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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.2012-10-21
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    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. http://www.wolframalpha.com/input/?i=graph+y%3D%28x%5E2%29%5Ex+from+-.001+to+.0012012-10-21
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    Have you checked the definition of derivative at 0?2012-10-21
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    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.2012-10-21
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    @coffeemath In this case, we define it to be 1.2012-10-21
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    Not only is `Potato`'s answer the usual approach to defining $0^0$ in professional mathematics (e.g. the recursive definition of exponentiation), it is also consistent with taking the limit of $\mathrm e^{x \ln x}$ as $x \to 0^+$. (In many theoretical applications of real analysis, such as entropies of probability distributions in information theory, one tacitly adopts $0 \ln(0) = 0$ as a harmless abuse of notation for precisely this reason.)2012-10-21
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    Yep. Just did the calculation and got 1 for $\lim e^{x \ln x}$. Should have done that before I chimed in. It's like when one speaks of the function $x \sin(1/x)$ without specifically mention one means $0$ when $x=0$.2012-10-22

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