# 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

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 sec2⁡x

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

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) $$\frac{a^x (ln\, ⁡a-1)}{(e^x)(e^x)}$$

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) ⁡$$\frac{9}{x^3}$$

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

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

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

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

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

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)

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

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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