# Question #9eb92

Use the Power Rule and Chain Rule to write $h ' \left(x\right) = \frac{4}{5} {\left(f \left(x\right)\right)}^{- \frac{1}{5}} \cdot f ' \left(x\right) = \frac{4 f ' \left(x\right)}{5 \sqrt[5]{f \left(x\right)}}$.
The Power Rule says that $\frac{d}{\mathrm{dx}} \left({x}^{n}\right) = n {x}^{n - 1}$ for any number $n$. The Chain Rule says that $\frac{d}{\mathrm{dx}} \left(g \left(f \left(x\right)\right)\right) = g ' \left(f \left(x\right)\right) \cdot f ' \left(x\right)$ when $f$ is differentiable at the value $x$ and $g$ is differentiable at the value $f \left(x\right)$.
For the problem at hand, $g \left(x\right) = \sqrt[5]{{x}^{4}} = {x}^{\frac{4}{5}}$ and therefore $g ' \left(x\right) = \frac{4}{5} {x}^{- \frac{1}{5}} = \frac{4}{5 \sqrt[5]{x}}$.
Therefore $h ' \left(x\right) = \frac{d}{\mathrm{dx}} \left(g \left(f \left(x\right)\right)\right) = \frac{4}{5} {\left(f \left(x\right)\right)}^{- \frac{1}{5}} \cdot f ' \left(x\right) = \frac{4 f ' \left(x\right)}{5 \sqrt[5]{f \left(x\right)}}$