# How do you find the derivative of f(x)=2/root3(x)+3cosx?

Dec 3, 2017

$- \frac{2}{3 {\sqrt[3]{x}}^{4}} - 3 \sin x$

#### Explanation:

First, we can write those square root/ other roots into a simple form ${x}^{n}$.
Then we can apply the power rule:
$y = {x}^{n}$
$\frac{\mathrm{dy}}{\mathrm{dx}} = y ' = n {x}^{n - 1}$

Let's start to solve it!
$f \left(x\right) = \frac{2}{\sqrt[3]{x}} + 3 \cos x$
$= 2 {x}^{- \frac{1}{3}} + 3 \cos x$

$f ' \left(x\right) = 2 \left(- \frac{1}{3}\right) {x}^{- \frac{1}{3} - 1} + 3 \left(- \sin x\right)$
$= - \frac{2}{3} {x}^{- \frac{4}{3}} - 3 \sin x$
$= - \frac{2}{3 {\sqrt[3]{x}}^{4}} - 3 \sin x$

Feel free to ask me if you have any questions.

Dec 3, 2017

#### Explanation:

Use the power rule and the derivative of cosine.

$\frac{d}{\mathrm{dx}} \left(\frac{2}{\sqrt[3]{x}}\right) = \frac{d}{\mathrm{dx}} \left(2 {x}^{- \frac{1}{3}}\right) = - \frac{2}{3} {x}^{- \frac{4}{3}} = - \frac{2}{3 {x}^{\frac{4}{3}}}$

$\frac{d}{\mathrm{dx}} \left(3 \cos x\right) = 3 \left(- \sin x\right) = - 3 \sin x$.

So,

$f ' \left(x\right) = - \frac{2}{3} {x}^{- \frac{4}{3}} - 3 \sin x$

Rewrite the first term to preference.