This set of Class 11 Maths Chapter 13 Multiple Choice Questions & Answers (MCQs) focuses on “Derivatives”.
1. Find the derivative of ex2.
a) ex2
b) 2x
c) 2ex2
d) 2xex2
View Answer
Explanation: We apply chain rule. First we differentiate x2.
\(\frac{d}{dx}\) (x2) = 2x
Now, we know that \(\frac{d}{dx}\) (ex) = ex
We differentiate ex2 in the same manner and then multiply with the derivative of x2
\(\frac{d}{dx}\) (ex2) = 2xex2
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 sec2x
View Answer
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
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
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 = ax and g = ex
\(\frac{d}{dx} (\frac{a^x}{e^x}) = \frac{e^x.\frac{d}{dx}(a^x)-a^x.\frac{d}{dx}(e^x)}{(e^x)(e^x)}\)
\(\frac{d}{dx} (\frac{a^x}{e^x}) = \frac{e^x a^x ln\, a-a^x e^x}{(e^x)(e^x)}\)
\(\frac{d}{dx} (\frac{a^x}{e^x}) = \frac{a^x (ln \,a-1)}{e^x}\)
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
Explanation: We use chain rule to find \(\frac{d}{dx}\)(ln \(\frac{3}{x}\)).
\(\frac{d}{dx}(ln \frac{3}{x}\)) = \(\frac{1}{\frac{3}{x}} \frac{d}{dx}(\frac{3}{x})\)
\(\frac{d}{dx}(ln \frac{3}{x}\)) = \(\frac{x}{3}(-\frac{3}{x^2})\)
\(\frac{d}{dx}(ln \frac{3}{x}\)) = \(\frac{-1}{x}\)
6. The derivative of ln ex = 1. Is the statement true or false?
a) True
b) False
View Answer
Explanation: We know that ln ex = x
\(\frac{d}{dx}\) (x) = 1
7. What is the value of \(\frac{d}{dx}\)(ex sinx + ex cos x)?
a) 0
b) 2 cosx
c) 2ex.sin x
d) 2ex.cos x
View Answer
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)
\(\frac{d}{dx}\) (ex sin x + ex cos x) = (ex.\(\frac{d}{dx}\) (sin x) + sin x.\(\frac{d}{dx}\) (ex)) + (ex.\(\frac{d}{dx}\) (cos x) + cos x.\(\frac{d}{dx}\) (ex))
\(\frac{d}{dx}\) (ex sin x + ex cos x) =(ex.cos x + sin x . ex) + (ex.(-sin x) + cos x.ex)
\(\frac{d}{dx}\) (ex sin x + ex cos x) = ex.cos x + sin x . ex – ex.sin x + cos x.ex
\(\frac{d}{dx}\) (ex sin x + ex cos x) = 2ex.cos x
8. What is the value of \(\frac{d}{dx}\) (ex tan x) at x = 0?
a) 0
b) 1
c) -1
d) 2
View Answer
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 = ex and g = tan x
\(\frac{d}{dx}\) (ex tan x) = tan x.\(\frac{d}{dx}\) (ex) + ex.\(\frac{d}{dx}\) (tan x)
\(\frac{d}{dx}\) (ex tan x) = tan x.ex + ex . sec2x
At x = 0 we get,
= tan 0.e0 + e0.sec20
= 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
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 \(\frac{d}{dx}(\frac{cosx}{secx \,tanx})\)?
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
Explanation: \(\frac{d}{dx}(\frac{cosx}{secx \,tanx}) = \frac{d}{dx}(\frac{cosx}{\frac{1}{cosx}\frac{sinx}{cosx}})\)
=\(\frac{d}{dx}(\frac{cos^3x}{sinx})\)
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
\(\frac{d}{dx}(\frac{cos^3x}{sinx})\) = \(\frac{sinx.\frac{d}{dx} (cos^3x)-cos^3x.\frac{d}{dx}(sinx)}{sin^2x}\)
\(\frac{d}{dx}(\frac{cos^3x}{sinx})\) = \(\frac{sinx (-3 cos^2 x \,sin x) – cos^3x(cosx)}{sin^2x}\)
\(\frac{d}{dx}(\frac{cos^3x}{sinx})\) = \(\frac{-3 sin^2x \,cos^2x – cos^4x}{sin^2x}\)
\(\frac{d}{dx}(\frac{cos^3x}{sinx})\) = -cot2 x(3 sin2 x + cos2 x)
11. What is the value of \(\frac{d}{dx}(\frac{secx}{cosec\, x \,tanx}\))?
a) 0
b) 1
c) cos x
d) sin x
View Answer
Explanation: \(\frac{d}{dx}(\frac{secx}{cosec\, x\, tanx})\) = \(\frac{d}{dx}(\frac{\frac{1} {cosx}}{\frac{1}{sin\, x}.\frac{sinx}{cosx}}\))
= \(\frac{d}{dx}\) (1)
= 0
Sanfoundry Global Education & Learning Series – Mathematics – Class 11.
To practice all chapters and topics of class 11 Mathematics, here is complete set of 1000+ Multiple Choice Questions and Answers.
