# f(x)=(x-sinx)^(1/3); f'(0)=? How do I solve this ?

## f(x)=(x-sinx)^(1/3); f'(0)=?

Mar 11, 2018

$\frac{0}{0}$

Undefined

#### Explanation:

First find the derivative.

Use the Chain Rule.

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

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

Use the fact that ${a}^{-} 1 = \frac{1}{a}$

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

Simplify the $\frac{1}{\frac{1}{3}}$:

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

You can rewrite the $x - \sin {x}^{\frac{2}{3}}$ with roots but I'll leave it like this.

Substitute:

$f ' \left(0\right) = \frac{1 - \cos \left(0\right)}{3 {\left(0 - \sin \left(0\right)\right)}^{\frac{2}{3}}}$

$\cos \left(0\right) = 1 , \sin \left(0\right) = 0$

$\therefore f ' \left(0\right) = \frac{1 - 1}{3 {\left(0 - 0\right)}^{\frac{2}{3}}}$

$= \frac{0}{0}$ → Undefined.