# Is $(x^2)^x$ differentiable at $0$?

Have to grade a midterm where one of the true/false questions boils down to whether or not $f(x)=(x^2)^x$ is differentiable at 0. I'm not sure of the answer.

For one thing, the continuity of $f(x)$ is author-dependent since it hinges on what $0^0$ is taken to be. Let's assume $0^0$ is defined as 1, so that $f(x)$ is continuous at $0$.

For all $x\not=0$, $f'(x)=(\ln(x^2)+2)(x^2)^x$. Thus, $\lim_{x\to 0}f'(x)=-\infty$. Can we somehow deduce that $f'(0)$ is nonexistent from this, e.g., some kind of result of the form "wherever f is differentiable, it is continuously differentiable"?

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Just use the definition of derivative. I don't think it's differentiable. – Potato Oct 21 '12 at 23:01
It's worth pointing out that Wolfram Alpha shows a graph which very strongly suggests differentiability. Then again, if I had a penny every time Wolfram Alpha produced nonsense, I'd be richer than the House of Saud. wolframalpha.com/input/… – anon Oct 21 '12 at 23:08
Have you checked the definition of derivative at 0? – Potato Oct 21 '12 at 23:19
Is $(x^2)^x$ even defined at $0$? Whether one just plugs in to get $0^0$ or uses the definition $(x^2)^x=e^{x \ln(x)}$ it seems not. – coffeemath Oct 21 '12 at 23:28
@coffeemath In this case, we define it to be 1. – Potato Oct 21 '12 at 23:29

More direct approach can be made. Let assume we removed the discontinuity at $x=0$ by letting $f(0)=1$. Then near $x=0$, $x \log|x| = o(1)$ and hence

$$(x^2)^x = \exp(2x\log|x|) = 1 + 2x\log|x| + O(x^2 \log^2 |x|).$$

Thus we have

$$\frac{(x^2)^x - 1}{x} = 2\log |x| + O(x \log^2 |x|) = 2\log|x| + o(1).$$

This implies that $f$ is still non-differentiable at $x=0$ even after the continuation.

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"...some kind of result of the form "wherever $f$ is differentiable, it is continuously differentiable"?"

THEOREM Suppose that $f$ is continuous at $x=a$ and that $f'$ is defined for every $x$ on some neighborhood of $a$, except possibly at $x=a$. Suppose that $\lim\limits_{x\to a}f'(x)$ exists. Then $f'(a)$ exists and $f'(a)=\lim\limits_{x\to a}f'(x)$.

Here, "exists" strictly means the limit is a real number.

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Do you need an $f'$ there in the first line? – Thomas Oct 21 '12 at 23:45
@Thomas Nope. The theorem would be then a trivial assertion. It would be a restatement of the definition of continuity at a point. – Pedro Tamaroff Nov 7 '12 at 0:34
So your original post saying that "Suppose that $f$ is continuous at $x=a$ and that $f$ is defined for" was correct? (In that case, I am sorry for the edit.) – Thomas Nov 8 '12 at 19:44
@Thomas Actually your edit is OK. I thought you meant the very first $f$ in the text. – Pedro Tamaroff Nov 8 '12 at 21:44

We use the fact that $\lim_{h\to0} \exp(h \ln(h^2)) = 1$ below, which you may show by showing $\lim_{h \to 0^+} h \ln(h) = 0$.

\begin{align*} \lim_{h\to0^+} \frac{f(h) - f(0)}{h} &= \lim_{h\to0^+} \frac{(h^{2})^h - 1}{h} \\&= \lim_{h\to0^+} \frac{\exp(h \ln(h^2)) - 1}{h} \\&= \lim_{h\to0^+} \frac{\exp(h \ln(h^2)) \cdot \bigl[ \ln(h^2) + h \cdot\tfrac{1}{h^2}\cdot 2h\bigr] - 0}{1} \tag{L'Hospital} \\&= \lim_{h\to0^+} \exp(h \ln(h^2)) \cdot \bigl[ 2 \ln|h| + 2\bigr] \;\longrightarrow\; -\infty \\[2em] \lim_{h\to0^-} \frac{f(h) - f(0)}{h} &= \lim_{h\to0^-} \frac{(h^{2})^h - 1}{h} \\&= \lim_{h\to0^-} \frac{\exp(h \ln(h^2)) - 1}{h} \\&= \lim_{h\to0^+} \frac{\exp(-h \ln(h^2)) - 1}{-h} \\&= \lim_{h\to0^+} \frac{\exp(-h \ln(h^2)) \cdot \bigl[ -\ln(h^2) - h \cdot\tfrac{1}{h^2}\cdot 2h\bigr] - 0}{-1} \tag{L'Hospital} \\&= \lim_{h\to0^+} \exp(-h \ln(h^2)) \cdot \bigl[ 2 \ln|h| + 2\bigr] \;\longrightarrow\; -\infty \end{align*}

We find that both the limit from below and above fail to exist, though they are consistent (the curve is continuous without a cusp, having infinite negative slope at zero).

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