Elegant generalization of power rule via limits

Question

Today I was thinking about how to generalize the derivative of $f(x) = x^n$. Of course, we know the following:

$$\frac{d}{dx}x^n = n\cdot x^{n-1}$$

But how can we show that the power rule behaves this way? I took to the limit process and came up with this:

$$\frac{d}{dx}x^n = \lim_{h\to 0}\frac{(x+h)^n-x^n}{h}$$

First I thought to express $(x+h)^n$ as a sum through the binomial theorem and rewrite the exression.

$$\frac{d}{dx}x^n = \lim_{h\to 0}\frac{1}{h}\left[\sum_{i=0}^n\binom{n}{i}x^nh^i\;-x^n\right]$$

Here's where I struggle to elegantly finish the generalization.

$$\frac{d}{dx}x^n = \lim_{h\to0}\frac{1}{h}\left[\binom{n}{0}x^n+\binom{n}{1}x^{n-1}h+\binom{n}{2}x^{n-2}h^2\;...\;-x^n\right]$$

Simplifying the first term:

$$\frac{d}{dx}x^n = \lim_{h\to0}\frac{1}{h}\left[{x^n}+\binom{n}{1}x^{n-1}h+\binom{n}{2}x^{n-2}h^2\;...\;-x^n\right]$$

Cancelling end terms:

$$\frac{d}{dx}x^n = \lim_{h\to0}\frac{1}{h}\left[\binom{n}{1}x^{n-1}h+\binom{n}{2}x^{n-2}h^2\;...\right]$$

Factoring out an $h$:

$$\frac{d}{dx}x^n = \lim_{h\to0}\;\frac{h}{h}\cdot\left[\binom{n}{1}x^{n-1}+\binom{n}{2}x^{n-2}h\;...\right]$$

Simplifying and using direct substitution:

$$\frac{d}{dx}x^n = \binom{n}{1}x^{n-1}+0+0\;...$$

We know this:

$$\binom{n}{1} = \frac{n!}{(n-1)!1!} = n$$

Thus:

$$\frac{d}{dx}x^n = n\cdot x^{n-1}$$

I know it may seem petty, but I hate having the points of ellipsis in a formal proof and overall it's a bit messy. How can I clean this up a bit so that it's more presentable?

Answer 1

If it helps, you could retain the summation notation, assuming $n\ge2$: $$\eqalign{ \Bigl(\sum_{i=0}^n\binom ni x^{n-i}h^i\Bigr)-x^n &=x^n+nx^{n-1}h+\Bigl(\sum_{i=2}^n\binom ni x^{n-i}h^i\Bigr)-x^n\cr &=nx^{n-1}h+h^2\sum_{i=2}^n\binom ni x^{n-i}h^{i-2}\cr}$$ and keep going from there.

Answer 2

It might be more elegant to prove the power rule for $n\in \mathbb{N}$ using induction. If we already know the product rule and that $\frac{\mathrm{d}x}{\mathrm{d}x} = 1$ (which serves as our base case), then we can deduce that if $\frac{\mathrm{d}}{\mathrm{d}x}x^{n-1} = (n-1)x^{n-2}$, then $$\frac{\mathrm{d}}{\mathrm{d}x}x^n = \frac{\mathrm{d}}{\mathrm{d}x}(x\cdot x^{n-1}) = 1\cdot x^{n-1}+x\cdot (n-1)x^{n-2} = nx^{n-1}$$ Similarly, if we know that $\frac{\mathrm{d}}{\mathrm{d}x}x^n = nx^{n-1}$, then by the quotient rule, $$\frac{\mathrm{d}}{\mathrm{d}x}x^{n-1} = \frac{\mathrm{d}}{\mathrm{d}x}\frac{x^n}{x} = \frac{nx^{n-1}\cdot x-1\cdot x^n}{x^2} = (n-1)x^{n-2}$$ which combined with the fact that $\frac{\mathrm{d}}{\mathrm{d}x}\frac{1}{x} = -\frac{1}{x^2}$ allows us to use the rule for all $n\in \mathbb{Z_{\neq 0}}$.

To generalize this for $f(x) = x^r$ where $r\in \mathbb{Q}$, we let $r = \frac{p}{q}$ for $p, q\in \mathbb{Z}$. Then, $f(x)^q = x^p$, which lets us use the chain rule to write $$qf(x)^{q-1}f'(x) = px^{p-1}\Rightarrow f'(x) = \frac{p}{q}\frac{x^{p-1}}{f(x)^{q-1}} = \frac{p}{q}x^{\frac{p}{q}-1}$$ Generalizing to the irrationals is a little trickier, but you can use the density of $\mathbb{Q}$ in $\mathbb{R}$ and the uniform convergence of derivatives on compact sets to argue that the power rule is the same for all $r\in \mathbb{R}$ (except for $0$).

Answer 3

$$\frac{d}{dx}x^n = \lim_{h\to 0}\frac{(x+h)^n-x^n}{h} = \lim_{h\to 0}\left(\frac{x^n((x+h)^n-x^n)}{x^n\log\left(\frac{(x+h)^n}{x^n}\right)}\cdot\frac{nx\log\left(\frac{x+h}{x}\right)}{xh}\right) = \lim_{h\to0}\left(\frac{x^n}{\log\left(1 + \frac{(x+h)^n-x^n}{x^n}\right)^\frac{x^n}{(x+h)^n-x^n}}\cdot\frac{n\log\left(1 + \frac{h}{x}\right)^\frac{x}{h}}{x}\right) = \frac{x^n}{\log e}\cdot\frac{n\log e}{x} = nx^{n-1}$$

Answer 4

A variation which might be useful

We obtain

\begin{align*} \frac{d}{dx}x^n&=\lim_{h\rightarrow 0}\frac{(x+h)^n-x^n}{h}\\ &=\lim_{h\rightarrow 0}\frac{1}{h}\left(\sum_{i=0}^n\binom{n}{i}x^{n-i}h^i-x^n\right)\tag{1}\\ &=\lim_{h\rightarrow 0}\sum_{i=1}^n\binom{n}{i}x^{n-i}h^{i-1}\tag{2}\\ &=\binom{n}{1}x^{n-1}\\ &=nx^{n-1} \end{align*}

Comment:

• In (1) we see that $x^n$ drops out ($i=0$) and we respect the factor $\frac{1}{h}$ by letting $h^i\rightarrow h^{i-1}$.

• In (2) we observe that only the first summand ($i=1$) is constant with respect to $h$.