Take the 2-minute tour ×  I can't get any of these on my own and
I am attempting to do $x^x$ I had it explained to me for about
an hour and I still can't do it on my own. I thought I was
supposed to make it into $x(\ln x)$ which should be equivalent
to the original term. Then the differentiation should be easy
for most people from here.  $$1(\ln x)+x(\frac1x) $$ or $$\ln
x+1$$  This is of course wrong but I do not know why.
share|improve this question Yikes,I failed a test in elementary
real analysis for screwing that one up......LOL – 
Mathemagician1234 Oct 14 '11 at 3:55 @Mathemagician1234 : What
is the point of that comment? Why should we care? –  Adam Smith
Oct 14 '11 at 4:54 add comment  2 Answers  up vote 5 down vote
accepted  Use the fact that $y=e^{\ln(y)}$ and that
$\ln(a^b)=b\ln(a)$ to get  $$ x^x=e^{\ln(x^x}=e^{x\ln(x)}. $$
Then the derivative you want will be the derivative of this:  $$
\frac{d}{dx}(x^x)=\frac{d}{dx}(e^{x\ln(x)}). $$  By chain rule
this derivative is then:  $$ \frac{d}{dx}(e^{x\