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

Answer: (C)

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