Question

If $\cos (x+y)=y\sin x$, then find $\frac{dy}{dx}.$

Answer 1

Given

$\cos (x+y)=y\sin x$

Differentiating both sides with respect to x, we get

$-\sin (x+y)\{1+\frac{dy}{dx}\}=y\cos x+\sin x\frac{dy}{dx}$

$-\sin (x+y)-\sin (x+y)\frac{dy}{dx}=y\cos x+\sin x\frac{dy}{dx}$

$\Rightarrow -\sin (x+y)\frac{dy}{dx}-\sin x\frac{dy}{dx}=y\cos x+\sin (x+y)$

$\Rightarrow \{-\sin (x+y)-\sin x\}\frac{dy}{dx}=y\cos x+\sin (x+y)$

$\Rightarrow \frac{dy}{dx}=\frac{y\cos x+\sin (x+y)}{\sin (x+y)-\sin x}$.