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

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

Aug 13, 2018

$\frac{d}{\mathrm{dx}} \left({f}^{-} 1 \left(x\right)\right) = - 3 {x}^{-} 4$

#### Explanation:

Here ,

$f \left(x\right) = y = {x}^{- \frac{1}{3}}$

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

So , the inverse function of $f \left(x\right)$ exists.**

$\therefore y = {x}^{- \frac{1}{3}} = {\left(\frac{1}{x}\right)}^{\frac{1}{3}} \to \left[\because {a}^{-} n = \frac{1}{a} ^ n\right]$

$\therefore {y}^{3} = {\left({\left(\frac{1}{x}\right)}^{\frac{1}{3}}\right)}^{3}$

$\therefore {y}^{3} = {\left(\frac{1}{x}\right)}^{\frac{1}{3} \times 3} \to \left[\because {\left({a}^{m}\right)}^{n} = {a}^{m n}\right]$

$\therefore {y}^{3} = \frac{1}{x}$

$\therefore x = \frac{1}{y} ^ 3$

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

We know that ,

$f \left(x\right) = y \implies x = {f}^{-} 1 \left(y\right) \to \left[\because \text{definition of inverse function}\right]$

$\therefore$:.d/(dx)(f^-1(x))=-3x^-4 x=f^-1(y)=y^-3

Now changing variable from $y \to x ,$

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

$\therefore \frac{d}{\mathrm{dx}} \left({f}^{-} 1 \left(x\right)\right) = \frac{d}{\mathrm{dx}} \left({x}^{-} 3\right)$

$\therefore \frac{d}{\mathrm{dx}} \left({f}^{-} 1 \left(x\right)\right) = - 3 {x}^{- 3 - 1}$

$\therefore \frac{d}{\mathrm{dx}} \left({f}^{-} 1 \left(x\right)\right) = - 3 {x}^{-} 4$