{d}{dx}(\ln x) = \frac{1}{x \cdot 1} = \frac{1}{x}$, which matches the natural logarithm derivative.

Some Standard Results on Derivatives (Consolidated Formulas)

Building upon the definition of the derivative from first principles, and using algebraic manipulations, standard limits, and differentiation rules, we have derived formulas for the derivatives of various elementary functions and rules for differentiating combinations of functions. This section consolidates these essential results.

Basic Differentiation Rules

These rules allow us to differentiate functions that are sums, differences, constant multiples, products, or quotients of simpler functions whose derivatives we know.

• Constant Rule: The derivative of a constant is zero.

  $\frac{d}{dx}(c) = 0$

• Power Rule: The derivative of $x$ raised to any real power $n$.

  $\frac{d}{dx}(x^n) = nx^{n-1}$

• Constant Multiple Rule:

  $\frac{d}{dx}[c \cdot f(x)] = c \cdot f'(x)$

• Sum Rule:

  $\frac{d}{dx}[f(x) + g(x)] = f'(x) + g'(x)$

• Difference Rule:

  $\frac{d}{dx}[f(x) - g(x)] = f'(x) - g'(x)$

• Product Rule:

  $\frac{d}{dx}[f(x) \cdot g(x)] = f(x)g'(x) + g(x)f'(x)$

• Quotient Rule: (Valid where $g(x) \neq 0$)

  $\frac{d}{dx}\left[\frac{f(x)}{g(x)}\right] = \frac{g(x)f'(x) - f(x)g'(x)}{[g(x)]^2}$

Standard Derivatives of Elementary Functions

These are the derivatives of common functions that serve as building blocks for differentiating more complex expressions.

• Derivative of $x$: A specific case of the Power Rule with $n=1$.

  $\frac{d}{dx}(x) = 1$

• Trigonometric Functions:

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

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

  $\frac{d}{dx}(\tan x) = \sec^2 x$

  $\frac{d}{dx}(\cot x) = -\text{cosec}^2 x$

  $\frac{d}{dx}(\sec x) = \sec x \tan x$

  $\frac{d}{dx}(\text{cosec } x) = -\text{cosec } x \cot x$

• Exponential Functions: (For $a>0, a\neq 1$)

  $\frac{d}{dx}(e^x) = e^x$

  $\frac{d}{dx}(a^x) = a^x \ln a$

• Logarithmic Functions: (For $x>0$, $a>0, a\neq 1$)

  $\frac{d}{dx}(\ln x) = \frac{1}{x}$

  $\frac{d}{dx}(\log_a x) = \frac{1}{x \ln a}$

These rules and standard formulas are indispensable tools for solving differentiation problems and are applied extensively in all areas of calculus and its applications. More advanced techniques, such as the Chain Rule (for composite functions), implicit differentiation, and logarithmic differentiation, build upon these fundamental results.