#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

Answer:

#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.