Question

If `f(x) =x^(-1/3)`, what is the derivative of the inverse of f(x)?

the answer is `-3x^-4`, but i don't know how to get there.

Answer 1

`d/(dx)(f^-1(x))=-3x^-4`

Explanation:

Here ,

`f(x)=y=x^(-1/3)`

**Let , the function f be one-one and onto.

So , the inverse function of `f(x)` exists.**

`:.y=x^(-1/3)=(1/x)^(1/3)to[because a^-n=1/a^n]`

`:.y^3=((1/x)^(1/3))^3`

`:.y^3=(1/x)^(1/3xx3) to[because(a^m)^n=a^(mn)]`

`:.y^3=1/x`

`:.x=1/y^3`

`:.color(red)(x=y^(-3)`

We know that ,

`f(x)=y=>x=f^-1(y)to[because "definition of inverse function"]`

`:.`:.d/(dx)(f^-1(x))=-3x^-4# x=f^-1(y)=y^-3#

Now changing variable from `yto x,`

`:.color(blue)(f^-1(x)=x^-3`

`:.d/(dx)(f^-1(x))=d/(dx)(x^-3)`

`:.d/(dx)(f^-1(x))=-3x^(-3-1)`

`:.d/(dx)(f^-1(x))=-3x^-4`