Mathematics Questions and Answers – Derivatives

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This set of Mathematics Multiple Choice Questions & Answers (MCQs) focuses on “Derivatives”.

1. Find the derivative of ex2.
a) ex2
b) 2x
c) 2ex2
d) 2xex2
View Answer

Answer: d
Explanation: We apply chain rule. First we differentiate x2.
\(\frac{d}{dx}\) (x2) = 2x
Now, we know that \(\frac{d}{dx}\) (ex) = ex
We differentiate ex2 in the same manner and then multiply with the derivative of x2
\(\frac{d}{dx}\) (ex2) = 2xex2
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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
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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 = 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 ln ⁡\(\frac{3}{x}\)?
a) ⁡\(\frac{2}{x^3}\)
b) ⁡\(\frac{-3}{x^3}\)
c) ⁡\(\frac{-9}{x^3}\)
d) ⁡\(\frac{9}{x^3}\)
View Answer

Answer: c
Explanation: We use chain rule to find \(\frac{d}{dx}\)(ln ⁡\(\frac{3}{x}\)).
\(\frac{d}{dx}(ln \frac{3}{x}\)) = \(\frac{1}{\frac{3}{x}} \frac{d}{dx}(\frac{3}{x})\)
\(\frac{d}{dx}(ln \frac{3}{x}\)) = \(\frac{3}{x}(-\frac{3}{x^2})\)
\(\frac{d}{dx}(ln \frac{3}{x}\)) = \(\frac{-9}{x^3}\)
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6. The derivative of ln⁡ ex = 1. Is the statement true or false?
a) True
b) False
View Answer

Answer: a
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 cos⁡x
c) 2ex.sin ⁡x
d) 2ex.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)
\(\frac{d}{dx}\) (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
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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

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 = 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 . 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 = 1?
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)
\(\frac{d}{dx}\) (ex tan⁡x) = tan⁡ x.(-2 cos⁡ x sin⁡ x) + cos2 ⁡x.sec2 x
At x = 1 we get,
= tan⁡0.(-2 cos⁡ 0 sin⁡ 0) + cos2⁡ 0.sec2 ⁡0
= 1
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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: \(\frac{d}{dx}(\frac{sec⁡x}{cosec\, x\, tan⁡x})\) = \(\frac{d}{dx}(\frac{{1} {cos⁡x}}{\frac{1}{sin⁡\, x}.\frac{sin⁡x}{cos⁡x}}\))
= \(\frac{d}{dx}\) (1)
= 0

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

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Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He is Linux Kernel Developer & SAN Architect and is passionate about competency developments in these areas. He lives in Bangalore and delivers focused training sessions to IT professionals in Linux Kernel, Linux Debugging, Linux Device Drivers, Linux Networking, Linux Storage, Advanced C Programming, SAN Storage Technologies, SCSI Internals & Storage Protocols such as iSCSI & Fiber Channel. Stay connected with him @ LinkedIn | Youtube | Instagram | Facebook | Twitter