A) \[0\]
B) \[1\]
C) \[e\]
D) \[-1\]
Correct Answer: B
Solution :
Given, \[x{{e}^{xy}}=y+{{\sin }^{2}}x\] ?..(i) On differentiating w. r. t .x, we get \[{{e}^{xy}}+x{{e}^{xy}}\left( y+x\frac{dy}{dx} \right)=\frac{dy}{dx}+2\sin x\,\cos x\] ?.(ii) On putting \[x=0,\] in Eq. (i), we get \[0({{e}^{0\times y}})=y+{{\sin }^{2}}(0)\] \[\Rightarrow \] \[0=y+0\Rightarrow y=0\] \[\therefore \] At point \[(0,0),\] from Eq. (ii), \[{{e}^{0}}+0{{e}^{0}}\left( 0+0\frac{dy}{dx} \right)=\frac{dy}{dx}+2\,\sin \,0\,\cos \,0\] \[\Rightarrow \] \[1+0=\frac{dy}{dx}+0\] \[\Rightarrow \] \[\frac{dy}{dx}=1\]
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