# How do you differentiate f(x)=cos3x^(1/3) using the chain rule?

Jul 16, 2016

$f ' \left(x\right) = - \frac{\sin \left(3 {x}^{\frac{1}{3}}\right)}{{x}^{\frac{2}{3}}}$

#### Explanation:

The chain rule for derivatives states that

$\frac{d}{\mathrm{dx}} \left[f \left(g \left(x\right)\right)\right] = f ' \left(g \left(x\right)\right) \cdot g ' \left(x\right)$

In our case, we have

$f \left(x\right) = \cos \left(3 {x}^{\frac{1}{3}}\right)$

Thus, by applying the chain rule, we get

$f ' \left(x\right) = \frac{d}{\mathrm{dx}} \left[\cos \left(3 {x}^{\frac{1}{3}}\right)\right] = - \sin \left(3 {x}^{\frac{1}{3}}\right) \cdot \frac{1}{\cancel{3}} \cdot \cancel{3} {x}^{- \frac{2}{3}}$

Simplifying this expression further yields

$f ' \left(x\right) = - \sin \left(3 {x}^{\frac{1}{3}}\right) \cdot {x}^{- \frac{2}{3}}$

$= - \frac{\sin \left(3 {x}^{\frac{1}{3}}\right)}{{x}^{\frac{2}{3}}}$