#### Answer

$$
S_w =\frac{z}{1+(wz)^2}, \quad S_z=\frac{w}{1+(wz)^2}.
$$

#### Work Step by Step

Recall that $(\tan^{-1} x)'=\dfrac{1}{1+x^2}$
Since $ S=\tan^{-1}(wz)$, then using the chain rule, we have
$$
S_w =\frac{z}{1+(wz)^2}, \quad S_z=\frac{w}{1+(wz)^2}.
$$