Question

If $\ln \left((e-1)e^{xy}+x^2\right)=x^2+y^2$, then $\frac{dy}{dx}$ at $(1,0)$ is equal to

• A. 0
• B. 1
• C. 2
• D. 3

Answer 1

Answer: (C)

Given, $(e-1)e^{xy}+x^2=e^{x^2+y^2}$
Differentiate both sides with respect to $x$, we get
$(e-1)\cdot e^{xy}\cdot \left(x\cdot \frac{dy}{dx}+y\right)+2x=e^{x^2+y^2}\cdot \left(2x+2y\cdot \frac{dy}{dx}\right)$
$\text{ put }x=-1,y=0;\text{ we get }(e-1)\cdot \left(\frac{dy}{dx}\right)+2={e(2+0)\Rightarrow \frac{dy}{dx}|}_{(1,0)}=2$