Question

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

• A. $2y^2-1$
• B. $2y$
• C. $y^2-y$
• D. $y^2+1$
• E. $y^2-1$

Answer 1

The correct option is E $y^2-1$

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