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2. Calculus — Differential Calculus

Derivative of composite function

The derivative of a composite function is governed universally by the Chain Rule, serving as the foundational engine for multivariable calculus, implicit differentiation, and coordinate transformations. This topic covers this principle.

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

The Chain Rule for Composite Functions
Let $g: I \to \mathbb{R}$ be differentiable at $c \in I$, and let $f: J \to \mathbb{R}$ be differentiable at $g(c)$, where $g(I) \subset J$. Then the composite function $f \circ g: I \to \mathbb{R}$ defined by $(f \circ g)(x) = f(g(x))$ is differentiable at $c$, and the derivative is given by $$(f \circ g)'(c) = f'(g(c)) \cdot g'(c)$$ or in Leibniz notation, $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$, where $u = g(x)$.

📐 Geometric Interpretation

Geometrically, the derivative of a composite function represents the scaling factor of slopes along successive transformations. The tangent slope of the composite curve at $x = c$ is the product of the inner function's slope at $c$ ($g'(c)$) and the outer function's slope evaluated at the intermediate point $g(c)$ ($f'(g(c))$).

⚙️ Algebraic & Structural Meaning

Algebraically, it establishes the functorial property of differentiation over function composition, allowing complex nested polynomial, trigonometric, and transcendental structures to be systematically decomposed into elementary linear approximations.

Conditions and hypotheses

Check each condition and what changes when it is omitted:

Hypothesis / Condition Domain / Interval Necessity What Fails If Condition Dropped
Inner Function Differentiability N/A Strictly Mandatory / Necessary If $g(x)$ fails to be differentiable at $c$ (e.g., has a sharp corner like $|x|$), the chain rule formula breaks down, and the composite function may be non-differentiable at $c$.
Outer Function Differentiability N/A Strictly Mandatory If $f(u)$ is not differentiable at the image point $u = g(c)$, the rate of change of the outer layer cannot be defined, rendering the chain product undefined.
Range Containment Condition N/A Essential for well-definedness If the outputs of the inner function fall outside the domain of the outer function, the composition $(f \circ g)(x)$ itself is undefined.

Useful results and shortcuts

Use these identities and results when their conditions apply:

01

Power of a Function Rule

$$ \frac{d}{dx} \left[ (f(x))^n \right] = n \left( f(x) \right)^{n-1} \cdot f'(x) $$

Rule: Bring down the power, reduce exponent by 1, and immediately multiply by the derivative of the base function inside.

Application: Directly applied in UP PGT/TGT questions involving large polynomial powers or radical expressions.
02

Exponential Chain Rule

$$ \frac{d}{dx} \left[ e^{f(x)} \right] = e^{f(x)} \cdot f'(x) $$

Rule: Retain the exponential expression entirely and append the derivative of the exponent as a coefficient factor.

03

Logarithmic Composite Rule

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

Rule: Place the derivative of the inner function in the numerator and the original inner function in the denominator.

Application: Standard pattern for logarithmic differentiation and inverse function derivatives.
04

Trigonometric Nesting Rule

$$ \frac{d}{dx} \left[ \sin(f(x)) \right] = \cos(f(x)) \cdot f'(x) $$

Rule: Differentiate the outer trigonometric function leaving the argument intact, then multiply by the derivative of the argument.

Application: Appears in multiple-angle and nested trigonometric differential equations.
05

Inverse Trigonometric Composite

$$ \frac{d}{dx} \left[ \sin^{-1}(f(x)) \right] = \frac{f'(x)}{\sqrt{1 - (f(x))^2}} $$

Rule: Substitute the inner function into the standard inverse formula and multiply by the inner derivative in the numerator.

06

Rapid Elimination via Logarithmic Pre-processing

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

Rule: For functions of the form $y = u^v$, take $\ln$ on both sides to convert exponentiation into multiplication before applying the chain rule.

Application: Eliminates lengthy algebraic expansion steps in variable-exponent problems.

Common errors and counterexamples

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

⚠️ COMMON ERROR #01

Forgetting to Multiply by the Inner Derivative

Common Pitfall: Candidates compute $\frac{d}{dx}[\sin(x^2)] = \cos(x^2)$, completely omitting the factor $2x$.

Counterexample: Let $y = \sin(x^2)$. Then $\frac{dy}{dx} \neq \cos(x^2)$.

Why it fails: Fails to account for the internal scaling of the domain variable $x^2$. The correct derivative is $2x \cos(x^2)$.

🛡️ Exam Defense: Always underline the inner function $g(x)$ before differentiating, and ensure its derivative $g'(x)$ is appended as a multiplier.
⚠️ COMMON ERROR #02

Confusing Composition with Multiplication (Leibniz Product Rule Mix-up)

Common Pitfall: Applying the product rule $(fg)' = f'g + fg'$ when given a composite function $f(g(x))$, or vice versa.

Counterexample: For $y = \sin(3x)$, treating it as a product of $\sin(x)$ and $3x$.

Why it fails: undefined

🛡️ Exam Defense: Check if one function is substituted inside the argument of another (Composition $\to$ Chain Rule) or if two independent functions are multiplied (Product Rule).

Comparison

Concept / TheoremHypotheses / AssumptionsKey Formula / OutcomeGeometric MeaningExam Identification Clue
Chain Rule (Composite)$g$ differentiable at $c$, $f$ differentiable at $g(c)$$(f \circ g)'(c) = f'(g(c))g'(c)$Product of successive tangent slopesNested functions like $\sin(x^2)$ or $e^{\tan x}$
Product Rule$f, g$ differentiable on interval $I$$(fg)' = f'g + fg'$Sum of two rectangular area expansion ratesAlgebraic product like $x^2 \sin x$
Quotient Rule$f, g$ differentiable, $g(x) \neq 0$$(\frac{f}{g})' = \frac{f'g - fg'}{g^2}$Rate of change of ratio of ordinatesFractional algebraic expressions
Inverse Function Rule$f$ bijective, continuous, $f'(g(x)) \neq 0$$(f^{-1})'(x) = \frac{1}{f'(f^{-1}(x))}$Reciprocal slope via reflection across $y=x$Questions asking for derivative of $\sin^{-1} x$ or $\ln x$

Solved examples

Each problem includes a full solution and a shorter method:

CANONICAL PROBLEM 01 Dual-Method Breakdown

If $y = \tan^{-1}\left(\frac{\sqrt{1+x^2} - 1}{x}\right)$, find $\frac{dy}{dx}$ at $x = 0$ (assuming appropriate limit interpretation).

(A) $\frac{1}{2}$(B) $1$(C) $0$(D) $-\frac{1}{2}$
✓ Correct Answer: (A)
📚 Method A: Formal Textbook Derivation
Step 1: Put $x = \tan \theta$. Then $\sqrt{1+x^2} = \sec \theta$. Step 2: Simplify argument to $\frac{\sec \theta - 1}{\tan \theta} = \frac{1 - \cos \theta}{\sin \theta} = \tan(\theta/2)$. Step 3: Thus $y = \theta/2 = \frac{1}{2}\tan^{-1}x$. Step 4: Differentiate to get $\frac{dy}{dx} = \frac{1}{2(1+x^2)}$. At $x=0$, value is $\frac{1}{2}$.
⚡ Method B: 25-Second Exam Shortcut
Substitute standard trigonometric substitution ($x = \tan\theta$) to collapse the composite inverse function into a linear term instantly.
💡 Tip: Whenever inverse trigonometric functions contain square roots of $1+x^2$ or $x^2-1$, trigonometric substitution is 3x faster than direct chain rule expansion.
CANONICAL PROBLEM 02 Dual-Method Breakdown

If $f(x) = \sqrt{sin(\sqrt{x})}$, what is the structure of its derivative?

(A) $\frac{\cos(\sqrt{x})}{4\sqrt{x}\sqrt{\sin(\sqrt{x})}}$(B) $\frac{\cos(\sqrt{x})}{2\sqrt{x}\sin(\sqrt{x})}$(C) $\frac{1}{4\sqrt{x}\cos(\sqrt{x})}$(D) $\frac{2\cos(\sqrt{x})}{\sqrt{sin(\sqrt{x})}}$
✓ Correct Answer: (A)
📚 Method A: Formal Textbook Derivation
Step 1: Identify three nested layers: Outer square root, middle sine, inner square root of $x$. Step 2: Apply chain rule sequentially: $\frac{d}{dx} [u^{1/2}] \cdot \frac{d}{dx}[\sin v] \cdot \frac{d}{dx}[x^{1/2}]$. Step 3: Multiply: $\frac{1}{2\sqrt{\sin(\sqrt{x})}} \cdot \cos(\sqrt{x}) \cdot \frac{1}{2\sqrt{x}}$. Step 4: Combine denominators to yield Option (A).
⚡ Method B: 25-Second Exam Shortcut
Count the number of nested square roots and radicals; each square root contributes a $2\sqrt{\text{term}}$ in the denominator.
💡 Tip: Triple-nested chain rule problems test patience. Work strictly from outside to inside.

Practice questions

Choose an answer to check it and see the solution:

Q1
MCQ 1: If $y = \sec(\tan(\sqrt{x}))$, then $\frac{dy}{dx}$ is equal to:
A $\frac{\sec(\tan(\sqrt{x})) \tan(\tan(\sqrt{x})) \sec^2(\sqrt{x})}{2\sqrt{x}}$
B $\frac{\sec^2(\tan(\sqrt{x}))}{2\sqrt{x}}$
C $\sec(\tan(\sqrt{x})) \tan(\sqrt{x})$
D $\frac{\tan(\tan(\sqrt{x}))}{2\sqrt{x}}$
Derivation & Analysis:

Apply chain rule across three layers: secant, tangent, and square root. Derivative of $\sec(u)$ is $\sec u \tan u$, derivative of $\tan(v)$ is $\sec^2 v$, and derivative of $\sqrt{x}$ is $\frac{1}{2\sqrt{x}}$. Multiplying them gives Option A.

Q2
MCQ 2: If $f(x) = e^{x^3}$, then $f'''(0)$ is:
A 0
B 3
C 6
D 18
Derivation & Analysis:

First derivative $f'(x) = 3x^2 e^{x^3}$. Second derivative using product/chain rule: $f''(x) = 6x e^{x^3} + 9x^4 e^{x^3}$. Third derivative evaluated at $x=0$ yields $6$, as terms with remaining powers of $x$ vanish.

Q3
MCQ 3: If $y = \log_e(\log_e x)$ for $x > 1$, then $\frac{dy}{dx}$ is:
A $\frac{1}{x \log_e x}$
B $\frac{1}{x}$
C $\frac{\log_e x}{x}$
D $\frac{1}{(\log_e x)^2}$
Derivation & Analysis:

Let $u = \log_e x$. Then $y = \log_e u$. By chain rule, $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = \frac{1}{u} \cdot \frac{1}{x} = \frac{1}{x \log_e x}.

Q4
MCQ 4: If $y = (x + \sqrt{1+x^2})^n$, then $(1+x^2)\left(\frac{dy}{dx}\right)^2$ is equal to:
A $n^2 y^2$
B $n y^2$
C $n^2 y$
D $\frac{n^2 y^2}{1+x^2}$
Derivation & Analysis:

Differentiate $y$ w.r.t $x$: $\frac{dy}{dx} = n(x+\sqrt{1+x^2})^{n-1} \cdot (1 + \frac{x}{\sqrt{1+x^2}}) = \frac{n y}{\sqrt{1+x^2}}$. Squaring both sides yields $(1+x^2)(\frac{dy}{dx})^2 = n^2 y^2$.

Q5
MCQ 5: The derivative of $f(x) = |x|^3$ at $x = 0$ is:
A 1
B 0
C Does not exist
D -1
Derivation & Analysis:

Expressed as a piecewise function, $f(x) = x^3$ for $x \ge 0$ and $-x^3$ for $x < 0$. Both left and right derivatives at $x=0$ evaluate to $0$, hence $f'(0) = 0$ despite the absolute value structure.

Q6
MCQ 6: If $y = \sin^{-1}(3x - 4x^3)$, then $\frac{dy}{dx}$ for $|x| < \frac{1}{2}$ is:
A $\frac{3}{\sqrt{1-x^2}}$
B $-\frac{3}{\sqrt{1-x^2}}$
C $\frac{3}{1+x^2}$
D $\frac{1}{\sqrt{1-x^2}}$
Derivation & Analysis:

Substitute $x = \sin \theta$. Then $3x - 4x^3 = \sin(3\theta)$, so $y = 3\theta = 3\sin^{-1}x$. Differentiating gives $\frac{3}{\sqrt{1-x^2}}$.

Q7
MCQ 7: If $f(g(x)) = x$ and $g'(x) = \frac{1}{1 + x^3}$, then $f'(x)$ evaluated at $g(x)$ is:
A $1 + [f(x)]^3$
B $1 + x^3$
C $\frac{1}{1 + x^3}$
D $3x^2$
Derivation & Analysis:

By the inverse function derivative property and chain rule, $f'(g(x)) \cdot g'(x) = 1 \implies f'(g(x)) = \frac{1}{g'(x)} = 1 + (g(x))^3$. Replacing $g(x)$ with $x$ gives $1 + f(x)^3$, matching standard inverse relations.

Q8
MCQ 8: If $y = \sqrt{\frac{1-\cos 2x}{1+\cos 2x}}$, then $\frac{dy}{dx}$ at $x = \frac{\pi}{4}$ is:
A 0
B 2
C 1
D Undefined
Derivation & Analysis:

Simplify using trigonometric identities: $\sqrt{\frac{2\sin^2 x}{2\cos^2 x}} = \tan x$ (for acute $x$). Thus $y = \tan x$, and $\frac{dy}{dx} = \sec^2 x$. At $x = \pi/4$, $\sec^2(\pi/4) = (\sqrt{2})^2 = 2$.

Q9
MCQ 9: If $y = \ln(\sec x + \tan x)$, then $\frac{dy}{dx}$ is:
A $\sec x$
B $\tan x$
C $\sec x \tan x$
D $\frac{1}{\sec x}$
Derivation & Analysis:

By chain rule, $\frac{dy}{dx} = \frac{\sec x \tan x + \sec^2 x}{\sec x + \tan x} = \frac{\sec x(\tan x + \sec x)}{\sec x + \tan x} = \sec x$.

Q10
MCQ 10: If $f(x) = \log_{10} x$, then $f'(x)$ is equal to:
A $\frac{1}{x}$
B $\frac{1}{x \log_e 10}$
C $\frac{\log_e 10}{x}$
D $\frac{10}{x}$
Derivation & Analysis:

Change base formula: $f(x) = \frac{\ln x}{\ln 10}$. Differentiating gives $\frac{1}{x \ln 10}$.

Common questions

Almost all advanced calculus problems in UP PGT/TGT examinations involve composite functions rather than isolated elementary functions. Without the chain rule, evaluating derivatives of nested trigonometric, exponential, and logarithmic functions would be algebraically intractable.

Candidates frequently miss intermediate composite layers (such as forgetting to differentiate the argument of an exponential or trigonometric term) or confuse the chain rule with the algebraic product rule.

Candidates should master standard trigonometric and logarithmic substitutions before differentiating, as simplifying the expression algebraically reduces the chain rule depth from three layers to one or two.

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