Question

Find \(\frac{dy}{dx}\), if x=tan⁡2θ and y=cos⁡2θ+sin^2⁡θ.

(a) –\(\frac{tan^2⁡2θ \,sin⁡2θ}{2}\)

(b) \(\frac{3 tan^2⁡2θ sin⁡2θ}{2}\)

(c) 0

(d) \(\frac{tan^2⁡2θ sin⁡2θ}{2}\)

I have been asked this question during an interview.

I want to ask this question from Derivatives of Functions in Parametric Forms topic in chapter Continuity and Differentiability of Mathematics – Class 12

Answer 1

The correct option is (a) –\(\frac{tan^2⁡2θ \,sin⁡2θ}{2}\)

The explanation is: Given that, x=tan⁡2θ and y=cos⁡2θ+sin^2⁡θ

\(\frac{dx}{dθ}\)=2 sec^2⁡2θ

\(\frac{dy}{dθ}\)=-2 sin⁡2θ+2 sin⁡θ cos⁡θ=-2 sin⁡2θ+sin⁡2θ=-sin⁡2θ

∴\(\frac{dy}{dx}\)=\(\frac{dy}{dθ}.\frac{dθ}{dx}=-\frac{sin⁡2θ}{2 sec^2⁡2θ}=-\frac{sin⁡2θ}{2 cos^2⁡2θ}.sin^2⁡2θ=-\frac{tan^2⁡2θ sin⁡2θ}{2}\)