# Class 12 Maths MCQ – Second Order Derivatives

This set of Class 12 Maths Chapter 5 Multiple Choice Questions & Answers (MCQs) focuses on “Second Order Derivatives”.

1. Find the second order derivative of y=9 log⁡ t3.
a) $$\frac{27}{t^2}$$
b) –$$\frac{27}{t^2}$$
c) –$$\frac{1}{t^2}$$
d) –$$\frac{27}{2t^2}$$

Explanation: Given that, y=9 log⁡t3
$$\frac{dy}{dx}=9.\frac{1}{t^3}.3t^2=\frac{27}{t}$$
$$\frac{d^2 y}{dx^2}=27(-\frac{1}{t^2})=-\frac{27}{t^2}$$.

2. Find $$\frac{d^2y}{dx^2}$$, if y=tan2⁡x+3 tan⁡x.
a) sec2⁡⁡x tan⁡x (2 tan⁡x+sec⁡x+3)
b) 2 sec2⁡⁡x tan⁡x (2 tan⁡x-sec⁡x+3)
c) 2 sec2⁡⁡x tan⁡x (2 tan⁡x+sec⁡x+3)
d) 2 sec2⁡⁡x tan⁡x (2 tan⁡x+sec⁡x-3)

Explanation: Given that, y=tan2⁡⁡x+3 tan⁡x
$$\frac{dy}{dx}$$=2 tan⁡x sec2⁡⁡x+3 sec2⁡x=sec2⁡⁡x (2 tan⁡x+3)
By using the u.v rule, we get
$$\frac{d^2 y}{dx^2}=\frac{d}{dx}$$ (sec2⁡⁡x).(2 tan⁡x+3)+$$\frac{d}{dx}$$ (2 tan⁡x+3).sec2⁡⁡x
$$\frac{d^2 y}{dx^2}$$=2 sec2⁡⁡x tan⁡x (2 tan⁡x+3)+sec2⁡⁡x (2 sec⁡x tanx)
=2 sec2⁡x tan⁡x (2 tan⁡x+sec⁡x+3).

3. If y=6x2+3, then $$\left (\frac{dy}{dx}\right )^2=\frac{d^2 y}{dx^2}$$.
a) True
b) False

Explanation: The given statement is false. Given that, y=6x2+3
$$\frac{dy}{dx}$$=12x
⇒$$\left (\frac{dy}{dx}\right )^2=(12x)^2=144x^2$$
$$\frac{d^2 y}{dx^2}=\frac{d}{dx}$$ (12x)=12
∴$$\left (\frac{dy}{dx}\right )^2≠\frac{d^2 y}{dx^2}$$

4. Find the second order derivative of y=2e2x-3 log⁡(2x-3).
a) 8e2x+$$\frac{1}{(2x-3)^2}$$
b) 8e2x–$$\frac{12}{(2x-3)^2}$$
c) e2x+$$\frac{12}{(2x-3)^2}$$
d) 8e2x+$$\frac{12}{(2x-3)^2}$$

Explanation: Given that, y=2e2x-3 log⁡(2x-3)
$$\frac{dy}{dx}$$=4e2x-3.$$\frac{1}{(2x-3)}$$.2=4e2x–$$\frac{6}{(2x-3)}$$
$$\frac{d^2 y}{dx^2}=\frac{d}{dx} (\frac{dy}{dx})$$=8e2x+$$\frac{12}{(2x-3)^2}$$

5. Find $$\frac{d^2 y}{dx^2}$$, if y=2 sin-1⁡(cos⁡x).
a) 0
b) sin-1$$(\frac{1}{cos⁡x})$$
c) 1
d) -1

Explanation: Given that, y=2 sin-1⁡(cos⁡x)
$$\frac{dy}{dx}=2.\frac{1}{\sqrt{1-cos^2⁡x}}$$.-sin⁡x=-2 (∵$$\sqrt{1-cos^2⁡x}$$=sin⁡x)
$$\frac{d^2 y}{dx^2}$$=$$\frac{d}{dx} (\frac{dy}{dx})=\frac{d}{dx}$$ (-2)=0
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6. If y=log⁡(2x3), find $$\frac{d^2 y}{dx^2}$$.
a) –$$\frac{2}{x^2}$$
b) $$\frac{3}{x^2}$$
c) $$\frac{2}{x^2}$$
d) –$$\frac{3}{x^2}$$

Explanation: Given that, y=log⁡(2x3)
$$\frac{dy}{dx}=\frac{1}{(2x^3)}.6x^2=\frac{3}{x}$$
$$\frac{d^2 y}{dx^2}=-\frac{3}{x^2}$$

7. Find $$\frac{d^2 y}{dx^2}$$-6 $$\frac{dy}{dx}$$ if y=4x4+2x.
a) $$(4x^2+8x-1)$$
b) $$12(4x^2+8x-1)$$
c) –$$12(4x^2+8x-1)$$
d) $$12(4x^2-8x-1)$$

Explanation: Given that, $$y=4x^4+2x$$
$$\frac{dy}{dx}$$=16x3+2
$$\frac{d^2 y}{dx^2}$$=48x2
$$\frac{d^2 y}{dx^2}$$-6 $$\frac{dy}{dx}=48x^2-96x^3-12$$
=12(4x2-8x-1)

8. Find the second order derivative y=e2x+sin-1⁡ex .
a) e2x+$$\frac{e^x}{(1-e^2x)^{3/2}}$$
b) 4e2x+$$\frac{1}{(1-e^2x)^{3/2}}$$
c) 4e2x–$$\frac{e^x}{(1-e^2x)^{3/2}}$$
d) 4e2x+$$\frac{e^x}{(1-e^2x)^{3/2}}$$

Explanation: Given that, y=e2x+sin-1⁡ex
$$\frac{dy}{dx}$$=2e2x+$$\frac{1}{\sqrt{1-e^{2x}}} e^x$$
$$\frac{d^2 y}{dx^2} = 4e^2x+\bigg(\frac{\frac{d}{dx} (e^x) \sqrt{1-e^{2x}} – \frac{d}{dx} (\sqrt{1-e^{2x}}).e^x}{(\sqrt{1-e^{2x}})^2}\bigg)$$
$$=4e^{2x}+\frac{(e^x \sqrt{1-e^{2x}})-e^x \left(\frac{1}{2\sqrt{1-e^{2x}}}.-2e^{2x}\right)}{1-e^{2x}}$$
$$=4e^{2x}+\frac{(e^x (1-e^{2x})+e^{3x})}{(1-e^{2x})^{\frac{3}{2}}}$$
$$=4e^{2x}+\frac{e^x (1-e^{2x}+e^{2x})}{(1-e^{2x})^{\frac{3}{2}}}$$
4e2x+$$\frac{e^x}{(1-e^2x)^{3/2}}$$.

9. Find the second order derivative of y=3x2 1 + log⁡(4x)
a) 3+$$\frac{1}{x^2}$$
b) 3-$$\frac{1}{x^2}$$
c) 6-$$\frac{1}{x^2}$$
d) 6+$$\frac{1}{x^2}$$

Explanation: Given that, y=3x2+log⁡(4x)
$$\frac{dy}{dx}=6x+\frac{1}{4x}.4=6x+\frac{1}{x}=\frac{6x^2+1}{x}$$
$$\frac{d^2 y}{dx^2}=\frac{\frac{d}{dx} (6x^2+1).(x)-\frac{d}{dx} (x).(6x^2+1)}{x^2} \Big(using\, \frac{d}{dx} (\frac{u}{v})=\frac{(\frac{d}{dx} (u).v-\frac{d}{dx} (v).u)}{v^2}\Big)$$
$$\frac{d^2 y}{dx^2}=\frac{(12x.x-6x^2-1)}{x^2}$$
$$\frac{d^2 y}{dx^2}=\frac{6x^2-1}{x^2} = 6-\frac{1}{x^2}$$.

10. Find the second order derivative if y=e2x2.
a) 4e2x2 (4x2+3)
b) 4e2x2 (4x2-1)
c) 4e2x2 (4x2+1)
d) e2x2 (4x2+1)

Explanation: Given that, y=e2x2
$$\frac{dy}{dx}$$=e2x2.4x
By using u.v rule, we get
$$\frac{d^2 y}{dx^2}=\frac{d}{dx} (e^{{2x}^2}).4x+\frac{d}{dx} (4x).e^{{2x}^2}$$
16x2 e2x2+4e2x2=4e2x2 (4x2+1)

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