Answer
$f'(x)=\frac{1}{2\sqrt x}-\frac{2}{\sqrt[3] {x^2}}$.
Work Step by Step
$f(x)=\sqrt x-6\sqrt[3] x=x^{\frac{1}{2}}-6x^{\frac{1}{3}}$
$f'(x)=(x^{\frac{1}{2}}-6x^{\frac{1}{3}})'$
$=(x^{\frac{1}{2}})'-(6x^{\frac{1}{3}})'$
$=\frac{1}{2}x^{\frac{1}{2}-1}-(6)(\frac{1}{3})(x^{\frac{1}{3}-1})$
$=\frac{1}{2\sqrt x}-\frac{2}{\sqrt[3] {x^2}}$.
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.
