1.

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»

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

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


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