Question

find the derivative.
$\frac{d}{dx}(sqrt3{x}-\frac{6}{x})$
$\frac{d}{dx}(sqrt3{x}-\frac{6}{x})=square$

Explanation:

Step1: Rewrite terms

Rewrite $\sqrt[3]{x}=x^{\frac{1}{3}}$ and $\frac{6}{x}=6x^{-1}$.

Step2: Apply derivative rules

The derivative of $x^n$ is $nx^{n - 1}$. So $\frac{d}{dx}(x^{\frac{1}{3}})=\frac{1}{3}x^{-\frac{2}{3}}$ and $\frac{d}{dx}(6x^{-1})=- 6x^{-2}$.

Step3: Calculate result

$\frac{d}{dx}(x^{\frac{1}{3}}-6x^{-1})=\frac{1}{3}x^{-\frac{2}{3}}+6x^{-2}$.

Answer:

$\frac{1}{3}x^{-\frac{2}{3}}+6x^{-2}$