Question

If $siny+e^{-xcosy}=e$, then $\frac{dy}{dx}$ at $(1$, $\pi )$ is equal to

• A. $siny$
• B. $-xcosy$
• C. $e$
• D. $siny-xcosy$

Answer 1

Answer: (C)

We have, $siny+e^{-xcosy}=e$
Differentiating both sides $w$.$r$.$t$. $x$, we get
$cosy\frac{dy}{dx}+e^{-xcosy}\left\{xsiny\frac{dy}{dx}-cosy\right\}=0$
$\Rightarrow \left(cosy+e^{-xcosy}\cdot xsiny\right)\frac{dy}{dx}=\left(e^{-xcosy}\right)cosy$
$\Rightarrow \frac{dy}{dx}=\frac{e^{-xcosy}\cdot cosy}{cosy+x\cdot e^{-xcosy}siny}$
$\therefore \frac{dy}{dx}{|}_{(1,\pi )}=\frac{e^{-cos\pi }\cdot cos\pi }{cos\pi +e^{-cos\pi }sin\pi }=e$