Question

Differentiate

$y = \sqrt[3]{\frac{tanx}{x}}$

Answer 1

$y'=\frac{1}{3}\left(\frac{\tan x}{x}\right)^{-\frac{2}{3}}\frac{x\sec^2 x-\tan x}{x^2}$.

Answer 2

Solution Explanation:
Let

$$u=\frac{\tan x}{x}\quad \text{so that} \quad y=u^{\frac{1}{3}}.$$

Differentiating using the chain rule:

$$y'=\frac{1}{3}u^{-\frac{2}{3}}\, u'.$$

Differentiate $u=\frac{\tan x}{x}$ via the quotient rule:

$$u'=\frac{x\cdot\sec^2 x-\tan x}{x^2}.$$

Thus, the derivative of $y$ is:

$$y'=\frac{1}{3}\left(\frac{\tan x}{x}\right)^{-\frac{2}{3}}\frac{x\sec^2 x-\tan x}{x^2}.$$