# How do you find the derivative of  ln(x+y) = x^2?

May 15, 2016

$y ' = 2 x {e}^{{x}^{2}} - 1$

#### Explanation:

The inverse relation is $x + y = {e}^{{x}^{2}}$.

Term-by-term differentiation gives

$1 + y ' = 2 x {r}^{{x}^{2}}$, applying function of function rule..

So, $y ' = 2 x {e}^{{x}^{2}} - 1$