# $\dfrac{\mathrm d}{\mathrm dx}x^x=$?

Let $f(x)=x^x$.

What is the derivative of $f$?

This function can't be treated by chain rule or product rule or $(e^x)'=e^x$

-
What is your definition of $x^x$? –  Chris Eagle Oct 10 '11 at 18:17
hint: $x^x = e^{\log(x^x)} = e^{x\log(x)}$ –  Andy Oct 10 '11 at 18:20

\begin{align*}\frac{d}{dx}(x^x) &= \frac{d}{dx}(e^{x \ln x}) & \textrm{(Using the fact that x^x = e^{x \ln x})}\\ & = e^{x \ln x} \frac{d}{dx}(x \ln x) & \textrm{(Using the chain rule)} \\ &=x^x (\ln x + 1) & \textrm{(Using the product rule)}\end{align*}
One can in fact use this technique to derive the more general formula for $f(x)^{g(x)}$. –  Ｊ. Ｍ. Oct 10 '11 at 22:49
let $y=x^x$ $\implies$ $\log y=x\log x$ Differentiate on both sides $\dfrac{dy}{dx}\left(\dfrac{1}{y}\right)=\log x+1$ $\implies$ $\dfrac{dy}{dx}=x^x(\log x+1)$
$f(x)=x^x=e^{x\ln x}$ from here you use the chain rule for $(e^g)'=e^g g'$.