# Proving theorems and special cases (Part 14): The Power Law of differentiation

In a recent class with my future secondary math teachers, we had a fascinating discussion concerning how a teacher should respond to the following question from a student:

Is it ever possible to prove a statement or theorem by proving a special case of the statement or theorem?

Usually, the answer is no. In this series of posts, we’ve seen that a conjecture could be true for the first 40 cases or even the first $10^{316}$ cases yet not always be true. We’ve also explored the computational evidence for various unsolved problems in mathematics, noting that even this very strong computational evidence, by itself, does not provide a proof for all possible cases.

However, there are plenty of examples in mathematics where it is possible to prove a theorem by first proving a special case of the theorem. For the remainder of this series, I’d like to list, in no particular order, some common theorems used in secondary mathematics which are typically proved by first proving a special case.

5. Theorem. For any rational number $r$, we have $\displaystyle \frac{d}{dx} x^r = r x^{r-1}$.

This theorem is typically proven using the Chain Rule (in the guise of implicit differentiation) and the following lemma:

Lemma. For any integer $n$, we have $\displaystyle \frac{d}{dx} x^n = n x^{n-1}$.

Clearly, the lemma is a special case of the main theorem. However, the lemma can be proven without using the main theorem:

Proof of Lemma (Case 1). If $n$ is a positive integer, then $\displaystyle \frac{d}{dx} x^n = \displaystyle \lim_{h \to 0} \frac{(x+h)^n - x^n}{h}$ $= \displaystyle \lim_{h \to 0} \frac{x^n + n x^{n-1} h + \frac{1}{2} n(n-1) x^{n-2} + \dots + h^n - x^n}{h}$ $= \displaystyle \lim_{h \to 0} \left[ n x^{n-1} + \frac{1}{2} n(n-1) x^{n-2} h + \dots + h^{n-1} \right]$ $= n x^{n-1} + 0 + \dots + 0$ $= n x^{n-1}$

Case 1 can also be proven using the Product Rule and mathematical induction.

Proof of Lemma (Case 2). If $n = 0$, then the theorem is trivially true since $x^0 = 1$, and the derivative of a constant is zero.

Proof of Lemma (Case 3). If $n$ is a negative integer, then write $n = -m$, where $m$ is a positive integer. Then, using the Quotient Rule, $\displaystyle \frac{d}{dx} x^n = \displaystyle \frac{d}{dx} \left( x^{-m} \right)$ $= \displaystyle \frac{d}{dx} \left( \frac{1}{x^m} \right)$ $= \displaystyle \frac{0 \cdot x^m - 1 \cdot m x^{m-1}}{x^{2m}}$ $= -m x^{-m - 1}$ $= n x^{n-1}$

QED

Now that the lemma has been proven, the main theorem can be proven using the lemma.

Proof of Theorem. Suppose that $r = p/q$, where $p$ and $q$ are integers. Suppose that $y = x^r = x^{p/q}$. Then: $y = x^{p/q}$ $y^q = \displaystyle \left[ x^{p/q} \right]^q$ $y^q = x^p$

Let’s now differentiate with respect to $x$: $q y^{q-1} \displaystyle \frac{dy}{dx} = p x^{p-1}$ $\displaystyle \frac{dy}{dx} = \displaystyle \frac{p x^{p-1}}{q y^{q-1}}$ $\displaystyle \frac{dy}{dx} = \displaystyle \frac{p}{q} \frac{x^{p-1}}{[x^{p/q}]^{q-1}}$ $\displaystyle \frac{dy}{dx} = \displaystyle \frac{p}{q} \frac{x^{p-1}}{x^{p - p/q}}$ $\displaystyle \frac{dy}{dx} = \displaystyle \frac{p}{q} x^{p-1 - (p-p/q)}$ $\displaystyle \frac{dy}{dx} = \displaystyle \frac{p}{q} x^{p - 1 - p + p/q}$ $\displaystyle \frac{dy}{dx} = \displaystyle \frac{p}{q} x^{p/q - 1}$ $\displaystyle \frac{dy}{dx} = r x^{r-1}$

QED