Question

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) \(\frac{1}{3(1-4t^2)}\)

The question was asked during an internship interview.

This interesting question is from Derivatives of Functions in Parametric Forms topic in section Continuity and Differentiability of Mathematics – Class 12

Answer 1

Right answer is (d) \(\frac{1}{3(1-4t^2)}\)

The explanation: Given that, x=6 sin^-1⁡2t and y=\(\frac{1}{\sqrt{1-4t^2}}\)

\(\frac{dx}{dt}\)=\(\frac{6}{\sqrt{1-4t^2}}.2=\frac{12}{\sqrt{1-4t^2}}\)

\(\frac{dy}{dt}\)=-\(\frac{1}{2(1-4t^2)^{3/2}}.(-8t)=\frac{4t}{(1-4t^2)^{3/2}}\)

\(\frac{dy}{dx}\)=\(\frac{dy}{dt}.\frac{dt}{dx}=\frac{4t}{(1-4t^2)^{3/2}}.\frac{\sqrt{1-4t^2}}{12}\)

\(\frac{dy}{dx}\)=\(\frac{t}{3(1-4t^2)}\)