Question

If $ x{{e}^{xy}}=y+{{\sin }^{2}}x, $ then $ \frac{dy}{dx} $ at $ x=0 $ is equal to

• A. $ 0 $
• B. $ 1 $
• C. $ e $
• D. $ -1 $

Answer 1

Answer: (B)

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 $