What is the Derivative of x^x

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I was browsing through my old textbook and I found this problem:

Find Derivative of $ x^x$

My work :

Haven't got a clue yet, where to start?

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You start by writing $x^x = e^{x\ln x}$. I hope you can progress from there by the chain rule, product rule, et cetera.

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If $f(x) = x^x = e^{\ln (x^x)} = e^{x\ln(x)}$

Now we can calculate the derivate of $e^{x\ln(x)}$.

So we get $f^{\prime}(x) = e^{x\ln(x)}\left(\ln(x) + \frac{x}{x}\right) = (\ln(x) + 1)x^x$.

Do you see why? I used the product rule and the fact that $x^x = e^{\ln(x^x)} = e^{x\ln(x)}$.

I hope this helps.