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

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

1 Answer
Mar 11, 2018

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.