## Calculus 10th Edition

$f'(x)=\frac{1}{2\sqrt x}-\frac{2}{\sqrt[3] {x^2}}$.
$f(x)=\sqrt x-6\sqrt[3] x=x^{\frac{1}{2}}-6x^{\frac{1}{3}}$ $f'(x)=(x^{\frac{1}{2}}-6x^{\frac{1}{3}})'$ $=(x^{\frac{1}{2}})'-(6x^{\frac{1}{3}})'$ $=\frac{1}{2}x^{\frac{1}{2}-1}-(6)(\frac{1}{3})(x^{\frac{1}{3}-1})$ $=\frac{1}{2\sqrt x}-\frac{2}{\sqrt[3] {x^2}}$. You rewrite the function as a sequence of $x$ terms raised to rational powers. By using the index rule $((x^n)'=nx^{n-1})$, differentiate the terms to get the overall derivative.