Question

If $\sin y+e^{-x\cos y}=e$ then $\frac{dy}{dx}$ at $(1,\pi )$ is equal to

• A. $\sin y$
• B. $-x\cos y$
• C. $e$
• D. $\sin y-x\cos y$

Answer 1

The correct option is C $e$

Given,

$\sin y+e^{-x\cos y}=e$

$\cos y\frac{dy}{dx}+e^{-x\cos y}[x\sin y\frac{dy}{dx}-\cos y]=0$

$\Rightarrow \cos y\frac{dy}{dx}+e^{-x\cos y}x\sin y\frac{dy}{dx}-\cos y{e}^{-x\cos y}=0$

$\frac{dy}{dx}[\cos y+x\sin y{e}^{-x\cos y}]=\cos y{e}^{-x\cos y}$

$\frac{dy}{dx}=\frac{\cos y{e}^{-x\cos y}}{\cos y+x\sin y{e}^{-x\cos y}}$

${\left[\frac{dy}{dx}\right]}_{(1,\pi )}=\frac{\cos \pi e^{-\cos \pi }}{\cos \pi +1\sin \pi e^{-\cos \pi }}$

${\left[\frac{dy}{dx}\right]}_{(1,\pi )}=\frac{-1\times e}{-1+0\times e}=e$