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), $<br>e^0+0e^0\left(0+0\frac{dy}{dx}\right)=\frac{dy}{dx}+2\sin 0\cos 0$
$\Rightarrow$ $1+0=\frac{dy}{dx}+0$
$\Rightarrow$ $\frac{dy}{dx}=1$