Proof of the derivative of $x^n$

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?

• this looks entirely correct Commented Jun 22, 2015 at 21:39
• Well, unless $n$ is not an integer... but it seems you're set for the simple version. Commented Jun 22, 2015 at 21:40
• The proof you give is correct only when $n$ is a positive integer. it is easy to extend the proof for negative integer $n$ as well. There are slight complications when proving the formula for rational $n$. The derivative formula amounts to the following standard limit $$\lim_{x \to a}\frac{x^{n} - a^{n}}{x - a} = na^{n - 1}$$ which I have proved in my post paramanands.blogspot.com/2013/11/… (see "Proof of Standard Limits"). Commented Jun 23, 2015 at 4:03

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:

For $n\ge 2$ there exists some polynomial $P(x,h)$ such that $$(x+h)^n=x^n+nhx^{n-1}+h^2P(x,h)$$

• Is rigor or understanding more important? Commented Jul 3, 2015 at 2:15
• @JacksonFitzsimmons Everybody understamds the set of all possible sets. Bur rigorously, it does not exist. Commented Dec 21, 2023 at 13:38

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):

$$y=x^a =e^{a \log x} \rightarrow y'=e^{a \log x} (a\log x)'=e^{a \log x} \dfrac{a}{x}=ax^{a-1}$$

• I think you skipped the part when $n$ is rational. When $n$ is rational the derivative of $x^{n}$ can and should be calculated via algebraical means rather than using the theory of exponential and logarithmic functions. Commented Jun 23, 2015 at 3:59
• @Paramanand: This proof is valid for any $a \in \mathbb{R}$, so also rational. We can give a proof for rational exponents ( positive or negative) without use of the exponential function, but it is more laborious. Note that the derivative of the exponential function can be found very quickly by the definition of derivative using the fact that $e^{x+h}=e^xe^h$. Commented Jun 23, 2015 at 7:54