Question

For the equation given below, evaluate y′ at the point (−2,2) xe^y−4y=2x−4−2e^2

Answer 1

Implicit differentiation
chain rule is important here

I'll show the steps partially
$e^y+xe^yy'-4y'=2$
$xe^yy'-4y'=2-e^y$
$y'(xe^y-4)=2-e^y$
$y'=\dfrac{2-e^y}{xe^y-4}$
now evaluate for (-2,2)
x=-2 and y=2
$y'=\dfrac{2-e^2}{-2e^2-4}$
$y'=\dfrac{2-e^2}{-2e^2-4}$
$y'=\dfrac{e^2-2}{2e^2+4}$
that's it, simplest form