Question

2e^-x + e^y = 3^x-y at 0,0

Answer 1

Given,

$2e^{-x} + e^y = 3^{x-y}$

Differentiate with respect to x-

$-2e^{-x}+e^y\dfrac{dy}{dx}=3^{x-y}[1-\dfrac{dy}{dx}]$

then Rearranging the equation and we get-

$\dfrac{dy}{dx}=\dfrac{2e^{-x}+3^{x-y}}{e^y+3^{x-y}}$

$\dfrac{dy}{dx}_{(0,0)}=\dfrac{2e^0+3^0}{e^0+3^0}=\dfrac{2+1}{1+1}=\dfrac{3}{2}$