Question

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

• A. $0$
• B. $1$
• C. $-1$
• D. None of these

Answer 1

The correct option is B $1$

Given $x{e}^{xy}=y+{\sin }^{2}x$ ...(1)
on putting $x=0$, we get $0.e^0=y+0\Rightarrow y=0$
on differentiating (1) both sides w.r.t $x$, we get
$1.e^{xy}+x.e^{xy}(x.\frac{dy}{dx}+y)=\frac{dy}{dx}+2\sin x\cos x$
On putting $x=0,y=0$, we get
$e^0+0(0+0)=\left[\frac{dy}{dx}\right]+2\sin 0$

$\Rightarrow \frac{dy}{dx}=1$