Completing the Derivative

Most students will memorise the differentiation result

$\displaystyle \frac{\mathrm d}{\mathrm d x} (x^n) = nx^{n-1}.$

Fairly enough, students apply it to the special cases $n = 2, 1/2, -1$ to obtain the crucial results

$\displaystyle \frac{\mathrm d}{\mathrm d x}(x^2) = 2x,\quad \frac{\mathrm d}{\mathrm d x}(\sqrt x) = \frac 1{2 \sqrt x},\quad \frac{\mathrm d}{\mathrm d x} \left( \frac 1x \right)= -\frac{1}{x^2} .$

But nowhere we have proven this result. Our goal is to formally prove these results, rather than believe them from on high. Our first step is to establish the result for non-negative integers $n$ .

We will first state the definition of the derivative.

Definition (Derivative). Let $f$ be a real-valued function defined on $(a-r,a+r)$ . The derivative of $f$ at $x=a$ , denoted $f'(a)$ , is defined by

$\displaystyle f'(a) := \lim_{x \to a} \frac{f(a+h) - f(a)}{h},$

whenever the right-hand side exists. If $f'$ is defined on $\mathbb R$ , we let

$\displaystyle f'(x) \equiv \frac{\mathrm d}{\mathrm d x}(f(x))$ denote the gradient function of $f.$

Later on, we will take advantage of this definition to establish useful differentiation techniques. For now, we will use this definition to prove our key result.

Theorem (Power Rule). For any positive integer $n$ , $\displaystyle \frac{\mathrm d}{\mathrm d x}(x^n) = nx^{n-1}$ .

Proof of Power Rule $(n=1)$ . Fix any nonnegative integer $n$ . Fix any real number $a$ . By definition, we hope to compute

$\displaystyle \frac{f(a+h) - f(a)}{h} = \frac{(a+h)^n - a^n}{h},$

then take $h \to 0$ . If $n = 1$ , then $f(x) = x^1$ , so that

$\displaystyle \frac{f(a+h) - f(a)}{h} = \frac{(a+h) - a}{h}=\frac{h}{h} = 1.$

Taking $h \to 0$ , this tells us that $f'(a) = 1$ . Since this holds for any real $a$ ,

$\displaystyle \frac{\mathrm d}{\mathrm d x}(x^1) = f'(x) = 1 = 1x^{1-1},$

where $x^0 = 1$ by convention. Thus, we have proven the result for $n=1$ .

What about general $n$ ? There are various approaches, but we will adopt the binomial theorem, which we will state without proof, but contextualised to the result we aim to establish.

Lemma (Binomial Theorem). For any positive integer $n$ ,

$\displaystyle {(a+h)}^n = \sum_{k=0}^n {n \choose k} a^{n-k} h^k ,$

which implies

$\displaystyle \frac{{(a+h)}^n -a^n}{h} = n a^{n-1} + h \cdot \sum_{k=1}^{n-1} {n \choose k+1} a^{n-k-1} h^{k-1}.$

The proof of the binomial theorem pertains a clever insight in combinatorics. We might discuss this in a future writeup in combinatorics. Ironically, this was my weakest topic in mathematics.

Proof of Power Rule (General $n$ ). Using the binomial theorem,

$\displaystyle \begin{aligned} \frac{f(a+h) - f(a)}{h} &= \frac{(a+h)^n - a^n}{h} \\ &= n a^{n-1} + \underbrace{h \cdot \sum_{k=1}^{n-1} {n \choose k+1} a^{n-k-1} h^{k-1}}_{P_{n-1}(h)} \\ &= na^{n-1} + P_{n-1}(h). \end{aligned}$

Since $n,a$ are constants, $P_{n-1}(h)$ is a polynomial in $h$ with degree $n-1$ . This, in fact, is the property of continuity. Therefore, $P_{n-1}(h) \to P_{n-1}(0) = 0$ as $h \to 0$ . Therefore, taking $h \to 0$ ,

$\displaystyle \lim_{h \to 0} \frac{f(a+h) - f(a)}{h} = na^{n-1} + 0.$

Since this holds for any real $a$ ,

$\displaystyle \frac{\mathrm d}{\mathrm d x}(x^n) = f'(x) = nx^{n-1}.$

Now this works for positive values of $n$ . But the result somehow should work for non-positive integers like $n = 0,-1$ .

In the $n = 0$ case, we use $x^0 = 1$ for any real $x \neq 0$ . In the $n = -1$ case, we will employ the calculation:

$\displaystyle \frac{\frac 1{a+h} - \frac 1a}{h} = -\frac 1{a(a+h)} \xrightarrow{h \to 0} -\frac 1{a^2} = (-1)a^{-2},\quad a \neq 0.$

Yet even more bizzarely, the works for non-integer rationals like $n = 1/2$ . What’s going on?

For these results, the most straightforward tool we will need the chain rule, which we will explore more pertinently when discussing differentiation.

—Joel Kindiak, 18 Oct 24, 1338H

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Completing the Derivative

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