Derivative of ${x^{x^2}}$ Studying past exam problems for my exam in ~$4$ weeks, and I came across this derivative as one of the questions. I actually have no idea how to solve it.
$$\frac{d}{dx} (x^{x^2})$$
Using the chain rule on it letting $x^2 = u$ led to me getting $2x^{x^2-2}$, which isn't right. The function acts like $e^x$ so I am thinking I have to convert it to this form.
So I took it to be:
$$\frac{d}{dx} (e^{{x^2}log(x)})$$
Not really sure where to go from here, or if I am going in the right direction.
 A: Hint:
$$\frac{d}{dx} (e^{{x^2}log(x)})=\frac{d}{dx}\left({{x^2}log(x)}\right)e^{{x^2}log(x)}$$

$$\frac{d}{dx} (e^{{x^2}log(x)})=\frac{d}{dx}\left({{x^2}log(x)}\right)e^{{x^2}log(x)}=\left(x+2x\log x \right)(e^{{x^2}log(x)})$$

A: First, we think back to when we learned how to differentiate $x^n$, and write $\frac{d}{dx} x^{x^2} = x^2 x^{x^2 -1} = x^{x^2+1}$.
But oh no!  Later we learned how to differentiate $c^x$ as $\operatorname{log}c \cdot c^x$!  So let's do it that way, making sure to remember the chain rule: $\frac{d}{dx} x^{x^2} = 2x \log x \cdot x^{x^2}$.
Still, we know that neither of those answers can be right.  So we'll just write them both down, and hope we get partial credit: $\frac{d}{dx} x^{x^2} = x^{x^2+1} + 2x \log x \cdot x^{x^2}$.
A: I also first try to express the exponentiation in $y = x^{x^2}$ with the exponential function, because I can remember the derivation rules here better. :-)
Taking the natural logarithm both sides of the equation, we get $\ln y = x^2 \ln x$ and then inverting it again, we have $y = e^{x^2 \ln x}$. That is how far you got. 
Now we use the chain rule:
$y' = e^{x^2 \ln x} \left(x^2 \ln x\right)'$ and then
$y' = e^{x^2 \ln x} \left(2 x \ln x + x^2 \frac{1}{x} \right)$ which gives
$y' = x^{x^2} \left(2 x \ln x + x \right)$ and finally
$y' = x^{x^2 + 1} \left(2 \ln x + 1 \right)$.
This is similiar to the answer by user T. Bongers, if you resubstitute $y = x^{x^2}$ in his answer and do a bit of combining the terms.
Using the übercool JavaScript port of Gnuplot at http://gnuplot.respawned.com/, where I paste 
set terminal svg enhanced size 400,300
set output 'out.svg'
set grid
plot [0:2][-0.5:1.5] exp(x*x*log(x)) title "y", exp((x*x+1) * log(x)) \
* (2 * log(x) + 1) title "y'", 2 * log(x) + 1 title "n"            

I get this nice graph:

If you read until here I thank you. For me it was my first post here and i am amazed about the expressive features (formulas, graphs) here. 
A: Yes, you've done it correctly so far, and could proceed by the chain rule.
On the other hand, set
$$y = x^{x^2}$$
Then $\ln{y} = x^2 \ln{x}$; taking a derivative on both sides and using the chain rule for the left leads to
$$\frac{y'}{y} = 2x \ln{x} + \frac{x^2}{x}$$
