# How do you differentiate yz = ln(x + z)?

$f ' \left(z\right) = \frac{1}{z + x}$
$\left(\ln \left(z\right)\right) ' = \frac{1}{z}$
$f \left(z\right) = \ln \left(z + x\right)$
$f ' \left(z\right) = \frac{1}{z + x}$