Derivatives - Class 11 Maths MCQ

This set of Class 11 Maths Chapter 13 Multiple Choice Questions & Answers (MCQs) focuses on “Derivatives”.

1. Find the derivative of e^x².
a) e^x²
b) 2x
c) 2e^x²
d) 2xe^x²
View Answer

Answer: d
Explanation: We apply chain rule. First we differentiate x².
\(\frac{d}{dx}\) (x²) = 2x
Now, we know that \(\frac{d}{dx}\) (e^x) = e^x
We differentiate e^x² in the same manner and then multiply with the derivative of x²
\(\frac{d}{dx}\) (e^x²) = 2xe^x²

2. What is the value of \(\frac{d}{dx}\) (sin⁡ x tan⁡ x)?
a) sin⁡ x + tan⁡ x sec⁡ x
b) cos⁡ x + tan⁡ x sec⁡ x
c) sin⁡ x + tan⁡ x
d) sin⁡ x + tan⁡ x sec²⁡x
View Answer

Answer: a
Explanation: We follow product rule \(\frac{d}{dx}\) (f.g) = g.\(\frac{d}{dx}\) (f) + f.\(\frac{d}{dx}\) (g)
Here, f = sin⁡ x and g = tan⁡ x
\(\frac{d}{dx}\) (sin⁡ x tan⁡ x) = cos⁡ x tan⁡ x + sec²⁡ x sinx
\(\frac{d}{dx}\) (sin⁡ x tan⁡ x) = sin⁡ x + tan⁡ x sec⁡ x

3. What is the value of \(\frac{d}{dx}\) (sin⁡ x³ cos⁡ x²)?
a) 3x² cos x² cos⁡ x³ + 2x sin⁡ x³ sin x²
b) 3x² cos⁡ x² cos⁡ x³ – 2x sin⁡ x³ sin x²
c) 2x cos x² cos⁡ x³ – 2x sin⁡ x³ sin x²
d) 2x cos x² cos⁡ x³ + 3x² sin⁡ x³ sin x²
View Answer

Answer: b
Explanation: We follow product rule\(\frac{d}{dx}\) (f.g)= g.\(\frac{d}{dx}\) (f)+ f.\(\frac{dy}{dx}\) (g)
Here f = sin⁡ x³ and g = cos⁡ x²
\(\frac{d}{dx}\) (f) = 3x² cos⁡ x³
\(\frac{d}{dx}\) (g) = -2x sin x²
We now substitute this in our main equation,
= cos⁡ x².3x² cos⁡ x³ + sin⁡ x³.(-2x sin x²)
= 3x² cos x² cos⁡ x³ – 2x sin⁡ x³ sin x²

4. What is the value of \(\frac{d}{dx}(\frac{a^x}{e^x})\)?
a) \(\frac{a^x (ln \,⁡a-a^x)}{e^x}\)
b) \(\frac{a^x (ln\, ⁡a-e^x)}{e^x}\)
c) \(\frac{a^x (ln\, ⁡a-1)}{e^x}\)
d) \(\frac{a^x (ln\, ⁡a-1)}{(e^x)(e^x)}\)
View Answer

Answer: c
Explanation: Using quotient rule, we know that, \(\frac{d}{dx} (\frac{f}{g}) = \frac{g.\frac{d}{dx} (f) – f.\frac{d}{dx}(g)}{g^2}\)
Here, f = a^x and g = e^x
\(\frac{d}{dx} (\frac{a^x}{e^x}) = \frac{e^x.\frac{d}{dx}(a^x)-a^x.\frac{d}{dx}(e^x)}{(e^x)(e^x)}\)
\(\frac{d}{dx} (\frac{a^x}{e^x}) = \frac{e^x a^x ln⁡\, a-a^x e^x}{(e^x)(e^x)}\)
\(\frac{d}{dx} (\frac{a^x}{e^x}) = \frac{a^x (ln⁡ \,a-1)}{e^x}\)

5. What is the value of \(\frac{d}{dx}\)(ln \(⁡\frac{3}{x}\))?
a) ⁡\(\frac{2}{x^3}\)
b) ⁡\(\frac{-3}{x^3}\)
c) ⁡\(\frac{-1}{x}\)
d) ⁡\(\frac{-9}{x^3}\)
View Answer

Answer: c
Explanation: We use chain rule to find \(\frac{d}{dx}\)(ln ⁡\(\frac{3}{x}\)).
\(\frac{d}{dx}(ln \frac{3}{x}\)) = \(\frac{1}{\frac{3}{x}} \frac{d}{dx}(\frac{3}{x})\)
\(\frac{d}{dx}(ln \frac{3}{x}\)) = \(\frac{x}{3}(-\frac{3}{x^2})\)
\(\frac{d}{dx}(ln \frac{3}{x}\)) = \(\frac{-1}{x}\)

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6. The derivative of ln⁡ e^x = 1. Is the statement true or false?
a) True
b) False
View Answer

Answer: a
Explanation: We know that ln⁡ e^x = x
\(\frac{d}{dx}\) (x) = 1

7. What is the value of \(\frac{d}{dx}\)(e^x sinx + e^x cos ⁡x)?
a) 0
b) 2 cos⁡x
c) 2e^x.sin ⁡x
d) 2e^x.cos⁡ x
View Answer

Answer: d
Explanation: We need to use product rule in both the terms to get the answer.
\(\frac{d}{dx}\) (f.g) = g.\(\frac{d}{dx}\) (f)+ f.\(\frac{dy}{dx}\) (g)
\(\frac{d}{dx}\) (e^x sin x + e^x cos ⁡x) = (e^x.\(\frac{d}{dx}\) (sin⁡ x) + sin ⁡x.\(\frac{d}{dx}\) (e^x)) + (e^x.\(\frac{d}{dx}\) (cos ⁡x) + cos⁡ x.\(\frac{d}{dx}\) (e^x))
\(\frac{d}{dx}\) (e^x sin x + e^x cos ⁡x) =(e^x.cos⁡ x + sin ⁡x . e^x) + (e^x.(-sin ⁡x) + cos ⁡x.e^x)
\(\frac{d}{dx}\) (e^x sin x + e^x cos ⁡x) = e^x.cos⁡ x + sin⁡ x . e^x – e^x.sin⁡ x + cos ⁡x.e^x
\(\frac{d}{dx}\) (e^x sin x + e^x cos ⁡x) = 2e^x.cos⁡ x

8. What is the value of \(\frac{d}{dx}\) (e^x tan x) at x = 0?
a) 0
b) 1
c) -1
d) 2
View Answer

Answer: b
Explanation: We need to use product rule in both the terms to get the answer.
\(\frac{d}{dx}\) (f.g) = g.\(\frac{d}{dx}\) (f)+ f.\(\frac{dy}{dx}\) (g)
Here f = e^x and g = tan ⁡x
\(\frac{d}{dx}\) (e^x tan x) = tan⁡ x.\(\frac{d}{dx}\) (e^x) + e^x.\(\frac{d}{dx}\) (tan ⁡x)
\(\frac{d}{dx}\) (e^x tan x) = tan⁡ x.e^x + e^x . sec²⁡x
At x = 0 we get,
= tan ⁡0.e⁰ + e⁰.sec²⁡0
= 0.(1) + 1.(1)
= 1

9. What is the value of \(\frac{d}{dx}\)(cos²⁡ x tan⁡ x) at x = 0?
a) -1
b) 0
c) -2
d) 1
View Answer

Answer: d
Explanation: We need to use product rule in both the terms to get the answer.
\(\frac{d}{dx}\) (f.g) = g.\(\frac{d}{dx}\) (f) + f.\(\frac{d}{dx}\) (g)
Here f = cos²⁡ x and g = tan ⁡x
To differentiate f, we need to use chain rule.
\(\frac{d}{dx}\) (cos² ⁡x tan⁡ x) = tan ⁡x.\(\frac{d}{dx}\) (cos² x) + cos² x.\(\frac{d}{dx}\) (tan⁡ x)
= tan⁡ x.(-2 cos⁡ x sin⁡ x) + cos² ⁡x.sec² x
= tan⁡ x.(-2 cos⁡ x sin⁡ x) + 1 (as we know (sec x) = (1/cos x))
At x = 0 we get,
= tan ⁡0.(-2 cos⁡ 0 sin⁡ 0) + 1
= 1

10. What is the value of \(\frac{d}{dx}(\frac{cos⁡x}{sec⁡x \,tan⁡x})\)?
a) -tan²⁡ x(3 sin²⁡ x – cos² x)
b) -cot² x(3 sin²⁡ x – cos² x)
c) -cot² x(3 sin²⁡ x + cos² x)
d) -tan²⁡ x(3 sin²⁡ x + cos² x)
View Answer

Answer: c
Explanation: \(\frac{d}{dx}(\frac{cos⁡x}{sec⁡x \,tan⁡x}) = \frac{d}{dx}(\frac{cos⁡x}{\frac{1}{cos⁡x}\frac{sin⁡x}{cos⁡x}})\)
=\(\frac{d}{dx}(\frac{cos^3⁡x}{sin⁡x})\)
Using quotient rule, we know that, \(\frac{d}{dx} (\frac{f}{g}) = \frac{g.\frac{d}{dx} (f) – f.\frac{d}{dx}(g)}{g^2}\)
Here, f = cos³⁡ x and g = sin ⁡x
\(\frac{d}{dx}(\frac{cos^3⁡x}{sin⁡x})\) = \(\frac{sin⁡x.\frac{d}{dx} (cos^3⁡x)-cos^3⁡x.\frac{d}{dx}(sinx)}{sin^2⁡x}\)
\(\frac{d}{dx}(\frac{cos^3⁡x}{sin⁡x})\) = \(\frac{sin⁡x (-3 cos^2⁡ x \,sin⁡ x) – cos^3⁡x(cos⁡x)}{sin^2⁡x}\)
\(\frac{d}{dx}(\frac{cos^3⁡x}{sin⁡x})\) = \(\frac{-3 sin^2⁡x \,cos^2⁡x – cos^4⁡x}{sin^2⁡x}\)
\(\frac{d}{dx}(\frac{cos^3⁡x}{sin⁡x})\) = -cot²⁡ x(3 sin²⁡ x + cos² x)

11. What is the value of \(\frac{d}{dx}(\frac{sec⁡x}{cosec\, x \,tan⁡x}\))?
a) 0
b) 1
c) cos ⁡x
d) sin ⁡x
View Answer

Answer: a
Explanation: \(\frac{d}{dx}(\frac{sec⁡x}{cosec\, x\, tan⁡x})\) = \(\frac{d}{dx}(\frac{\frac{1} {cos⁡x}}{\frac{1}{sin⁡\, x}.\frac{sin⁡x}{cos⁡x}}\))
= \(\frac{d}{dx}\) (1)
= 0

Sanfoundry Global Education & Learning Series – Mathematics – Class 11.

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