2. Calculus — Differential Calculus

Successive differentiations

Successive differentiation involves finding the higher-order derivatives ($y_2, y_3, \dots, y_n$) of a function by repeated application with respect to an independent variable. This topic covers standard formulae, Leibniz's theorem for the product of functions, and implicit differentiation.

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

📖 Topic progress

○ Mark as complete

Definition and statement

Leibniz's Theorem for $n$-th Derivative of a Product

If $u$ and $v$ are two functions of $x$ such that their $n$-th derivatives $u_n$ and $v_n$ exist, then the $n$-th derivative of their product $uv$ is given by $$(uv)_n = u_n v + {n \choose 1} u_{n-1} v_1 + {n \choose 2} u_{n-2} v_2 + \dots + {n \choose r} u_{n-r} v_r + \dots + u v_n = \sum_{r=0}^{n} {n \choose r} u_{n-r} v_r$$

📐 Geometric Interpretation

Extends the concept of tangent slopes and curvature to higher rates of change, characterizing inflection points, concavity, and local oscillatory behavior of composite curves.

⚙️ Algebraic & Structural Meaning

Mirrors the binomial theorem expansion, bridging combinatorial coefficients with differential operators, and facilitates reduction formulae for differential equations.

Conditions and hypotheses

Check each condition and what changes when it is omitted:

Hypothesis / Condition	Domain / Interval	Necessity	What Fails If Condition Dropped
$n$-th Differentiability of factors	`(a, b) (Open interval)`	Strictly Mandatory / Necessary	If either function fails to possess continuous derivatives up to order $n$, Leibniz's expansion collapses and higher-order terms are undefined.
Existence of lower-order derivatives	`[a, b] (Closed interval)`	Strictly Mandatory	Discontinuities in any intermediate derivative $u_k$ ($k < n$) invalidate the summation structure of Leibniz's theorem.
Appropriate choice of $u$ and $v$	`Domain of definition`	Essential for simplification	Choosing $v$ such that its higher-order derivatives do not terminate (e.g., non-polynomial functions) leads to infinite series instead of a finite closed-form solution.

Useful results and shortcuts

Use these identities and results when their conditions apply:

Standard $n$-th Derivative of Exponential-Trigonometric Products

$$ \frac{d^n}{dx^n}[e^{ax} \sin(bx + c)] = r^n e^{ax} \sin(bx + c + n\theta) $$ where $r = \sqrt{a^2 + b^2}$ and $\theta = \tan^{-1}\left(\frac{b}{a}\right)$

Rule: Substitute $a, b, c$ directly into the amplitude-phase form without applying product rule $n$ times.

$n$-th Derivative of $e^{ax} \cos(bx + c)$

$$ \frac{d^n}{dx^n}[e^{ax} \cos(bx + c)] = r^n e^{ax} \cos(bx + c + n\theta) $$ where $r = \sqrt{a^2 + b^2}$ and $\theta = \tan^{-1}\left(\frac{b}{a}\right)$

Rule: Directly replace sine with cosine in the amplitude-phase formula.

$n$-th Derivative of $(ax + b)^m$ when $m = n$

$$ \frac{d^n}{dx^n}[(ax + b)^n] = n! \cdot a^n $$

Rule: If degree equals order of differentiation, the result is a constant equal to $n!$ times $a^n$. For $m < n$, the derivative is $0$.

Application: Quick evaluation trick for polynomial expansion terms.

$n$-th Derivative of $\log(ax + b)$

$$ \frac{d^n}{dx^n}[\log(ax + b)] = \frac{(-1)^{n-1} (n-1)! a^n}{(ax + b)^n} $$

Rule: Memorize the sign alternation $(-1)^{n-1}$, factorial $(n-1)!$, and power of $a$ for rapid option matching.

Application: Directly appears in UP PGT paper matching questions.

$n$-th Derivative of $\frac{1}{ax + b}$

$$ \frac{d^n}{dx^n}\left[\frac{1}{ax + b}\right] = \frac{(-1)^n n! a^n}{(ax + b)^{n+1}} $$

Rule: Note the $(-1)^n$ factor and $n!$ numerator for rational function shortcuts.

Maclaurin Series Coefficient Extraction

$$ f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!} x^n $$

Rule: To find $f^{(n)}(0)$, expand the function using standard binomial or exponential series and read off the coefficient of $x^n$, then multiply by $n!$.

Application: Saves 3-4 minutes on Taylor/Maclaurin derivative evaluation questions.

Common errors and counterexamples

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

⚠️ COMMON ERROR #01

Confusing Sine and Cosine Phase Angles in Exponential Products

Common Pitfall: Applying the cosine formula when the function is sine, or forgetting to add $n\theta$ to the phase angle.

Counterexample: For $y = e^x \sin x$, $y_1 = \sqrt{2} e^x \sin\left(x + \frac{\pi}{4}\right)$, not $\cos\left(x + \frac{\pi}{4}\right)$.

Why it fails: Misidentifying trigonometric phase shifts alters the correct quadrant evaluation.

🛡️ Exam Defense: Always check function type: sine retains sine with phase addition $n\theta$; cosine retains cosine.

⚠️ COMMON ERROR #02

Sign Alternation Errors in Logarithmic and Reciprocal Derivatives

Common Pitfall: Dropping or misplacing the $(-1)^{n-1}$ or $(-1)^n$ factor in logarithmic and rational derivatives.

Counterexample: For $\frac{d^2}{dx^2}[\log x] = -\frac{1}{x^2}$, candidate writes $+\frac{1}{x^2}$.

Why it fails: Leads to sign mismatch with multiple-choice options.

🛡️ Exam Defense: Test for $n=1$ or $n=2$ mentally to verify the sign of $(-1)^{n-1}$ before finalizing the formula.

Comparison

Concept / Theorem	Hypotheses / Assumptions	Key Formula / Outcome	Geometric Meaning	Exam Identification Clue
Leibniz's Theorem	Functions $u, v$ possess $n$-th derivatives	$(uv)_n = \sum {n \choose r} u_{n-r}v_r$	Higher-order curvature of products	Product of polynomial and trigonometric/exponential
Standard Exponential-Trigo	Real constants $a, b$, valid on $\mathbb{R}$	$r^n e^{ax} \sin(bx + c + n\theta)$	Scaled and phase-shifted oscillations	Expression contains $e^{ax} \sin(bx)$ or $e^{ax} \cos(bx)$
Logarithmic Expansion	$ax + b > 0$	$\frac{(-1)^{n-1}(n-1)!a^n}{(ax+b)^n}$	Decaying rate of logarithmic growth	Logarithmic functions requiring $n$-th derivative
Maclaurin Series Method	Function analytic at $x = 0$	$f^{(n)}(0) = n! \times (\text{coeff of } x^n)$	Polynomial approximation near origin	Evaluation of $f^{(n)}(0)$ for composite functions

Solved examples

Each problem includes a full solution and a shorter method:

CANONICAL PROBLEM 01 Dual-Method Breakdown

If $y = e^{a \sin^{-1} x}$, then the value of $(1 - x^2)y_2 - x y_1$ is:

(A) $a^2 y$(B) $-a^2 y$(C) $a y$(D) 0

📚 Method A: Formal Textbook Derivation

Step 1: Differentiate $y = e^{a \sin^{-1} x}$ once to get $y_1 = \frac{a e^{a \sin^{-1} x}}{\sqrt{1 - x^2}}$. Step 2: Cross-multiply: $\sqrt{1 - x^2} y_1 = a y$. Step 3: Square both sides: $(1 - x^2)y_1^2 = a^2 y^2$. Step 4: Differentiate with respect to $x$: $(1 - x^2)2y_1 y_2 - 2x y_1^2 = a^2 2y y_1$. Step 5: Cancel $2y_1$ to obtain $(1 - x^2)y_2 - x y_1 = a^2 y$.

⚡ Method B: 25-Second Exam Shortcut

Recognize this as the standard recurrence relation for Chebyshev-type differential equations where $(1-x^2)y_2 - xy_1 = n^2 y$. Here $n=a$, hence result is $a^2 y$.

💡 Tip: Memorize standard recurrence relations for $e^{a \sin^{-1} x}$ and $(\sin^{-1} x)^2$ to solve.

CANONICAL PROBLEM 02 Dual-Method Breakdown

The $n$-th derivative of $y = \frac{1}{x^2 - a^2}$ at $x = 0$ (for even $n$) is:

(A) $\frac{n!}{a^{n+2}}$(B) $\frac{n!}{a^n}$(C) 0(D) $\frac{2 n!}{a^{n+2}}$

📚 Method A: Formal Textbook Derivation

Step 1: Resolve into partial fractions: $y = \frac{1}{2a}\left[\frac{1}{x-a} - \frac{1}{x+a}\right]$. Step 2: Write the $n$-th derivative: $y_n = \frac{(-1)^n n!}{2a} \left[\frac{1}{(x-a)^{n+1}} - \frac{1}{(x+a)^{n+1}}\right]$. Step 3: Evaluate at $x=0$: $y_n(0) = \frac{(-1)^n n!}{2a} \left[\frac{1}{(-a)^{n+1}} - \frac{1}{a^{n+1}}\right]$. Step 4: For even $n$, $(-1)^n = 1$ and $(-a)^{n+1} = -a^{n+1}$, leading to cancellation yielding 0.

⚡ Method B: 25-Second Exam Shortcut

Since $\frac{1}{x^2-a^2}$ is an even function, its odd derivatives at $x=0$ are non-zero and even derivatives at $x=0$ are zero.

💡 Tip: Apply parity (even/odd function property) before performing tedious expansions.

Practice questions

Choose an answer to check it and see the solution:

MCQ 1: If $y = \sin(ax + b)$, then the $n$-th derivative $y_n$ is given by:

A $a^n \sin(ax + b)$

B $a^n \sin\left(ax + b + \frac{n\pi}{2}\right)$

C $n! \sin(ax + b)$

D $a^n \cos(ax + b)$

Derivation & Analysis:

Since $\frac{d}{dx}[\sin(ax+b)] = a \cos(ax+b) = a \sin(ax+b+\frac{\pi}{2})$, repeating $n$ times gives $a^n \sin(ax+b+\frac{n\pi}{2})$.

MCQ 2: If $y = (x^2 - 1)^n$, then the $n$-th derivative at $x = 1$, i.e., $y_n(1)$, by Rodrigues' formula is:

A $2^n n!$

B $n!$

C 0

D $n^n$

Derivation & Analysis:

Rodrigues' formula for Legendre polynomials gives $P_n(x) = \frac{1}{2^n n!} \frac{d^n}{dx^n}[(x^2-1)^n]$, and evaluating at specific points yields $2^n n!$ for the $n$-th derivative.

MCQ 3: The $n$-th derivative of $\log(ax + b)$ is:

A $\frac{(n-1)! a^n}{(ax+b)^n}$

B $\frac{(-1)^n (n-1)! a^n}{(ax+b)^n}$

C $\frac{(-1)^{n-1} (n-1)! a^n}{(ax+b)^n}$

D $\frac{n! a^n}{(ax+b)^{n+1}}$

Derivation & Analysis:

First derivative is $\frac{a}{ax+b}$, second is $-\frac{a^2}{(ax+b)^2}$, third is $\frac{2! a^3}{(ax+b)^3}$, generalizing to $\frac{(-1)^{n-1}(n-1)!a^n}{(ax+b)^n}$.

MCQ 4: If $y = e^{mx} + e^{-mx}$, then $y_2 - m^2 y$ equals:

A $m^2 y$

B $2m^2 y$

C 0

D $m y$

Derivation & Analysis:

Here $y_1 = m(e^{mx} - e^{-mx})$ and $y_2 = m^2(e^{mx} + e^{-mx}) = m^2 y$. Thus $y_2 - m^2 y = 0$.

MCQ 5: If $x = \sin t$ and $y = \sin pt$, then $(1-x^2)y_2 - x y_1$ is equal to:

A $p^2 y$

B $-p^2 y$

C $p y$

D 0

Derivation & Analysis:

This is a standard transformation leading to $p^2 y$ with negative sign due to $\frac{d^2 y}{dx^2} = -p^2 y$ under substitution.

MCQ 6: The $n$-th derivative of $x^n$ is:

A $0$

B $n!$

C $(n-1)!$

D $nx^{n-1}$

Derivation & Analysis:

Differentiating $x^n$ exactly $n$ times reduces the polynomial to a constant $n!$.

MCQ 7: If $y = \tan^{-1} x$, then $y_1(0)$ and $y_2(0)$ respectively are:

A $1, 0$

B $0, 1$

C $-1, 0$

D $0, -1$

Derivation & Analysis:

At $x=0$, $\tan^{-1}(0) = 0$, $y_1 = \frac{1}{1+x^2} \implies y_1(0) = 1$, and $y_2 = \frac{-2x}{(1+x^2)^2} \implies y_2(0) = 0$.

MCQ 8: If $y = \frac{x}{x^2 + a^2}$, then the $n$-th derivative at $x = 0$ for even $n$ is:

A $0$

B $\frac{n!}{a^{n+1}}$

C $(-1)^{n/2} \frac{n!}{a^n}$

D $\frac{(-1)^{n/2} n!}{a^{n+1}}$

Derivation & Analysis:

The function $f(x) = \frac{x}{x^2+a^2}$ is an odd function. Hence all its even-order derivatives at $x=0$ are zero.

MCQ 9: Using Leibniz theorem, if $u = x^2$ and $v = e^x$, the 3rd derivative $(uv)_3$ is:

A $(x^2 + 6x + 6)e^x$

B $(x^2 + 3x + 3)e^x$

C $(x^3 + 3x^2 + 6)e^x$

D $e^x(x^2 + 2x)$

Derivation & Analysis:

Applying Leibniz: $u_3 v + 3 u_2 v_1 + 3 u_1 v_2 + u v_3$. Here $u=x^2$, $u_1=2x$, $u_2=2$, $u_3=0$, and $v_k = e^x$. Summing gives $0 \cdot e^x + 3(2)e^x + 3(2x)e^x + x^2 e^x = (x^2 + 6x + 6)e^x$.

Q10

MCQ 10: If $y = a \cos(\log x) + b \sin(\log x)$, then $x^2 y_2 + x y_1$ is:

A $y$

B $-y$

C $0$

D $y^2$

Derivation & Analysis:

Differentiating $y = a \cos(\log x) + b \sin(\log x)$ twice and multiplying by $x^2$ and $x$ respectively yields $x^2 y_2 + x y_1 = -y$.

Common questions

Always choose $u$ to be the algebraic or polynomial function whose higher-order derivatives eventually become zero, and $v$ to be the transcendental function (like $e^x$, $\sin x$, or $\cos x$) whose derivatives follow a repeating cyclic pattern.

Yes. Standard relations involving $y_n, y_{n+1}, y_{n+2}$ derived from functions like $e^{a \sin^{-1} x}$ or $(\sin^{-1} x)^2$ appear in at least 2 to 3 questions every year.

Memorize standard $n$-th derivative formulae for exponential-trigonometric products, logarithmic functions, and rational fractions, and utilize parity (even/odd function properties) to eliminate options instantly.

← Previous Topic

Rolle's theorem

Next Topic →

Equation of tangent and normal