## Calculus 10th Edition

$h'(x)=\dfrac{3}{\sqrt{x}}+\dfrac{1}{\sqrt[3]{x^2}}.$
Using the Power Rule and the Sum Rule: $\dfrac{d}{dx}(h(x))=\dfrac{d}{dx}(6\sqrt{x})+\dfrac{d}{dx}(3\sqrt[3]{x})\rightarrow$ $h'(x)=6(\dfrac{1}{2\sqrt{x}})+3(\dfrac{1}{3\sqrt[3]{x^2}})=\dfrac{3}{\sqrt{x}}+\dfrac{1}{\sqrt[3]{x^2}}.$