# Question #8d476

Mar 2, 2016

${f}_{x} = z {e}^{x z} + y$

${f}_{y} = x$

#### Explanation:

When we differentiate with respect to a given variable, all other variables are treated as constants. Thus, given
$f \left(x , y , z\right) = {e}^{x z} + x y$

$\text{ }$

${f}_{x} = \frac{d}{\mathrm{dx}} \left({e}^{x z} + x y\right)$

$= \left(\frac{d}{\mathrm{dx}} {e}^{x z}\right) + \left(\frac{d}{\mathrm{dx}} \left(x y\right)\right)$

$= z {e}^{x z} + y$

and

${f}_{y} = \frac{d}{\mathrm{dy}} \left({e}^{x z} + x y\right)$

$= \left(\frac{d}{\mathrm{dy}} {e}^{x z}\right) + \left(\frac{d}{\mathrm{dy}} x y\right)$

$= x$