Is $|X|^{|X|}$ differentiable at $X_0 = 0$

Question

Let $f:\mathbb{R} \mapsto \mathbb{R}$, $f(x) = |x|^{|x|}$
We also assume that $f(0) = 1$

is $f$ differentiable at $x_0 = 0$ ?

I tried the following :

$$f'(0) = \lim_{h\to 0} \frac{f(h) - f(0)}{h} = \frac{f(h) - 1}{h}$$

I also tried taking the $\lim_{h\to 0^{+}}$ and $\lim_{h\to 0^{-}}$ but I end up dividing by zero.

EDIT:

So following some of the tips in the answers, I did the following :

$$f(x)=|x|^{|x|}$$ $$f'(x) = |x|^{|x|} \cdot (\frac{ln(|x|) \cdot x}{|x|} + 1)$$
Would someone please confirm if what I did is correct or not.

Answer 1

You can prove $\frac{d}{dh}h^h=h^h\ln(eh)$ by logarithmic differentiation, so by L'Hôpital's rule$$\lim_{h\to0^+}\frac{h^h-1}{h}=\lim_{h\to0^+}\underbrace{h^h}_{\to1}\underbrace{\ln(eh)}_{\to-\infty}=-\infty.$$Since the derivative of an even function is odd, the $h\to0^-$ limit is $+\infty$, so there is no $h\to0$ limit.

Answer 2

When working with expressions of the form $f(x)^{g(x)}$ where it is known that $f(x)>0$ it is often comfortable to note that $f(x)^{g(x)}=e^{\ln(f(x)^{g(x)})} = e^{f(x)\ln g(x)}$. In particular, when $x\ne 0 $ you know that $|x|>0$.

You get for $x>0$ that $|x|^{|x|} = e^{x \ln(x)}$, and using the chain law shows that for $x>0$ it holds that $$\frac{d}{dx}e^{x \ln(x)} = x\ln(x)e^{x\ln(x)}(\ln(x)+1)$$

In particular, the right derivative at $0$ only exists if the following limit exists, and then it is equal this limit:

$$\lim_{x\to0^+} x\ln(x)e^{x\ln(x)}(\ln(x)+1)$$

The trick to calculating this limit is to first note, using L'hopital's theorem, that

$$\lim_{x\to0^{+}}x\ln\left(x\right)=\lim_{x\to0^{+}}\frac{\ln\left(x\right)}{1/x}\overset{Lh}{=}\lim_{x\to0^{+}}\frac{1/x}{-1/x^{2}}=\lim_{x\to0^{+}}-x=0$$

The continuity of $e^x$ at $0$ implies that $\lim_{x\to0^{+}}e^{x\ln x}=1$, so if the limit $\lim_{x\to0^{+}}x\ln\left(x\right)\left(\ln\left(x\right)+1\right)$ exists and equal to $L$ you get from limit arithmetic that the right derivative is $L$. However, $x\ln\left(x\right)\left(\ln\left(x\right)+1\right)=x\ln(x)+x\ln(x)\ln(x)$ and we already know that $\lim_{x\to0^{+}}x\ln\left(x\right)=0$ so it remains to calculate $\lim_{x\to0^{+}}x\ln\left(x\right)\ln(x)$.

It follows from another aplication of L'hopital that this limit is also $0$, so the upshot of this whole operation is that the right derivative of $|x|^{|x|}$ is $0$.

To finish the exercise, repeat the procedure to figure out whether the left derivative exists and equals the right derivative.