Why can you compute $\dfrac{d}{dx}{x^x} $ by holding constant the base + exponent? Pls discuss just the method below, and refrain from other methods. The alternate solution below looks wrong to me, because both the exponent and base are $x$! So how can you possibly pretend or playact that they're not, and treat the exponent and base separately as constants?

I once asked students to find the derivative of $x^x$ (with respect to $x$). One student figured that if the exponent were a constant then the answer would be $xx^{x-1}$ which is to say $x^x$, while if the base were constant the answer would be $x^x\log x$, so she added the two together to get $x^x+x^x\log x$. I was just about to mark the answer as wrong, when I realized that she had arrived at the correct answer – and, later, realized that it wasn't a coincidence, her unorthodox method actually works in a more general setting. 

 A: If $f(x) = x^x$, the student is unwittingly defining $g(x,y) = x^y$, noting that $f(x) = g(x,x)$ and computing $$f'(x) = \frac{\partial g}{\partial x}(x,x) + \frac{\partial g}{\partial y}(x,x)$$with a multivariable chain rule.
A: More generally$$\frac{d}{dx}f[g(x),\,h(x)]=\frac{\partial f}{\partial g}\frac{dg}{dx}+\frac{\partial f}{\partial h}\frac{dh}{dx}.$$Because each term vanishes if either $g$ or $h$ accordingly is constant, this result is the sum of two wrong-assumption results. For example,$$\frac{d}{dx}u^a=au^{a-1}u^\prime,\,\frac{d}{dx}a^v=a^v\ln a\cdot v^\prime,\,\frac{d}{dx}u^v=vu^{v-1}u^\prime+u^v\ln u\cdot v^\prime=u^v(vu^\prime/u+v^\prime\ln u).$$
A: For any function like that you can take two $x$s as different variable. For example $x^2, x\text{ln}(x)....$.
Why?
See, $x^x=x^y$ when $y=I(x)$, $I(x)$ is the identity function. Obviously they are the same thing and ultimately becomes a single variable function.
And then you can perform partial derivative and replace$y$ by $x$, so $\frac{dy}{dx}=1$.
If you want to check whether it actually is like partial derivative, then
$\frac{dx^x}{dx}=\lim_{h \to 0} \frac{(x+h)^{(x+h)}-x^x}{h}$
or, 
$\frac{dx^x}{dx}=\lim_{h \to 0} \frac{(x+h)^{(x+h)}-(x+h)^x+(x+h)^x-x^x}{h}=\lim \limits_{h \to 0} (\frac{d x^y}{dx})|_{y=x}+(\frac{d (x+h)^y}{dy})|_{y=x} =(\frac{d x^y}{dx})|_{y=x}+(\frac{d (x)^y}{dy})|_{y=x}$
Note that : in this special example $\frac{1}{h}=\frac{1}{\delta x}=\frac{1}{\delta y}$ as $y=x$. In general $h=\delta x =\delta y(\frac{dy}{dx})^{-1}$
