A) \[\sin y\]
B) \[-x\cos y\]
C) \[e\]
D) \[\sin y-x\cos y\]
E) \[\sin y+x\cos y\]
Correct Answer: C
Solution :
\[\sin y+{{e}^{-x\cos y}}=e\] On differentiating w.r.t.\[x,\]we get \[\cos y\frac{dy}{dx}-{{e}^{-x\cos y}}\] \[\times \left( -x\sin y\frac{dy}{dx}+\cos y \right)=0\] \[\Rightarrow \]\[(\cos y+x{{e}^{-x\cos y}}\sin y)\frac{dy}{dx}\] \[={{e}^{-x\cos y}}\cos y\] At\[(1,\pi )\] \[\cos \pi {{\left( \frac{dy}{dx} \right)}_{(1,\pi )}}={{e}^{-\cos \pi }}.\cos \pi \] \[\Rightarrow \] \[{{\left( \frac{dy}{dx} \right)}_{(1,\pi )}}=e\]
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