Question

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

• A. 0
• B. 1
• C. 2
• D. 3

Answer 1

Answer: (C)

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