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Given $f(x)=\sqrt[3]{x}+\cfrac{1}{x\sqrt{x}}$, find $f'(x)$.

$f'(x)=\cfrac{1}{3\sqrt{x^3}}-\cfrac{3}{2\sqrt[5]{x^2}}$

$f'(x)=\cfrac{1}{3\sqrt[3]{x^2}}-\cfrac{3}{2\sqrt{x^5}}$

$f'(x)=\cfrac{1}{3\sqrt[3]{x^2}}-\cfrac{3}{2\sqrt{x}}$

$f'(x)=\cfrac{3}{\sqrt[3]{x^2}}-\cfrac{2}{3\sqrt{x}}$