Calculus

Differentiation of logarithmic functions

The differentiation of logarithmic functions forms the analytical cornerstone of calculus, allowing the linearization of complex products, quotients, and powers via logarithmic differentiation.

For: UP PGT Mathematics UP TGT Mathematics DSSSB PGT / TGT KVS / NVS PGT LT Grade

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Definition and statement

Derivative of Natural and General Logarithmic Functions

Let $f(x) = \log_a(x)$, where $a > 0$ and $a \neq 1$, and $x \in (0, \infty)$. Then $f'(x) = \frac{1}{x \ln a}$. In particular, for the natural logarithm where $a = e$, $\frac{d}{dx}(\ln x) = \frac{1}{x}$. For composite functions, if $y = \ln[u(x)]$, then $\frac{dy}{dx} = \frac{1}{u(x)} \cdot u'(x).$

📐 Geometric Interpretation

The derivative $\frac{1}{x}$ represents the instantaneous rate of change of the logarithmic curve, showing that as $x$ increases, the slope of the tangent line decreases monotonically, reflecting the concave-down nature of the logarithmic graph ($f''(x) = -\frac{1}{x^2} < 0$).

⚙️ Algebraic & Structural Meaning

Logarithms convert multiplicative structures into additive structures ($\log(uv) = \log u + \log v$), which simplifies polynomial powers and variable exponents through logarithmic differentiation before applying standard derivative rules.

Conditions and hypotheses

Check each condition and what changes when it is omitted:

Hypothesis / Condition	Domain / Interval	Necessity	What Fails If Condition Dropped
Domain Positivity ($x > 0$)	`(0, \infty) (Open interval)`	Strictly Mandatory / Necessary	The argument of a real logarithm must be strictly positive; non-positive values result in undefined real logarithmic outputs.
Base Constraints ($a > 0, a \neq 1$)	`Discrete parameter constraint`	Strictly Mandatory	A base of 1 is undefined for division by $\ln(1) = 0$, and negative bases lead to complex multi-valued functions.
Differentiability of Inner Function $u(x)	`Set where $u(x) > 0$ and $u'(x)$ exists`	Essential for chain rule application	If $u(x)$ fails to be differentiable or becomes zero or negative, the composite derivative $\frac{u'(x)}{u(x)}$ is invalid.

Useful results and shortcuts

Use these identities and results when their conditions apply:

Power-Function Differentiation via Logarithms

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

Rule: Direct formula for functions of the form $y = u(x)^{v(x)}$ without performing full logarithmic differentiation from scratch every time.

Application: Frequently asked in UP PGT and DSSSB PGT for functions like $x^x$, $(\sin x)^{\cos x}$, etc.

Logarithmic Differentiation for Products/Quotients

$$ \frac{d}{dx} \ln \left( \prod_{i=1}^{n} f_i(x)^{k_i} \right) = \sum_{i=1}^{n} k_i \frac{f_i'(x)}{f_i(x)} $$

Rule: Taking $\ln$ on both sides transforms products and quotients into sums and differences of simple fractions.

Derivative of Infinite Logarithmic Towers

$$ \text{If } y = x^{x^{x^{\dots}}}, \text{ then } \frac{dy}{dx} = \frac{y^2}{x(1 - y \ln x)} $$

Rule: Substitute $y$ for the infinite tail and differentiate implicitly after taking natural logarithms.

Application: Direct pattern recognition for UP TGT/PGT repeating tower problems.

Implicit Logarithmic Derivative Shortcut

$$ \frac{dy}{dx} = -\frac{\frac{\partial F}{\partial x}}{\frac{\partial F}{\partial y}} $$

Rule: Use partial derivatives for implicit equations containing logarithmic terms like $\ln(xy) + x^2 y = 5$.

Logarithmic Derivative of Modulus Arguments

$$ \frac{d}{dx}(\ln |f(x)|) = \frac{f'(x)}{f(x)} \quad \forall \, f(x) \neq 0 $$

Rule: The absolute value sign inside a logarithm can be ignored during differentiation because the derivative is identical wherever $f(x) \neq 0$.

Application: Tested in trick questions involving domains spanning negative numbers.

Rapid Elimination via Log-Slope Testing

$$ \left. \frac{d}{dx} \ln(1+x) \right|_{x=0} = 1 $$

Rule: Test $x=0$ or $x=1$ on options to eliminate incorrect derivative expressions immediately.

Common errors and counterexamples

These examples show how a conclusion can fail when a condition is missing:

⚠️ COMMON ERROR #01

Ignoring Domain Restrictions of Logarithms

Common Pitfall: Applying derivative formulas without checking if the domain of the resulting expression matches the original function.

Counterexample: f(x) = \ln(x^2) \implies f'(x) = \frac{2x}{x^2} = \frac{2}{x}

Why it fails: While $\ln(x^2)$ is defined for all $x \neq 0$, the simplified derivative $\frac{2}{x}$ has the exact same domain, but rewriting it as $2 \ln|x|$ changes the domain perspective if not careful.

🛡️ Exam Defense: Always simplify using properties of logarithms ($\ln(x^n) = n \ln|x|$) *before* differentiating to avoid domain discrepancies.

⚠️ COMMON ERROR #02

Misapplying Chain Rule to General Logarithms

Common Pitfall: Forgetting the $\ln a$ term in the denominator when differentiating $\log_a x$.

Counterexample: \frac{d}{dx}(\log_{10} x) = \frac{1}{x} \text{ (Incorrect omission of } \ln 10 \text{)}

Why it fails: Base change formula states $\log_a x = \frac{\ln x}{\ln a}$, hence the derivative must include $\frac{1}{\ln a}$.

🛡️ Exam Defense: Always convert non-natural logs using change of base: $\log_a u = \frac{\ln u}{\ln a}$ before differentiation if unsure.

Comparison

Concept / Theorem	Hypotheses / Assumptions	Key Formula / Outcome	Geometric Meaning	Exam Identification Clue
Natural Logarithm	$x > 0$	${\frac{d}{dx}(\ln x) = \frac{1}{x}}$	Decreasing positive slope, concave down	Standard calculus problems
General Logarithm	$x > 0, a > 0, a \neq 1$	${\frac{d}{dx}(\log_a x) = \frac{1}{x \ln a}}$	Scaled slope by factor of $\frac{1}{\ln a}$	Base explicitly given as $2, 10$, etc.
Logarithmic Differentiation	$f(x) > 0$	${\frac{y'}{y} = \frac{d}{dx}(\ln y)}$	Slope of compounded growth rate	Variable raised to variable power ($u(x)^{v(x)}$)
Composite Logarithm	$u(x) > 0, u'(x)$ exists	${\frac{d}{dx}(\ln u) = \frac{u'}{u}}$	Chain rule application on logarithmic scale	Functions of functions like $\ln(\sin x)$

Solved examples

Each problem includes a full solution and a shorter method:

CANONICAL PROBLEM 01 Dual-Method Breakdown

Find the derivative of $y = x^{\sin x}$ with respect to $x$.

(A) $x^{\sin x} \left(\frac{\sin x}{x} + \cos x \ln x\right)$(B) $x^{\sin x} (\cos x \ln x)$(C) $\sin x \cdot x^{\sin x - 1}$(D) $x^{\cos x} (\sin x \ln x)$

📚 Method A: Formal Textbook Derivation

Step 1: Take natural log on both sides: $\ln y = \sin x \ln x$. Step 2: Differentiate implicitly w.r.t $x$: $\frac{1}{y} \frac{dy}{dx} = \cos x \ln x + \sin x \cdot \frac{1}{x}$. Step 3: Multiply by $y = x^{\sin x}$ to get $\frac{dy}{dx} = x^{\sin x} \left(\frac{\sin x}{x} + \cos x \ln x\right)$.

⚡ Method B: 25-Second Exam Shortcut

Apply Shortcut 1 directly: $\frac{d}{dx}[u^v] = u^v [v' \ln u + v \frac{u'}{u}]$. Here $u=x, v=\sin x$.

💡 Tip: Standard variable-power format frequently tested in UP PGT.

CANONICAL PROBLEM 02 Dual-Method Breakdown

If $y = \log_7(\log_3 x)$, find $\frac{dy}{dx}$ at $x = 3$.

(A) $\frac{1}{3 \ln 7}$(B) $\frac{1}{3 \ln 3 \ln 7}$(C) $\frac{1}{\ln 21}$(D) $\frac{1}{7 \ln 3}$

📚 Method A: Formal Textbook Derivation

Step 1: Rewrite using change of base: $y = \frac{\ln(\ln x)}{\ln 7}$. Step 2: Differentiate using chain rule: $\frac{dy}{dx} = \frac{1}{\ln 7} \cdot \frac{1}{\ln x} \cdot \frac{1}{x}$. Step 3: Substitute $x = 3$: $\left.\frac{dy}{dx}\right|_{x=3} = \frac{1}{3 \ln 3 \ln 7}$.

⚡ Method B: 25-Second Exam Shortcut

Chain rule from outside in: $\frac{1}{(\ln 3) \ln 7} \times \frac{1}{3}$.

💡 Tip: Watch out for base conversion constants like $\ln 7$ in denominators.

Practice questions

Choose an answer to check it and see the solution:

MCQ 1: What is the derivative of $\ln(\sec x + \tan x)$ with respect to $x$?

A $\sec x$

B $\csc x$

C $\tan x$

D $\cos x$

Derivation & Analysis:

Let $y = \ln(\sec x + \tan x)$. Then $\frac{dy}{dx} = \frac{1}{\sec x + \tan x}(\sec x \tan x + \sec^2 x) = \frac{\sec x(\tan x + \sec x)}{\sec x + \tan x} = \sec x$.

MCQ 2: If $y = \log_x(e)$, then $\frac{dy}{dx}$ is equal to:

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

B $\frac{1}{x \ln x}$

C $-\frac{1}{x^2}$

D 0

Derivation & Analysis:

Using change of base, $y = \frac{\ln e}{\ln x} = \frac{1}{\ln x} = (\ln x)^{-1}$. Differentiating using chain rule: $\frac{dy}{dx} = -1(\ln x)^{-2} \cdot \frac{1}{x} = -\frac{1}{x(\ln x)^2}.

MCQ 3: If $y = x^x$, the value of $\frac{dy}{dx}$ at $x = 1$ is:

A 0

B 1

C $e$

D $1 + e$

Derivation & Analysis:

For $y = x^x$, $\frac{dy}{dx} = x^x(1 + \ln x)$. At $x = 1$, $\frac{dy}{dx} = 1^1(1 + \ln 1) = 1(1 + 0) = 1$.

MCQ 4: If $f(x) = \log_10(5x^2 + 3)$, then $f'(x)$ is:

A $\frac{10x}{(5x^2 + 3)\ln 10}$

B $\frac{10x}{5x^2 + 3}$

C $\frac{1}{(5x^2 + 3)\ln 10}$

D $\frac{5x}{(5x^2 + 3)\ln 10}$

Derivation & Analysis:

Derivative of $\log_a u$ is $\frac{u'}{u \ln a}$. Here $u = 5x^2 + 3$, so $u' = 10x$ and $a = 10$. Thus $f'(x) = \frac{10x}{(5x^2 + 3)\ln 10}$.

MCQ 5: The derivative of $\ln(\sqrt{x})$ with respect to $x$ is:

A $\frac{1}{x}$

B $\frac{1}{2x}$

C $\frac{2}{x}$

D $\frac{1}{\sqrt{x}}$

Derivation & Analysis:

Simplify first: $\ln(\sqrt{x}) = \frac{1}{2} \ln x$. Differentiating gives $\frac{1}{2} \cdot \frac{1}{x} = \frac{1}{2x}$.

MCQ 6: If $y = \sqrt{\ln x + \sqrt{\ln x + \sqrt{\ln x + \dots_\infty}}}$, then $\frac{dy}{dx}$ is:

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

B $\frac{1}{x \ln x}$

C $\frac{y}{x}$

D $\frac{1}{2y}$

Derivation & Analysis:

Equation reduces to $y = \sqrt{\ln x + y} \implies y^2 = \ln x + y$. Differentiating implicitly: $2y y' = \frac{1}{x} + y' \implies y'(2y - 1) = \frac{1}{x} \implies y' = \frac{1}{x(2y - 1)}$.

MCQ 7: If $y = x^{\ln x}$, then $\frac{dy}{dx}$ at $x = e$ is:

A $2$

B $e$

C $1$

D $2e$

Derivation & Analysis:

Taking $\ln$: $\ln y = (\ln x)^2$. Differentiating: $\frac{1}{y} y' = 2(\ln x) \cdot \frac{1}{x} \implies y' = x^{\ln x} \left(\frac{2 \ln x}{x}\right)$. At $x = e$, $e^{\ln e} \left(\frac{2 \ln e}{e}\right) = e \left(\frac{2(1)}{e}\right) = 2$.

MCQ 8: Assertion (A): $\frac{d}{dx}(\ln |x|) = \frac{1}{x}$ for all $x \neq 0$. Reason (R): The domain of $\ln |x|$ is $\mathbb{R} \setminus \{0\}$.

A Both (A) and (R) are true and (R) is the correct explanation of (A)

B Both (A) and (R) are true but (R) is not the correct explanation of (A)

C (A) is true but (R) is false

D (A) is false but (R) is true

Derivation & Analysis:

Both statements are mathematically rigorous and correct. The derivative holds across the entire domain of $\ln|x|$ which is $\mathbb{R} \setminus \{0\}$.

MCQ 9: If $y = \ln \left( \frac{e^x}{1 + e^x} \right)$, then $\frac{dy}{dx}$ is:

A $\frac{1}{1 + e^x}$

B $-\frac{1}{1 + e^x}$

C $\frac{1}{1 - e^x}$

D $1$

Derivation & Analysis:

Using log properties: $y = \ln(e^x) - \ln(1 + e^x) = x - \ln(1 + e^x)$. Differentiating: $\frac{dy}{dx} = 1 - \frac{e^x}{1 + e^x} = \frac{1 + e^x - e^x}{1 + e^x} = \frac{1}{1 + e^x}$... Wait, let's re-verify: $1 - \frac{e^x}{1+e^x} = \frac{1}{1+e^x}$. Correct option is A.

Q10

MCQ 10: What is $\lim_{h \to 0} \frac{\ln(x + h) - \ln x}{h}$?

A $\frac{1}{x}$

B $\ln x$

C $x$

D $0$

Derivation & Analysis:

This is the first principle definition of the derivative of $\ln x$, which evaluates directly to $\frac{1}{x}$.

Common questions

Logarithmic differentiation is mandatory when dealing with functions of the form $f(x)^{g(x)}$ (variable to the power variable) or when a product/quotient contains more than three algebraic factors, as it converts cumbersome multiplications and divisions into easily manageable sums and differences.

Absolute value signs ensure the argument of the logarithm remains strictly positive, thereby expanding the valid domain of the resulting expression to include negative numbers where the original function or its integral is valid (e.g., $\int \frac{1}{x} dx = \ln|x| + C$).

Memorize the direct shortcut formula: $\frac{d}{dx}[u^v] = u^v \left( v' \ln u + v \frac{u'}{u} \right)$. This bypasses taking logs and doing implicit differentiation steps manually, saving precious seconds.