Question

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

• A. $-1$
• B. 0
• C. 1
• D. 2
• E. $-2$

Answer 1

Answer: (C)

Give that, $x{e}^{xy}=y+si{n}^{2}x$ at $x=0,y=0$ On differentiating w.r.t. $x,$ we get $x{e}^{xy}\left(x\frac{dy}{dx}+y\right)+e^{xy}=\frac{dy}{dx}+2\sin x\cos x$ $\Rightarrow$ $x^2{e}^{xy}\frac{dy}{dx}-\frac{dy}{dx}=2\sin x\cos x$ $-e^{xy}-x{e}^{xy}y$ $\Rightarrow$ $\frac{dy}{dx}=\frac{\sin 2x-e^{xy}(1+xy)}{(x^2{e}^{xy}-1)}$ $\therefore$ ${\frac{dy}{dx}|}_{x=0}=\frac{0-1(1)}{(-1)}=1$