## Calculus (3rd Edition)

$$z_x= y^x\ln y,\quad z_y= \frac{x}{y}y^x.$$
Recall the product rule: $(uv)'=u'v+uv'$ Recall that $(\ln x)'=\dfrac{1}{x}$ Since $z=y^x$, applying $\ln$ on both sides, we get $$\ln z=x\ln y.$$ Now, by using the product rule, we have $$\frac{z_x}{z}=\ln y \Longrightarrow z_x=z\ln y=y^x\ln y,\\ \frac{z_y}{z}=\frac{x}{y} \Longrightarrow z_y=z \frac{x}{y}=\frac{x}{y}y^x.$$