Why can't I take the derivative of $x^x$ as $x(x^{x-1})$? Why can't I take the derivative of $x^x$ as  $x(x^{x-1})$?
I don't understand why I have to convert it to $e^{x\ln(x)}$ first.
 A: The chain rule can be applied directly, if we consider $x^x=\left.y^x\vphantom{\frac{\mathrm{d}}{\mathrm{d}x}}\right|_{y=x}$ and set $\frac{\mathrm{d}y}{\mathrm{d}x}=1$.
The part where we take the partial with respect to $x$ is missing if we just use the rule cited in the question (which is used when taking the partial with respect to $y$).
$$
\begin{align}
\frac{\mathrm{d}}{\mathrm{d}x}x^x
&=\overbrace{\left.\frac{\partial}{\partial y}y^x\frac{\mathrm{d}y}{\mathrm{d}x}\right|_{y=x}}^{\text{used in question}}+\overbrace{\left.\frac{\partial}{\partial x}y^x\right|_{y=x}}^{\substack{\text{missing in}\\\text{question}}}\\
&=\left.xy^{x-1}\frac{\mathrm{d}y}{\mathrm{d}x}\right|_{y=x}+\left.\log(y)\,y^x\vphantom{\frac{\partial}{\partial x}}\right|_{y=x}\\
&=x^x+\log(x)\,x^x
\end{align}
$$
A: You want to differentiate $x\mapsto x^x$ with respect to "$x$".
You can not use the rule
$$\frac{\mathrm d}{\mathrm d x}(x^n)=nx^{n-1}$$
because here, "$n$" is depending on $x$.
What you are stated is correct, you have to use the identity 
$$x^x=e^{x\ln(x)}$$
and then use the chain rule.
A: Taking the basis as variable, the derivative is $x\cdot x^{x-1}$, taking the exponent as such the derivative is $x^x\cdot\ln(x)$.  Adding both gives the derivative of $x^x$, namely
$$\left(x^x\right)'=x^x+x^x\cdot\ln(x).$$
 Astonishing at the first glance, isn't it?
A: I think you are asking why you can't differentiate $x^x$ in the same way you would differentiate $x^3$?
The shortcut that you use when differentiating $x^r$ where $r$ is an real number, only hold when $r$ is a real number. In the case of $x^x$, the exponent is a variable so the shortcut doesn't apply.
A: just hint
It doesn't work cause the exponent is not constant.
we cannot factor 
$x^x-x_0^{x_0}$
it is not like 
$x^n-x_0^n$
A: Some people have given good hints, but I would like to elaborate on their responses.
Students get confused with the power rule very often. I've seen these at all levels. The key to remember is that the power rule allows us to look at derivatives of polynomials - specifically, those functions whose powers are fixed numbers.
We know, and can show through induction (or even the binomial theorem; do it!) that
$$ \frac{d}{dx} [x^n] = nx^{n-1} $$
where $n \in \mathbb{R}$. What you have is a function whose base and power both vary. To use a rule we know, we must do the typical $e$ trick: recall that
$$ x^n = e^{n \ln x} $$
and so we just write $ x^x = e^{x \ln x}$ and use our typical differentiation rules.
A: This is a valid assumption many students make, but the reason is because $x^x$ is not like the power rule we have seen. The power rule applies to functions like $5x^4, 3x^2, x^{3.2}... etc$. Notice the pattern? The $exponent$ is a $number$ that stays $constant$. However, something like $x^x$, is not. The $x$ in the exponent is always $changing$, that's what $x$ does. So you cannot. 
A: As people have said, you cannot use the power rule to differentiate non-constant powers; it only works for functions resembling $f(x)^c$ where $c$ is a constant.
To differentiate $y = x^x$...
$ln(y) = ln (x^x)$
$ln(y) = x ln(x)$
$\frac{1}{y} dy = (ln(x) + x(\frac{1}{x}))dx$
$\frac{1}{y} dy = (ln(x) + 1)dx$
$\frac{dy}{dx} = (ln(x)+1)y$
$\frac{dy}{dx} = (ln(x)+1)x^x$
$\frac{dy}{dx} = x^xln(x)+x^x$
