Question

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

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

Answer 1

`0/0`

Undefined

Explanation:

First find the derivative.

Use the Chain Rule.

`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.