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

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

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Answer 1 · 2018-03-11T08:36:55

$0/0$

Undefined

First find the derivative.

$d/dx (x-sinx)^(1/3)=1/3(x-sinx)^(-2/3)*d/dx(x-sinx)$

$=1/3(x-sinx)^(-2/3)*(1-cosx)$

Use the fact that $a^-1=1/a$

$=(1-cosx)/(1/3(x-sinx)^(2/3))$

Simplify the $1/(1/3)$ :

$=(1-cosx)/(3(x-sinx)^(2/3))$

You can rewrite the $x-sinx^(2/3)$ with roots but I'll leave it like this.

Substitute:

$f'(0)= (1-cos(0))/(3(0-sin(0))^(2/3))$

$cos(0)= 1, sin(0)=0$

$therefore f'(0)= (1-1)/(3(0-0)^(2/3))$

$=0/0$ → Undefined.