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2. Calculus — Differential Calculus

Leibnitz theorem

Leibnitz's theorem (or Leibniz rule) provides a systematic formula for the $n$-th derivative of the product of two functions, generalizing the product rule through binomial coefficients.

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

Leibnitz's Theorem for $n$-th Derivative of a Product
Let $u(x)$ and $v(x)$ be two functions possessing derivatives up to order $n$ in a given interval. Then the $n$-th derivative of their product $y = uv$ is given by $$\frac{d^n}{dx^n}(uv) = uv_n + \binom{n}{1}u_1v_{n-1} + \binom{n}{2}u_2v_{n-2} + \dots + \binom{n}{r}u_rv_{n-r} + \dots + u_nv$$, where subscripts denote the order of differentiation, i.e., $u_r = \frac{d^r u}{dx^r}$ and $v_r = \frac{d^r v}{dx^r}$, with $u_0 = u$ and $v_0 = v$.

📐 Geometric Interpretation

While primarily an algebraic formula for successive differentiation, it forms the analytical backbone for establishing asymptotic expansions of curves, curvature vectors, and differential equations governing oscillatory motion.

⚙️ Algebraic & Structural Meaning

It establishes a direct isomorphism between the index shift operator in differential rings and the binomial expansion coefficients, facilitating the reduction of linear differential equations with polynomial coefficients into algebraic recurrence relations.

Conditions and hypotheses

Check each condition and what changes when it is omitted:

Hypothesis / Condition Domain / Interval Necessity What Fails If Condition Dropped
Existence of $n$-th derivatives ($u^{(n)}$ and $v^{(n)}$) (a, b) (Open interval) Strictly Mandatory If either function fails to be $n$ times differentiable at any point in the interval, the formula yields undefined terms or false values.
Continuity of lower-order derivatives [a, b] (Closed interval) Strictly Mandatory Discontinuities in derivatives $u_r$ or $v_r$ (for $r < n$) violate the foundational premises of successive application of product rules.
Domain Compatibility Common Domain Subspace Essential for conclusion If the domains of $u$ and $v$ do not overlap smoothly, the product function is not differentiable to order $n$ over the intended interval.

Useful results and shortcuts

Use these identities and results when their conditions apply:

01

Polynomial-Exponential Product ($x^n e^{ax}$)

$$ y_n = e^{ax} \sum_{r=0}^{n} \binom{n}{r} \frac{d^r}{dx^r}(x^n) a^{n-r} $$

Rule: Choose $u = e^{ax}$ and $v = x^n$ because successive derivatives of polynomials terminate rapidly.

Application: Frequently asked in UP PGT to find $y_n$ at $x=0$.
02

Polynomial-Trigonometric Product ($x^n \sin(ax)$ or $x^n \cos(ax)$)

$$ \text{If } y = x^n \sin(ax), \text{ use } v = x^n \text{ and } u = \sin(ax) \text{ where } u_n = a^n \sin\left(ax + \frac{n\pi}{2}\right) $$

Rule: Isolate the trigonometric part as $u$ since its general $n$-th derivative has a closed-form cyclic formula.

Application: Standard pattern in DSSSB PGT Mathematics.
03

Reduction to Second-Order Differential Equations

$$ (1-x^2)y_{n+2} - (2n+1)xy_{n+1} - (n^2 - m^2)y_n = 0 $$

Rule: Applied directly to Chebyshev and Legendre related forms like $y = \cos(m \cos^{-1} x)$ after differentiating twice and applying Leibnitz theorem.

04

Evaluation at Origin ($x = 0$)

$$ \left. \frac{d^n}{dx^n}(uv) \right|_{x=0} = \sum_{r=0}^{n} \binom{n}{r} u_r(0) v_{n-r}(0) $$

Rule: Many terms vanish when $x=0$ due to polynomial properties (e.g., $x^k$ derivatives at $x=0$ are zero for $k < n$).

05

Logarithmic and Inverse Trigonometric Products

$$ \text{For } y = (\sin^{-1} x)^2, \text{ show } (1-x^2)y_2 - xy_1 - 2 = 0 \text{ and apply Leibnitz to get } (1-x^2)y_{n+2} - (2n+1)xy_{n+1} - (n^2)y_n = 0 $$

Rule: Differentiate once or twice, clear denominators, and apply Leibnitz theorem immediately to generate recurrence relations.

06

Rapid Option Elimination via Degree Checking

$$ \text{Degree of polynomial coefficients in Leibnitz recurrence matches highest derivative indices.} $$

Rule: Check the coefficient of $y_{n+2}$ (usually $1-x^2$ or $x^2$) and indices of $y_{n+1}, y_n$ to eliminate 2 incorrect options instantly.

Common errors and counterexamples

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

⚠️ COMMON ERROR #01

Incorrect Assignment of $u$ and $v$

Common Pitfall: Assigning $u$ to a polynomial that does not terminate in $n$ derivatives, making calculation infinitely long.

Counterexample: \text{Let } y = x^n \ln x, \text{ taking } u = \ln x \text{ instead of } v = \ln x

Why it fails: Logarithmic derivatives do not terminate, whereas polynomial derivatives do. Always assign the polynomial to $v$ when its degree is $\le n$.

🛡️ Exam Defense: Always assign the terminating polynomial to $v$ and the transcendental function to $u$.
⚠️ COMMON ERROR #02

Index Shift Error in Recurrence Relations

Common Pitfall: Forgetting that applying Leibnitz theorem to $y_{k}$ increases all derivative subscripts by $n$.

Counterexample: \text{Writing } y_{n+1} \text{ instead of } y_{n+2} \text{ when differentiating } (1-x^2)y_1 - xy = 0 \text{ $n$ times.}

Why it fails: Results in mismatched recurrence indices and wrong answers.

🛡️ Exam Defense: Remember: differentiating $y_1$ $n$ times gives $y_{n+1}$, and differentiating $(1-x^2)y_1$ gives $(1-x^2)y_{n+1} - 2nx y_n - n(n-1)y_{n-1}$.

Comparison

Concept / TheoremHypotheses / AssumptionsKey Formula / OutcomeGeometric MeaningExam Identification Clue
Leibnitz TheoremFunctions $u, v$ $n$-times differentiable$(uv)_n = \sum \binom{n}{r} u_r v_{n-r}$Higher-order derivative product expansionKeywords: 'Find $n$-th derivative of product'
Taylor's Theorem$f(x)$ $n$-times differentiable on $(a, b)$$f(x) = \sum \frac{f^{(k)}(a)}{k!}(x-a)^k + R_n$Polynomial approximation of curvesKeywords: 'Expansion about point a'
Maclaurin's Theorem$f(x)$ differentiable at neighborhood of $0$$f(x) = f(0) + x f'(0) + \frac{x^2}{2!} f''(0) + \dots$Polynomial approximation around originKeywords: 'Series expansion of function'
Rolle's TheoremContinuous on $[a,b]$, Differentiable on $(a,b)$, $f(a)=f(b)$$f'(c) = 0$ for some $c \in (a,b)$Horizontal tangent existenceKeywords: 'Derivative equals zero between roots'

Solved examples

Each problem includes a full solution and a shorter method:

CANONICAL PROBLEM 01 Dual-Method Breakdown

If $y = e^{m \sin^{-1} x}$, prove that $(1-x^2)y_{n+2} - (2n+1)xy_{n+1} - (n^2 - m^2)y_n = 0$. What is the value of $y_n(0)$ when $n$ is even?

(A) $m^2(m^2-2^2)(m^2-4^2)\dots(m^2-(n-2)^2)$(B) $0$(C) $m^n$(D) $m(m^2-1^2)\dots(m^2-(n-1)^2)$
✓ Correct Answer: (A)
📚 Method A: Formal Textbook Derivation
Step 1: Differentiate $y = e^{m \sin^{-1} x}$ once to get $\sqrt{1-x^2} y_1 = m e^{m \sin^{-1} x}$. Step 2: Square both sides: $(1-x^2)y_1^2 = m^2 y^2$. Step 3: Differentiate again to get $(1-x^2)y_2 - xy_1 - m^2 y = 0$. Step 4: Apply Leibnitz theorem $n$ times to obtain the recurrence relation. Step 5: Substitute $x=0$ to find $y_n(0)$ for even $n$.
⚡ Method B: 25-Second Exam Shortcut
Recognize the standard Chebyshev-type recurrence relation and immediately extract the product structure for $x=0$ where odd derivatives vanish.
💡 Tip: Memorize the recurrence relations for $y = \sin(m \sin^{-1}x)$ and $y = \cos(m \cos^{-1}x)$ as they appear verbatim in UP PGT.
CANONICAL PROBLEM 02 Dual-Method Breakdown

If $y = (x^2 - 1)^n$, then the value of $y_n(1)$ (the $n$-th derivative at $x=1$) is:

(A) $2^n n!$(B) $n!$(C) $0$(D) $2^n$
✓ Correct Answer: (A)
📚 Method A: Formal Textbook Derivation
Step 1: Write $y = (x-1)^n (x+1)^n$. Step 2: Apply Leibnitz theorem to find the $n$-th derivative at $x=1$. Step 3: All terms containing $(x-1)$ to a positive power vanish at $x=1$. Step 4: The only non-vanishing term corresponds to differentiating $(x-1)^n$ $n$ times and evaluating $(x+1)^n$ at $x=1$, yielding $n! \cdot 2^n$.
⚡ Method B: 25-Second Exam Shortcut
Rodrigues' formula standard result: for $y = (x^2-1)^n$, $y_n(1) = 2^n n!$.
💡 Tip: Direct textbook result frequently used in Legendre polynomial properties.

Practice questions

Choose an answer to check it and see the solution:

Q1
If $y = x^3 \log x$, then $y_4$ is equal to:
A $\frac{6}{x}$
B $\frac{2}{x}$
C $\frac{24}{x}$
D 6x
Derivation & Analysis:

Using Leibnitz theorem with $u = \log x$ and $v = x^3$. Note that $v_1 = 3x^2, v_2 = 6x, v_3 = 6, v_4 = 0$. Also $u_1 = 1/x, u_2 = -1/x^2, u_3 = 2/x^3, u_4 = -6/x^4$. Summing terms gives $\frac{6}{x}$.

Q2
If $y = \sin(m \sin^{-1} x)$, then $(1-x^2)y_{n+2} - (2n+1)xy_{n+1}$ equals:
A $(n^2 - m^2)y_n$
B $-(n^2 - m^2)y_n$
C $(m^2 - n^2)y_n$
D 0
Derivation & Analysis:

This is the standard Leibnitz reduction formula for trigonometric inverse substitutions, resulting in $(n^2 - m^2)y_n$.

Q3
The $n$-th derivative of $x^n \log x$ at $x=0$ does not exist because:
A Logarithm is undefined at $0$
B Polynomial vanishes
C Derivatives of order $\ge 1$ of $\log x$ blow up at $0$
D It is indeterminate
Derivation & Analysis:

Derivatives of $\log x$ contain negative powers of $x$ (e.g., $-1/x^2, 2/x^3$), which diverge as $x \to 0$.

Q4
If $y = e^{ax} \cos(bx)$, what is $y_1$?
A $e^{ax}$[a $\cos$(bx) - b $\sin$(bx)]
B $e^{ax}$[a $\cos$(bx) + b $\sin$(bx)]
C $ae^{ax} \cos(bx)$
D $be^{ax} \sin(bx)$
Derivation & Analysis:

Direct application of product rule / Leibnitz theorem of order 1.

Q5
In Leibnitz formula for $(uv)_n$, the coefficient of $u_{n-1}v_1$ is:
A n
B $\binom{n}{2}$
C 1
D n!
Derivation & Analysis:

By binomial theorem expansion matching, the coefficient is $\binom{n}{1} = n$.

Q6
If $y = (x^2 - 1)^n$, then $y_{2n}$ is:
A (2n)!
B n!
C $2^n n!$
D 0
Derivation & Analysis:

Degree of $y$ is $2n$, so its $(2n)$-th derivative is a constant equal to $(2n)!$.

Q7
If $y = x^n$, then $y_{n+1}$ is:
A n!
B 0
C 1
D x
Derivation & Analysis:

The $(n+1)$-th derivative of a polynomial of degree $n$ is identically zero.

Q8
Leibnitz theorem is an extension of which standard differentiation rule?
A Chain rule
B Product rule
C Quotient rule
D L'Hopital's rule
Derivation & Analysis:

Leibnitz theorem generalizes the product rule $(uv)' = u'v + uv'$ to higher-order derivatives.

Q9
If $y = x e^x$, then $n$-th derivative $y_n(0)$ is:
A n
B n!
C $n 2^{n-1}$
D 1
Derivation & Analysis:

Using Leibnitz theorem: $y_n = e^x x + n e^x (1) = e^x(x + n)$. At $x = 0$, $y_n(0) = n$.

Q10
If $(1-x^2)y_2 - xy_1 = 0$, then by applying Leibnitz theorem $n$ times, the term corresponding to $y_{n+1}$ has coefficient:
A -(2n+1)x
B -2nx
C -xy
D -(n+1)x
Derivation & Analysis:

Differentiating $-xy_1$ $n$ times gives $-\left[ x y_{n+1} + \binom{n}{1}(1)y_n \right] = -xy_{n+1} - ny_n$, combining with the derivative of $(1-x^2)y_2$ yields total coefficient $-(2n+1)x$ for $y_{n+1}$.

Common questions

Always choose $v$ to be the algebraic polynomial function whose derivatives eventually vanish (if it is of finite degree), and $u$ to be the transcendental function (like $e^{ax}$, $\sin ax$, or $\log x$) whose $n$-th derivative follows a predictable general formula.

Arithmetic slips in binomial coefficients $\binom{n}{r}$ and incorrect index tracking when differentiating recurrence relations $n$ times.

No. To apply Leibnitz theorem to order $n$, both functions must possess derivatives up to order $n$ in the specified interval.

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