Differentiation Question 315
Question: If $ \sin y+{e^{-x\cos y}}=e, $ then $ \frac{dy}{dx} $ at $ (1,\pi ) $ is
[Kerala (Engg.) 2002]
Options:
A) $ \sin y $
B) $ -x\cos y $
C) $ e $
D) $ \sin y-x\cos y $
Show Answer
Answer:
Correct Answer: C
Solution:
$ \sin y+{e^{-x\cos y}}=e, $
Therefore $ \cos y\frac{dy}{dx}+{e^{-x\cos y}}{ (-x)( -\sin y\frac{dy}{dx} )+\cos y(-1) }=0 $
Therefore $ \cos y\frac{dy}{dx}+x\sin y{e^{-x\cos y}}\frac{dy}{dx}-\cos y{e^{-x\cos y}}=0 $
Therefore $ \frac{dy}{dx}=\frac{\cos y{e^{-x\cos y}}}{\cos y+x\sin y{e^{-x\cos y}}} $
Therefore $ {{( \frac{dy}{dx} )} _{(1,\pi )}}=\frac{\cos \pi {e^{-\cos \pi }}}{\cos \pi +\sin \pi {e^{-\cos \pi }}} $ = $ \frac{(-1)e}{-1+0}=e $ .
BETA
