Show that $x^x$ is differentiable Define $f\colon\mathbb R_{+} \to \mathbb R, x\mapsto x^x$.
Show that $x^x$ is differentiable.

$x^x = e^{x\ln(x)}$
$\lim_{h \rightarrow 0} \frac{e^{(x+h)\ln(x+h)} - e^{x\ln(x)}}{h} = \dots$
I calculated the derivative, which should be $e^{x\ln(x)}(\ln(x)+1) = x^x\cdot(\ln(x)+1)$. But how do I get here using the differential equation?
Edit:
The full question was: Show that $x^x$ is differentiable and calculate its derivative. I already calculated its derivative so I didn't mention it in the question.
It's not required to use the limit. I tried using it since it was the definition of the derivative in the text book.Any other suggestions would be helpful.
 A: Let $f: \mathbb{R}_{+}\to \mathbb{R}$, defined by $f(x)=x^{x}$. Since that $$f(x)=x^{x}=\exp\left(x\ln(x)\right),$$is the composition of differentiable functions, hence $f$ is a differentiable function.
On the other hand,  setting $y:=x^{x}$, so we have that
$$\ln(y)=\ln(x^{x})$$
$$\ln(y)=x\ln(x)$$
$$\frac{1}{y}\frac{dy}{dx}=x\frac{1}{x}+\ln(x)$$
$$\frac{dy}{dx}=x^{x}(1+\ln(x)).$$
Therefore, $$f'(x)=x^x(1+\ln(x)).$$
A: Let us first simplify the exponent of the first term in the numerator:
$$(x+h)\ln (x+h) = x\ln x + h \ln x + x \ln (1 + \frac{h}{x}) + h \ln (1 + \frac{h}{x})$$
Using the expansion of $\ln (1 + x) = x - \frac{x^2}{2} + \ldots$, we get
$$(x+h)\ln (x+h) = x\ln x + h \ln x + \left(h - \frac{h^2}{2x} + \ldots\right) + \left(\frac{h^2}{x} - \frac{h^3}{2x^2} + \ldots\right)$$
Now, using this in the limit and taking $e^{x\ln x}$ common from both the terms in numerator we have
$$\lim_{h\to 0}\frac{e^{(x+h)\ln (x+h)} - e^{x \ln x}}{h} = x^x\left( \frac{e^{h \ln x +h + \frac{h^2}{2x}} - 1}{h}\right)$$
Now, we know that $e^x = 1 + x + \frac{x^2}{2!} + \ldots$
Therefore,
$$\lim_{h\to 0}\frac{e^{(x+h)\ln (x+h)} - e^{x \ln x}}{h} = x^x\left( \frac{(1 + h\ln x + h + \frac{h^2}{2x} + \ldots ) - 1}{h}\right)$$
which gives
$$\lim_{h\to 0}\frac{e^{(x+h)\ln (x+h)} - e^{x \ln x}}{h} = x^x(\ln x + 1)$$
A: Using well known theorems on differentiable maps
$f(x)=x^x=e^{x \ln x}$ is the composition of $x \mapsto e^x$ and $x \mapsto x \ln x$, which is differentiable as the product of two differentiable maps. Therefore $f$ is differentiable.
And you computed the derivative.
