TITLE: Proof of the derivative of $x^n$
QUESTION [10 upvotes]: 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?

REPLY [8 votes]: 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):
$$
y=x^a =e^{a \log x} \rightarrow y'=e^{a \log x} (a\log x)'=e^{a \log x} \dfrac{a}{x}=ax^{a-1}
$$