What is the Derivative of x^x

Question

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?

huge hint: $\ x^x=e^{\ln(x^x)}$ – Mosk Apr 11 '15 at 16:37 — Mosk, Apr 11 '15 at 16:37
Set $y = x^{x}$, take logs of both sides, differentiate.. – Matthew Cassell Apr 11 '15 at 16:37 — Matthew Cassell, Apr 11 '15 at 16:37
Thanks @Mosk , really helped – Connor Verlekar Apr 11 '15 at 16:38 — Connor Verlekar, Apr 11 '15 at 16:38

score 3 · Answer 1 · answered Apr 11 '15 at 16:37

3

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.

answered Apr 11 '15 at 16:37

Rolf Hoyer

score 1 · Answer 2 · edited Apr 11 '15 at 17:26

1

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.

edited Apr 11 '15 at 17:26

K_user

answered Apr 11 '15 at 16:40

Nescrio

2 Answers2