Question

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

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

Answer 1

The correct option is A

$e$

Explanation for the correct option

Step 1: Differentiate with respect to $x$

Given information

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

$\therefore \cos y\frac{dy}{dx}+e^{-x\cos y}\left(-\cos y+x\mathrm{si}ny\frac{dy}{dx}\right)=0$

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

Step 2:Compare both sides and take the differential on one side

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

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

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

$\mathit{A}\mathit{t}\mathbf{}\mathbf{(}\mathbf{1}\mathbf{,}\mathit{\pi }\mathbf{)}\mathbf{}\mathbf{}\mathbf{\Rightarrow }$

$\Rightarrow \frac{dy}{dx}=\frac{\cos y{e}^{-\cos \pi }}{\left(\cos \pi +e^{-\cos \pi }\sin \pi \right)}$

$\Rightarrow \frac{dy}{dx}=e^{-1}\left(-1\right)\left(-1\right)$

$\Rightarrow \frac{dy}{dx}=e$

Therefore $At(1,\pi )$$\frac{dy}{dx}$ is $e$

Hence option (A) is the correct answer.