# Class 12 Maths MCQ – Derivatives of Functions in Parametric Forms

This set of Class 12 Maths Chapter 5 Multiple Choice Questions & Answers (MCQs) focuses on “Derivatives of Functions in Parametric Forms”.

1. Find $$\frac{dy}{dx}$$, if x=3a2 cos2⁡θ and y=4a sin2⁡θ.
a) $$\frac{3}{4a}$$
b) –$$\frac{4}{3a}$$
c) $$\frac{4}{3a}$$
d) –$$\frac{3}{4a}$$

Explanation: Given that, x=3a2 cos2⁡θ and y=4a sin2⁡θ
Then, $$\frac{dx}{dθ}$$=3a2.(2 cos⁡θ)(-sin⁡θ)=-3a2 sin⁡2θ
$$\frac{dy}{dθ}$$=4a(2 sin⁡θ)(cos⁡θ)=4a sin⁡2θ
$$\frac{dy}{dx}$$=$$\frac{dy}{dθ}×\frac{dθ}{dx}=-\frac{4a \,sin⁡2θ}{3a^2 \,sin⁡2θ}=-\frac{4}{3a}$$

2. Find $$\frac{dy}{dx}$$, if x=9t4 and y=t.
a) $$\frac{1}{36t^3}$$
b) $$\frac{1}{36t^2}$$
c) $$\frac{-1}{36t^3}$$
d) $$\frac{1}{32t^3}$$

Explanation: Given that, x=9t4 and y=t
$$\frac{dx}{dt}$$=36t3
$$\frac{dy}{dt}$$=1
∴$$\frac{dy}{dx}$$=$$\frac{dy}{dt}.\frac{dt}{dx}=\frac{1}{36t^3}$$

3. Find $$\frac{dy}{dx}$$, if x=sin⁡3t and y=t2 tan⁡2t.
a) $$\frac{3t(tan⁡2t+tsec^2 2t)}{4 cos⁡3t}$$
b) $$\frac{(tan⁡2t+tsec^2 2t)}{3 cos⁡3t}$$
c) $$\frac{-2t(tan⁡2t+tsec^2 2t)}{3 cos⁡3t}$$
d) $$\frac{2t(tan⁡2t+tsec^2 2t)}{3 cos⁡3t}$$

Explanation: Given that, x=sin⁡3t and y=t2 tan⁡2t
$$\frac{dx}{dt}$$=3 cos⁡3t
By using u.v rule, we get
$$\frac{dy}{dt}$$=$$\frac{d}{dx} \,(t^2) \,tan⁡2t+\frac{d}{dx} \,(tan⁡2t)$$ t2
$$\frac{dy}{dt}$$=2t tan⁡2t+2t2 sec2⁡2t
$$\frac{dy}{dx}$$=$$\frac{dy}{dt}.\frac{dt}{dx}=\frac{(2t \,tan⁡2t+2t^2 \,sec^2⁡2t)}{3 \,cos⁡3t}$$
∴$$\frac{dy}{dx}=\frac{2t(tan⁡2t+tsec^2 \,2t)}{3 \,cos⁡3t}$$

4. Find $$\frac{dy}{dx}$$, if x=log⁡t2 and y=$$\frac{1}{t}$$.
a) $$\frac{1}{2t}$$
b) –$$\frac{t}{2}$$
c) –$$\frac{1}{2t}$$
d) $$\frac{t}{2}$$

Explanation: Given that, x=log⁡t2 and y=$$\frac{1}{t}$$
$$\frac{dx}{dt}$$=$$\frac{1}{t^2}.2t=\frac{2}{t}$$
$$\frac{dy}{dt}$$=-$$\frac{1}{t^2}$$
∴$$\frac{dy}{dx}$$=$$\frac{dy}{dt}.\frac{dt}{dx}=-\frac{1}{t^2}.\frac{t}{2}=-\frac{1}{2t}$$

5. Find $$\frac{dy}{dx}$$, if x=6 sin-1⁡2t and y=$$\frac{1}{\sqrt{1-4t^2}}$$.
a) $$\frac{t}{1-4t^2}$$
b) –$$\frac{1}{3(1-4t^2)}$$
c) –$$\frac{t}{3(1-4t^2)}$$
d) $$\frac{1}{3(1-4t^2)}$$

Explanation: Given that, x=6 sin-1⁡2t and y=$$\frac{1}{\sqrt{1-4t^2}}$$
$$\frac{dx}{dt}$$=$$\frac{6}{\sqrt{1-4t^2}}.2=\frac{12}{\sqrt{1-4t^2}}$$
$$\frac{dy}{dt}$$=-$$\frac{1}{2(1-4t^2)^{3/2}}.(-8t)=\frac{4t}{(1-4t^2)^{3/2}}$$
$$\frac{dy}{dx}$$=$$\frac{dy}{dt}.\frac{dt}{dx}=\frac{4t}{(1-4t^2)^{3/2}}.\frac{\sqrt{1-4t^2}}{12}$$
$$\frac{dy}{dx}$$=$$\frac{t}{3(1-4t^2)}$$

6. Find $$\frac{dy}{dx}$$, if x=2t2 and y=6t6.
a) -9t4
b) 9t4
c) t4
d) 9t3

Explanation: Given that, x=2t2 and y=6t6
$$\frac{dx}{dt}$$=4t
$$\frac{dy}{dt}$$=36t5
$$\frac{dy}{dx}$$=$$\frac{dy}{dt}.\frac{dt}{dx}=\frac{36t^5}{4t}=9t^4$$

7. Find $$\frac{dy}{dx}$$, if x=2et and y=log⁡t
a) $$\frac{1}{2te^t}$$
b) –$$\frac{1}{2te^t}$$
c) $$\frac{1}{te^t}$$
d) $$\frac{1}{e^t}$$

Explanation: Given that, x=2et and y=log⁡t
$$\frac{dx}{dt}$$=2et
$$\frac{dy}{dt}$$=1/t
∴$$\frac{dy}{dx}$$=$$\frac{dy}{dt}.\frac{dt}{dx}=\frac{1}{t}.\frac{1}{2e^t}=\frac{1}{2te^t}$$.

8. Find $$\frac{dy}{dx}$$, if x=tan⁡2θ and y=cos⁡2θ+sin2⁡θ.
a) –$$\frac{tan^2⁡2θ \,sin⁡2θ}{2}$$
b) $$\frac{3 tan^2⁡2θ sin⁡2θ}{2}$$
c) 0
d) $$\frac{tan^2⁡2θ sin⁡2θ}{2}$$

Explanation: Given that, x=tan⁡2θ and y=cos⁡2θ+sin2⁡θ
$$\frac{dx}{dθ}$$=2 sec2⁡2θ
$$\frac{dy}{dθ}$$=-2 sin⁡2θ+2 sin⁡θ cos⁡θ=-2 sin⁡2θ+sin⁡2θ=-sin⁡2θ
∴$$\frac{dy}{dx}$$=$$\frac{dy}{dθ}.\frac{dθ}{dx}=-\frac{sin⁡2θ}{2 sec^2⁡2θ}=-\frac{sin⁡2θ}{2 cos^2⁡2θ}.sin^2⁡2θ=-\frac{tan^2⁡2θ sin⁡2θ}{2}$$

9. Find $$\frac{dy}{dx}$$, if x=log⁡(tan⁡t) and y=log⁡(sin⁡t).
a) 2 cos2⁡t
b) cos2⁡2t
c) cos2t
d) -cos2t

Explanation: Given that, x=log⁡(tan⁡t) and y=log⁡(sin⁡t)
$$\frac{dx}{dt}$$=$$\frac{1}{tan⁡t}.sec^2⁡t=cot⁡t sec^2⁡t$$
$$\frac{dy}{dt}=\frac{1}{sin⁡ \,t}.cos⁡ \,t=cot⁡ \,t$$
∴$$\frac{dy}{dx}$$=$$\frac{dy}{dt}.\frac{dt}{dx}=\frac{cot⁡\,t}{cot\,⁡t sec^2⁡t}=\frac{1}{sec^2⁡t}=cos^2⁡t$$.

10. Find $$\frac{dy}{dx}$$, if x=a2 t2 cotθ and y=at sin⁡θ.
a) $$\frac{tan⁡θ \,sin⁡θ}{at}$$
b) $$\frac{tan⁡θ \,sin⁡θ}{2at}$$
c) $$\frac{tan⁡θ \,sin⁡θ}{2t}$$
d) $$\frac{tan⁡θ \,sin⁡θ}{2a}$$

Explanation: Given that, x=a2 t2 cotθ and y=at sin⁡θ
$$\frac{dx}{dt}$$=2ta2 cot⁡θ
$$\frac{dy}{dt}$$=asin⁡θ
$$\frac{dy}{dx}$$=$$\frac{asin⁡θ}{2ta^2 \,cot⁡θ}=\frac{a sin⁡θ}{2a^2 t cos⁡θ}.sin⁡θ=\frac{tan⁡θ \,sin⁡θ}{2at}$$

Sanfoundry Global Education & Learning Series – Mathematics – Class 12.

To practice all chapters and topics of class 12 Mathematics, here is complete set of 1000+ Multiple Choice Questions and Answers.