A) \[\sin y\]
B) \[-x\cos y\]
C) \[e\]
D) \[\sin y-x\,\cos y\]
Correct Answer: C
Solution :
\[\sin y+{{e}^{-x\cos y}}=e,\] Þ \[\cos y\frac{dy}{dx}+{{e}^{-x\cos y}}\left\{ (-x)\,\left( -\sin y\frac{dy}{dx} \right)+\cos y(-1) \right\}\,=0\] Þ \[\cos y\frac{dy}{dx}+x\sin y\,\,{{e}^{-x\cos y}}\frac{dy}{dx}-\cos y{{e}^{-x\cos y}}=0\] Þ \[\frac{dy}{dx}=\frac{\cos y\,\,{{e}^{-x\cos y}}}{\cos y+x\sin y\,\,{{e}^{-x\cos y}}}\] Þ \[{{\left( \frac{dy}{dx} \right)}_{(1,\,\pi )}}=\frac{\cos \pi \,\,{{e}^{-\cos \pi }}}{\cos \pi +\sin \pi \,\,{{e}^{-\cos \pi }}}\]= \[\frac{(-1)e}{-1+0}=e\].
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