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Differentiation and Higher Order Derivatives PYQs for GATE DA

Solve 2+ Differentiation and Higher Order Derivatives previous year questions for GATE DA with answers and detailed solutions. Free sample questions below.

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Question 1

2025 PYQ

Level 3: Exam Standard

Let f (x) = \frac{e ^{x} - e ^{- x}}{2}, x \in R . Let f^{(k)} (a) denote the k^{t h} derivative of f evaluated at a . What is the value of f^{(10)} (0) ? (Note: ! denotes factorial)

Question 2

2024 PYQ

Level 3: Exam Standard

Let f : R \to R be the function f (x) = \frac{1}{1 + e ^{- x}} . The value of the derivative of f at x where f (x) = 0.4 is ______(rounded off to two decimal places). Note: R denotes the set of real numbers.

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Differentiation and Higher Order Derivatives PYQs for GATE DA

Solve 2+ Differentiation and Higher Order Derivatives previous year questions for GATE DA with answers and detailed solutions. Free sample questions below.

Chapter Roadmap: Differentiation and Higher Order Derivatives

Chapter Journey

Derivative Evaluation

Current Topic • Foundation

Rules of Differentiation

Product, Quotient, Chain Rule

Higher Order Derivatives

Leibniz Theorem, Patterns

Applications

Maxima/Minima, Mean Value Theorems

Goal: Memorize and intuitively understand the derivatives of basic building blocks (

e^{x}

sin x

ln x

) so you never hesitate when applying complex rules later.

The Hero Concept: What is a Derivative Really?

Intuition: The Instantaneous Slope

The derivative of a function $f (x)$ , denoted as $f^{'} (x)$ or $\frac{d y}{d x}$ , represents:

Geometric Meaning: The slope of the tangent to the curve $y = f (x)$ at point $x$ .
Physical Meaning: The rate of change of $y$ with respect to $x$ .

f^{'} (x) = h \to 0 lim \frac{f ( x + h ) - f ( x )}{h}

Why memorize standard forms?
In exams like GATE DA, you rarely use the limit definition directly. You use standard results as building blocks. If you know

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

by heart, you save critical seconds during complex chain rule problems.

Differentiation and Higher Order Derivatives: Solved Questions with Step-by-Step Explanations (2 Problems)

Question 1 · Calculus and Optimization · 2025 MCQ

Let $f (x) = \frac{e ^{x} - e ^{- x}}{2}$ , $x \in R$ . Let $f^{(k)} (a)$ denote the $k^{t h}$ derivative of $f$ evaluated at $a$ . What is the value of $f^{(10)} (0)$ ? (Note: ! denotes factorial)

A.
$0$
B.
$1$
C.
$\frac{1}{10 !}$
D.
$\frac{2}{10 !}$

Correct Answer:

Step-by-Step Solution

Insight: The function is the hyperbolic sine, $sinh (x)$ , whose derivatives cycle every two steps.

Exam route: Recognize $f (x) = sinh (x)$ . The $n$ -th derivative of $sinh (x)$ is $sinh (x)$ for even $n$ and $cosh (x)$ for odd $n$ . Since 10 is even, $f^{(10)} (x) = sinh (x)$ . Evaluate at $x = 0$ to get $sinh (0) = 0$ .

Learning route:

Write $f (x) = \frac{1}{2} e^{x} - \frac{1}{2} e^{- x}$ .

Differentiate 10 times using the rule $\frac{d ^{n}}{d x ^{n}} e^{a x} = a^{n} e^{a x}$ .

$f^{(10)} (x) = \frac{1}{2} (1)^{10} e^{x} - \frac{1}{2} (- 1)^{10} e^{- x} = \frac{1}{2} e^{x} - \frac{1}{2} e^{- x} = f (x)$ .

Substitute $x = 0$ : $f^{(10)} (0) = f (0) = \frac{e ^{0} - e ^{0}}{2} = 0$ .

Question 2 · Calculus and Optimization · 2024 NAT

Let

f : R \to R

be the function

f (x) = \frac{1}{1 + e ^{- x}}

.
The value of the derivative of

f

x

where

f (x) = 0.4

is ______
(rounded off to two decimal places).
Note:

R

denotes the set of real numbers.

Correct Answer:

0.24

Step-by-Step Solution

Insight: The derivative of the sigmoid function $f (x)$ can be expressed entirely in terms of $f (x)$ itself: $f^{'} (x) = f (x) (1 - f (x))$ .

Exam route: Use the identity $f^{'} (x) = f (x) (1 - f (x))$ . Substitute $f (x) = 0.4$ to get $0.4 \times 0.6 = 0.24$ .

Learning route:

Given $f (x) = (1 + e^{- x})^{- 1}$ .

Using the chain rule, $f^{'} (x) = - 1 (1 + e^{- x})^{- 2} \cdot (- e^{- x}) = \frac{e ^{- x}}{( 1 + e ^{- x} ) ^{2}}$ .

Rewrite this as $\frac{1}{1 + e ^{- x}} \cdot \frac{e ^{- x}}{1 + e ^{- x}}$ .

Notice that $\frac{e ^{- x}}{1 + e ^{- x}} = \frac{1 + e ^{- x} - 1}{1 + e ^{- x}} = 1 - \frac{1}{1 + e ^{- x}} = 1 - f (x)$ .

Thus, $f^{'} (x) = f (x) (1 - f (x))$ .

Given $f (x) = 0.4$ , we have $f^{'} (x) = 0.4 (1 - 0.4) = 0.4 \times 0.6 = 0.24$ .

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Unlock the whole course

Differentiation and Higher Order Derivatives PYQs for GATE DA

Chapter Roadmap: Differentiation and Higher Order Derivatives

Chapter Journey

The Hero Concept: What is a Derivative Really?

Intuition: The Instantaneous Slope

Differentiation and Higher Order Derivatives: Solved Questions with Step-by-Step Explanations (2 Problems)

Step-by-Step Solution

Step-by-Step Solution

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