Question

Differentiate ${\sin }^{2}x$ w. r. t. $e^{\cos x}$.

Answer 1

Let $P=\sin x$ ...(1)

$Q=e^{\cos x}$ ...(2)

On differentiating eq.(1) w.r.t. we get

$\frac{dp}{dx}=\frac{d}{dx}\left({\sin }^{2}x\right)=2\sin x\cos x$ ...(3)

On differentiating eq.(2) w.r.t we get

$\frac{d\theta }{dx}=\frac{d}{dx}(e^{\cos x)}=-e^{\cos x}\sin x$ ...(4)

dividing eq.(3) and (4)

$\frac{\frac{dp}{dx}}{\frac{d\theta }{dx}}=\frac{2\sin x\cos x}{-e^{\cos x}\sin x}=\frac{-2\cos x}{e^{\cos x}}$

$[\frac{\frac{dp}{dx}}{\frac{d\theta }{dx}}=\frac{-2\cos x}{e^{\cos x}}]$