Derivative of inverse of a function $f(x) = x^x$

Question

$f(x) = x^x, x$ belongs to $(0,∞)$ and I need to find the derivative of $f^{-1}(x)$

I have tried taking $\log$ on both sides (like this $\log f(x)=x\log x$ ) and now I can't decide what to do next.

Thanks in advance :)

Answer 1

Let $\displaystyle y = f(x) = x^x$

Then $\displaystyle x = f^{-1}(y)$. What you want to do is to find $\displaystyle \displaystyle \frac{dx}{dy}$ and express that in terms of $\displaystyle y$.

$\displaystyle \ln y = x\ln x$

$\displaystyle \frac 1y \cdot \frac{dy}{dx} = \ln x + 1$

$\displaystyle \frac{dx}{dy} = \frac 1{x^x(1+\ln x)}$.

$\displaystyle \frac{d}{dy}(f^{-1}(y)) = \frac 1{y(1+\ln x)}$

The problem is getting $\displaystyle x$ in terms of $\displaystyle y$, i.e. the inverse function of $\displaystyle y = x^x$. For this, you need the Lambert W function.

Start with $\displaystyle \ln y = x\ln x = (\ln x)e^{(\ln x)}$.

The Lambert W function $\displaystyle W(x)$ is the inverse function of $\displaystyle y = xe^x$. By comparing with the above, it should be evident that $\displaystyle W(\ln y) = \ln x$. (From this, we get that $\displaystyle x = e^{W(\ln y)} = \frac{\ln y}{W(\ln y)}$, the inverse of $\displaystyle y = x^x$, but we can end at the prior step for our purposes here).

Putting that back into our previous expression for the derivative gives $\displaystyle \frac{d}{dy}(f^{-1}(y)) = \frac 1{y(1+W(\ln y))}$

And finally, replacing $\displaystyle y$ with $\displaystyle x$ throughout to get the usual functional form, we arrive at:

$\displaystyle \frac{d}{dx}(f^{-1}(x)) = \frac 1{x(1+W(\ln x))}$ which is the final answer.