I am proving $(x^n)'=nx^{n-1}$ by the definition of the derivative:
\begin{align} (x^n)'&=\lim_{h \to 0} {(x+h)^n-x^n\over h}\\ &=\lim_{h \to 0} {x^n+nx^{n-1}h+{n(n-1)\over 2}x^{n-2}h^2+\cdots+h^n-x^n\over h} \\ &=\lim_{h \to 0} \left[ nx^{n-1}+{n(n-1)\over 2}x^{n-2}h+\cdots+h^{n-1} \right] \end{align}
Because polynomial is continuous for every $x$, we can conclude that $\lim_{x_0\to 0}(x_0)^n=0$. Therefore $$\lim_{h \to 0} \left[ nx^{n-1}+{n(n-1)\over 2}x^{n-2}h+\dots+h^{n-1} \right]= nx^{n-1}$$
Is this proof valid?
$\endgroup$ 32 Answers
$\begingroup$It is correct, but too many ellipses tend to obscure a bit the proof. And ellipses are not very rigurous, if you ask me.
I'd rewrite it using this:
$\endgroup$ 1 $\begingroup$For $n\ge 2$ there exists some polynomial $P(x,h)$ such that $$(x+h)^n=x^n+nhx^{n-1}+h^2P(x,h)$$
The proof in OP is correct if $n$ is a positive integer. For a generic real exponent $a$ we can start from the derivative of the exponential function $y=e^x \rightarrow y'=e^x$.
From the inverse function differentiation rule we find $y=\log x \rightarrow y'=\dfrac{1}{x}$ and (using the chain rule):
$$ y=x^a =e^{a \log x} \rightarrow y'=e^{a \log x} (a\log x)'=e^{a \log x} \dfrac{a}{x}=ax^{a-1} $$
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