Question

What is the second derivative of `f(x)=(x^2-x^3)^(1/3)`?

Answer 1

`1/3(2-6x)(x^2-x^3)^(-2/3){1-((4-6x))/((3x-3x^2))}`

Explanation:

here,
`f(x)=(x^2-x^3)^(1/3)`

so, the first derivative is,

`fprime(x)=d/(dx)(f(x))`

`=d/(dx)(x^2-x^3)^(1/3)`

`=1/3(x^2-x^3)^(1/3-1)d/(dx)(x^2-x^3)`

`=1/3(x^2-x^3)^(-2/3)(2x-3x^2)`

so the second derivative is,

`fprimeprime(x)=d/(dx)(fprime(x))`

`=d/(dx)(1/3(x^2-x^3)^(-2/3)(2x-3x^2))`

`=1/3d/(dx)((2x-3x^2)(x^2-x^3)^(-2/3))`

`=1/3{(2x-3x^2)d/(dx)(x^2-x^3)^(-2/3)+(x^2-x^3)^(-2/3)d/(dx)(2x-3x^2)}`

`=1/3{-2/3(2x-3x^2)(x^2-x^3)^(-5/3)(2-6x)+(x^2-x^3)^(-2/3)(2-6x)}`

`=1/3(2-6x)(x^2-x^3)^(-2/3){-2/3((2x-3x^2)/(x^2-x^3))+1}`

`=1/3(2-6x)(x^2-x^3)^(-2/3){-2/3((cancel(x)(2-3x))/(cancel(x)(x-x^2))+1}`

`=1/3(2-6x)(x^2-x^3)^(-2/3){1-((4-6x))/((3x-3x^2))}`