# How do you find the derivative of  z = e^[(x^2)+xy]?

Mar 26, 2018

$z ' = \left(2 x + y\right) {e}^{\left({x}^{2}\right) + x y} \mathmr{and} z ' = x {e}^{\left({x}^{2}\right) + x y}$

#### Explanation:

$z = {e}^{\left({x}^{2}\right) + x y}$

The derivative of $z$ in terms of $x$ is:

$\left({e}^{u}\right) ' = u ' \cdot {e}^{u}$
$u = \left({x}^{2}\right) + x y$
$u ' = 2 x + y$

$z ' = \left(2 x + y\right) {e}^{\left({x}^{2}\right) + x y}$

In terms of y:

$z ' = x {e}^{\left({x}^{2}\right) + x y}$