Differentiation and Higher Order Derivatives Practice Questions for GATE DA

Solve 79+ Differentiation and Higher Order Derivatives practice 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 (10 Problems)

Question 1 · Calculus and Optimization NAT

How many of the following functions have a derivative that is a rational function (a ratio of two polynomials)?

f (x) = x^{- 2}

g (x) = x^{1/2}

h (x) = x^{- 3/2}

k (x) = x^{3}

Correct Answer:

2.00

Step-by-Step Solution

Key idea: A rational function must only have integer powers of $x$ in the numerator and denominator.

Step 1: Derivative of $f (x) = x^{- 2}$

$f^{'} (x) = - 2 x^{- 3} = \frac{- 2}{x ^{3}}$ . This is a ratio of polynomials. (Yes)

Step 2: Derivative of $g (x) = x^{1/2}$

$g^{'} (x) = \frac{1}{2} x^{- 1/2} = \frac{1}{2 x}$ . The denominator contains a square root, so it is not a polynomial. (No)

Step 3: Derivative of $h (x) = x^{- 3/2}$

$h^{'} (x) = - \frac{3}{2} x^{- 5/2} = \frac{- 3}{2 x ^{2} x}$ . Contains a square root. (No)

Step 4: Derivative of $k (x) = x^{3}$

$k^{'} (x) = 3 x^{2} = \frac{3 x ^{2}}{1}$ . This is a polynomial, which is a rational function. (Yes)

Count: Functions 1 and 4 have rational derivatives.

Answer: 2.00

Question 2 · Calculus and Optimization MCQ

Let $f (x) = x^{3/2}$ . Which of the following values is IMPOSSIBLE for its derivative $f^{'} (x)$ ?

A.
1.50
B.
3.00
C.
-1.50
D.
0.00

Correct Answer:

Step-by-Step Solution

Key idea: Find the derivative and determine its range based on the domain of the original function.

Step 1: Compute the derivative

$f (x) = x^{3/2}$

$f^{'} (x) = \frac{3}{2} x^{1/2} = \frac{3}{2} x$

Step 2: Analyze the domain and range

The domain of $f (x) = x^{3/2}$ is $x \geq 0$ (for real-valued functions).

For $x \geq 0$ , the square root $x \geq 0$ .

Therefore, $f^{'} (x) = \frac{3}{2} x \geq 0$ .

Step 3: Check the options

The derivative must be greater than or equal to 0.

1.50 is possible (at $x = 1$ )
3.00 is possible (at $x = 4$ )
-1.50 is IMPOSSIBLE (negative)
0.00 is possible (at $x = 0$ )

Answer: -1.50

Question 3 · Calculus and Optimization NAT

What is the minimum value of $f^{'} (x)$ where $f (x) = x^{3} - 3 x$ on the interval $[- 1, 1]$ ?

Correct Answer:

-3.00

Step-by-Step Solution

Key idea: Find the derivative using the power rule and determine its minimum on the given closed interval.

Step 1: Compute the derivative

$f (x) = x^{3} - 3 x$

$f^{'} (x) = 3 x^{2} - 3$

Step 2: Analyze $f^{'} (x)$ on $[- 1, 1]$

The function $f^{'} (x) = 3 x^{2} - 3$ is a parabola opening upwards.

Its vertex is at $x = 0$ .

Since $x = 0$ is within the interval $[- 1, 1]$ , the minimum occurs at the vertex.

Step 3: Evaluate at the minimum

$f^{'} (0) = 3 (0)^{2} - 3 = - 3$

(Check endpoints just in case: $f^{'} (- 1) = 3 (1) - 3 = 0$ , $f^{'} (1) = 3 (1) - 3 = 0$ . The minimum is indeed -3.)

Answer: -3.00

Question 4 · Calculus and Optimization MCQ

Match each function in List I with its correct derivative in List II.

List I
P.

f (x) = 2^{x}

g (x) = e^{- x}

h (x) = x^{e}

k (x) = e^{x}

List II
1.

e^{x}

- e^{- x}

2^{x} ln (2)

e x^{e - 1}

Select the correct matching:

A.
P→3, Q→2, R→4, S→1
B.
P→1, Q→2, R→4, S→3
C.
P→3, Q→1, R→4, S→2
D.
P→4, Q→2, R→3, S→1

Correct Answer:

Step-by-Step Solution

Key idea: Identify whether the variable is in the base or the exponent to apply the correct rule.

Step 1: $f (x) = 2^{x}$

Variable $x$ is in the exponent. Use the exponential rule: $\frac{d}{d x} (a^{x}) = a^{x} ln (a)$ .

$f^{'} (x) = 2^{x} ln (2)$ . Matches 3.

Step 2: $g (x) = e^{- x}$

Use the chain rule with the exponential rule: $\frac{d}{d x} (e^{u}) = e^{u} \cdot u^{'}$ .

$g^{'} (x) = e^{- x} \cdot (- 1) = - e^{- x}$ . Matches 2.

Step 3: $h (x) = x^{e}$

Variable $x$ is in the base. Use the power rule: $\frac{d}{d x} (x^{n}) = n x^{n - 1}$ .

$h^{'} (x) = e x^{e - 1}$ . Matches 4.

Step 4: $k (x) = e^{x}$

The natural exponential is its own derivative.

$k^{'} (x) = e^{x}$ . Matches 1.

Answer: P→3, Q→2, R→4, S→1

Question 5 · Calculus and Optimization MCQ

Let $f (x) = x^{- 3}$ . Which of the following values is IMPOSSIBLE for its derivative $f^{'} (x)$ ?

A.
-3.00
B.
-0.75
C.
0.00
D.
-12.00

Correct Answer:

Step-by-Step Solution

Key idea: Find the derivative and determine its range based on the domain of the function.

Step 1: Compute the derivative

$f (x) = x^{- 3}$

$f^{'} (x) = - 3 x^{- 4} = \frac{- 3}{x ^{4}}$

Step 2: Analyze the domain and range

The domain of $f (x)$ is all real $x \neq = 0$ .

For any $x \neq = 0$ , $x^{4} > 0$ .

Therefore, the numerator is $- 3$ (negative) and the denominator is positive.

This means $f^{'} (x) < 0$ for all $x$ in the domain.

Step 3: Check the options

The derivative must be strictly negative.

-3.00 is possible (at $x = 1$ )
-0.75 is possible (at $x = 2$ )
0.00 is IMPOSSIBLE (it can never be zero)
-12.00 is possible (at $x = 1/ 2$ )

Answer: 0.00

Question 6 · Calculus and Optimization MCQ

Let

f_{1} (x) = sin (x)

f_{2} (x) = cos (x)

f_{3} (x) = tan (x)

, and

f_{4} (x) = cot (x)

. Evaluate their derivatives at

x = \frac{π}{4}

and rank the values from MOST NEGATIVE to MOST POSITIVE:

f_{1}^{'} (\frac{π}{4})

f_{2}^{'} (\frac{π}{4})

f_{3}^{'} (\frac{π}{4})

f_{4}^{'} (\frac{π}{4})

A.
S, Q, P, R
B.
S, P, Q, R
C.
Q, S, P, R
D.
P, Q, R, S

Correct Answer:

Step-by-Step Solution

Key idea: Compute each derivative and evaluate at $x = π /4$ using exact trigonometric values.

Step 1: $f_{1} (x) = sin (x)$

$f_{1}^{'} (x) = cos (x)$

$f_{1}^{'} (π /4) = cos (π /4) = \frac{1}{2} \approx 0.707$

Step 2: $f_{2} (x) = cos (x)$

$f_{2}^{'} (x) = - sin (x)$

$f_{2}^{'} (π /4) = - sin (π /4) = - \frac{1}{2} \approx - 0.707$

Step 3: $f_{3} (x) = tan (x)$

$f_{3}^{'} (x) = sec^{2} (x)$

$f_{3}^{'} (π /4) = sec^{2} (π /4) = (2)^{2} = 2$

Step 4: $f_{4} (x) = cot (x)$

$f_{4}^{'} (x) = - csc^{2} (x)$

$f_{4}^{'} (π /4) = - csc^{2} (π /4) = - (2)^{2} = - 2$

Step 5: Rank from most negative to most positive

S: $- 2$

Q: $- 0.707$

P: $0.707$

R: $2$

Order: S, Q, P, R

Answer: S, Q, P, R

Question 7 · Calculus and Optimization MCQ

Consider the following two statements regarding the derivative of $f (x) = sec (x)$ :

Assertion (A): The derivative of the function is $f^{'} (x) = sec (x) tan (x)$ .

Reason (R): The secant function is a co-function, and its derivative follows the same negative sign pattern as other co-functions like cosine and cotangent.

Which of the following is correct?

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 false but R is true
D.
A is true but R is false

Correct Answer:

Step-by-Step Solution

Key idea: Verify the standard derivative formula and check the sign pattern for co-functions.

Step 1: Analyze Assertion (A)

The standard derivative of $sec (x)$ is indeed $sec (x) tan (x)$ .

Thus, A is TRUE.

Step 2: Analyze Reason (R)

The reason claims that secant follows the "negative sign pattern" of co-functions.

While it is true that $cos (x)$ and $cot (x)$ have negative derivatives, $sec (x)$ is an exception. Its derivative $sec (x) tan (x)$ is positive in the first quadrant.

Thus, R is FALSE.

Step 3: Conclusion

A is true, but R is false.

Answer: A is true but R is false.

Question 8 · Calculus and Optimization NAT

How many of the following functions have a derivative that is ALWAYS positive for all

x

in their domain?

f (x) = tan (x)

g (x) = cot (x)

h (x) = sec (x)

k (x) = csc (x)

m (x) = tan (x) + x

Correct Answer:

2.00

Step-by-Step Solution

Key idea: Compute the derivative of each function and check if it is strictly positive over its entire domain.

Step 1: $f (x) = tan (x)$

$f^{'} (x) = sec^{2} (x)$ . Since $sec^{2} (x) > 0$ for all $x$ in the domain of $tan (x)$ , this is ALWAYS positive. (Yes)

Step 2: $g (x) = cot (x)$

$g^{'} (x) = - csc^{2} (x)$ . This is always negative. (No)

Step 3: $h (x) = sec (x)$

$h^{'} (x) = sec (x) tan (x)$ . This changes sign depending on the quadrant (e.g., positive in Q1, negative in Q2). (No)

Step 4: $k (x) = csc (x)$

$k^{'} (x) = - csc (x) cot (x)$ . This also changes sign depending on the quadrant. (No)

Step 5: $m (x) = tan (x) + x$

$m^{'} (x) = sec^{2} (x) + 1$ . Since $sec^{2} (x) > 0$ , the sum is always $> 1$ , so it is ALWAYS positive. (Yes)

Count: Functions 1 and 5 have always-positive derivatives.

Answer: 2.00

Question 9 · Calculus and Optimization NAT

What is the minimum value of $f^{'} (x)$ where $f (x) = sin (x) - x$ on the interval $[0, π]$ ?

Correct Answer:

-2.00

Step-by-Step Solution

Key idea: Find the derivative using standard rules and determine its minimum on the given closed interval.

Step 1: Compute the derivative

$f (x) = sin (x) - x$

$f^{'} (x) = cos (x) - 1$

Step 2: Analyze $f^{'} (x)$ on $[0, π]$

We need to find the minimum of $g (x) = cos (x) - 1$ on $[0, π]$ .

On the interval $[0, π]$ , the function $cos (x)$ decreases from 1 to -1.

Therefore, the minimum value of $cos (x)$ on this interval is -1 (which occurs at $x = π$ ).

Step 3: Find the minimum of $f^{'} (x)$

Since the minimum of $cos (x)$ is -1, the minimum of $cos (x) - 1$ is:

$- 1 - 1 = - 2$

Answer: -2.00

Question 10 · Calculus and Optimization MCQ

Consider the following two statements regarding the derivative of $f (x) = x^{2}$ at $x = 3$ :

Assertion (A): The derivative can be evaluated using the limit $lim_{h \to 0} \frac{( 3 + h ) ^{2} - 3 ^{2}}{h}$ .

Reason (R): The limit definition of a derivative is a fallback method used only when the standard power rule fails.

Which of the following is correct?

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 false but R is true
D.
A is true but R is false

Correct Answer:

Step-by-Step Solution

Key idea: The limit definition is the fundamental definition of a derivative, not a fallback.

Step 1: Analyze Assertion (A)

The limit $lim_{h \to 0} \frac{f ( x + h ) - f ( x )}{h}$ is the first-principles definition of $f^{'} (x)$ .

For $f (x) = x^{2}$ at $x = 3$ , this limit correctly evaluates to $2 (3) = 6$ . Thus, A is TRUE.

Step 2: Analyze Reason (R)

The reason claims the limit definition is only used when the power rule fails.

This is FALSE. The limit definition is universally applicable and is the foundation from which the power rule is derived. We use the power rule for speed, not because the limit definition is invalid.

Step 3: Conclusion

A is true, but R is false.

Answer: A is true but R is false.

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Differentiation and Higher Order Derivatives Practice Questions 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 (10 Problems)

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

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