It is well known that the formula $$\int x^n dx = \frac{x^{n+1}}{n+1}+C$$
breaks down at $n=-1$ (we get a logarithm).
My question is, does there exist a value of $n \in \mathbb{Z}$ for which the formula $$\frac{d}{dx}x^n = nx^{n-1}$$
breaks down?
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3Don't think so. In fact, it should work for any $n \in \Bbb R$. – Adriano Sep 09 '13 at 08:06
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The difference is that with integration you divide by zero, while you only multiply with zero by differentiation. the latter case is no problem – Dominic Michaelis Sep 09 '13 at 08:33
2 Answers
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Per definition we know that for $x>0$ $$x^n = \exp( n \cdot \log(x))$$
Furthermore we only need the case $x>0$ as for $x<0 $ we can write $x=(-1)\cdot (-x)$ and $-x$ is greater than zero.
By chain rule we know that $$\frac{d}{dx} \exp(n\cdot \log(x))= n \cdot \frac{1}{x} \cdot \exp(n\cdot \log(x))$$ Simplifcation leads to $$n\cdot x^{n-1}$$ so it works for all $n\in \mathbb{R}$ where $x^n$ is defined.
Dominic Michaelis
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In a subtle way it somewhat fails for $n=0$ as $0\cdot x^{-1}$ is undefined at $x=0$. Apart from that (or if we agree to ignore isolated singularities) the rule holds for all $n\in\mathbb Z$ (and more, as Adriano comments).
Hagen von Eitzen
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