Part 4: Proof of the Power Rule

Now we want a way to figure out the speed for different position functions. One important rule for doing that is the power rule. In Part 2, the example $x(t) = t^2$ suggested that the speed of $t^n$ should be $nt^{n-1}$. Now we can prove that pattern in general.

The short-time statement we want is

$$(t+h)^n = t^n + nt^{n-1}h + o(h).$$

If this holds, then the coefficient of $h$ is $nt^{n-1}$, so the speed of $x(t) = t^n$ is $nt^{n-1}$. If you want to recall the setup from Part 3, open the definition of o notation or the exact local speed statement below.

We will prove the formula by induction on $n$. When $n = 1$, we have

$$t+h = t + 1 \cdot h + o(h),$$

because 0 is certainly $o(h)$.

Now assume that for some $n$ we already know

$$(t+h)^n = t^n + nt^{n-1}h + o(h).$$

We want to prove the same kind of formula for $n+1$. Start with

$$(t+h)^{n+1} = (t+h)^n(t+h).$$

The only tools we need from Part 3 are the sum rule, the constant multiple rule, and the extra-factor rule.

Substitute the induction hypothesis.

$$(t+h)^{n+1} = \bigl(t^n + nt^{n-1}h + o(h)\bigr)(t+h).$$

Now expand.

$$\begin{aligned} (t+h)^{n+1} &= t^{n+1} + t^n h + nt^n h + nt^{n-1}h^2 + t\,o(h) + h\,o(h) \\ &= t^{n+1} + (n+1)t^n h + \bigl(nt^{n-1}h^2 + t\,o(h) + h\,o(h)\bigr). \end{aligned}$$

Now look at the bracketed terms. By the extra-factor rule, both $h^2$ and $h\,o(h)$ are $o(h)$. By the constant multiple rule, $nt^{n-1}h^2$ and $t\,o(h)$ are both $o(h)$. Then the sum rule says that the whole bracket is still $o(h)$.

That gives

$$(t+h)^{n+1} = t^{n+1} + (n+1)t^n h + o(h).$$

So if the formula is true for $n$, then it is also true for $n+1$. Since it is true for $n = 1$, it follows that it is true for every positive integer $n$.

Therefore the speed of $x(t) = t^n$ is

$$v(t) = nt^{n-1}.$$

That is the power rule.