If `f(x)=|x|^3` , show that `f"(x)` exists for all real `x` and find it.

Explanation

`f(x)=|x|^(3)`
when `x ge 0` then `|x| =x`
`f(x)=|x|^(3)=x^(3)`
` :. (d)/(dx)f(x)=(d)/(dx)x^(3)impliesf'(x)=3x^(2)`
Agian ` (d)/(dx)f'(x)=3(d)/(dx)x^(2)impliesf''(x)=3(2x)=6x`
When `x lt 0 " then "|x|= -x`
`f(x)=|x|^(3)=(-x)^(3)=-x^(3)`
` :. (d)/(dx)f(x)=(d)/(dx)(-x^(3))impliesf'(x)= -3x^(2)`
Again `(d)/(dx)f'(x)=(d)/(dx)(-3x^(2))impliesf''(x)= -6x`
` :. f''(x)={(" "6x","xge0),(-6x","xlt0):}`