KindiakMath

The Nuclear Weapon of Differentiation

In differentiation, three results are used repeatedly to establish a slew of commonly-used derivatives: the chain rule, the product rule, and the quotient rule.

The chain rule is the key result that establishes the other two, and even helps us differentiate x^r for rational r. For real r, we require a few more notions in our toolkit.

Recall the fundamental definition of the derivative of a function.

Definition. Let f be a real-valued function defined on an open interval (a,b) \subseteq \mathbb R. We say that f is differentiable at c \in (a,b) with derivative f'(c) if the limit of the following equation on the right-hand side exists:

\displaystyle f'(c) := \lim_{h \to 0} \frac{f(c+h) - f(c)}{h}.

We say that f is differentiable on (a,b) if f is differentiable at every c \in (a,b).

We have already seen that f'(x) = nx^{n-1} whenever f(x) = x^n, when n \geq -1 is a positive integer, and in all cases, when x \neq 0.

The line of attack is this: we will establish a useful differentiation lemma that helps us prove the chain rule. From the chain rule, we will have all we need to prove the product and quotient rules, in sequence.

For preliminaries, let’s verify that all differentiable functions are continuous.

Proposition. Let f : (a,b) \to \mathbb R be a function. For any c \in (a,b), if f is differentiable at c, then f is continuous at c.

Proof. Take x \to c and apply limit laws on the decomposition

\displaystyle f(x) = \frac{f(x) - f(c)}{x-c} \cdot (x-c) + f(c).

So all differentiable functions are continuous. Are all continuous functions differentiable? Clearly not, since |x| is continuous at 0, but not differentiable there. Carathéodory’s theorem below answers this question.

Lemma (Carathéodory’s Theorem). Let f : (a,b) \to \mathbb R be continuous. Then f is differentiable at c\in (a,b) if and only if the function \varphi : (a,b) \to \mathbb R defined by

\displaystyle \varphi(x) := \begin{cases} \frac{f(x) - f(c)}{x - c},& x\neq c,\\ f'(c), & x = c. \end{cases}

is continuous.

As an aside, this also helps us create differentiable functions using continuous functions.

Corollary. Let f : (a,b) \to \mathbb R be continuous. Fix c \in (a,b). Suppose the function \varphi : (a,b) \backslash\{c\} \to \mathbb R defined by

\displaystyle \varphi(x) := \frac{f(x) - f(c)}{x - c}

has a limit L at x = c. Then f is differentiable at c and f'(c) = L.

Proof. Extend the definition of \varphi by \varphi(c) := L, then apply Carathéodory’s theorem.

Carathéodory’s theorem is the key superpower to prove that the chain rule works as expected.

Theorem 1 (Chain Rule). Let f,g: \mathbb R \to \mathbb R be functions. Suppose f is differentiable at c and g is differentiable at f(c). Then g \circ f is differentiable at c with derivative (g \circ f)'(c) = g'(f(c))f'(c).

Proof. Since g is differentiable at f(c), we can apply Carathéodory’s theorem to find a continuous function \varphi such that

g(y) - g(f(c)) = \varphi(y) (y - f(c)),

and \varphi(f(c)) = g'(f(c)).

Setting y = f(x),

g(f(x)) - g(f(c)) = \varphi(f(x)) (f(x) - f(c)).

Apply Carathéodory’s theorem a second time to find some continuous function \psi such that

f(x) - f(c) = \psi(x)(x-c)

and \psi(c) = f'(c). Substituting,

(g \circ f)(x) - (g \circ f)(c) = ((\varphi \circ f) \cdot \psi)(x) (x-c).

By Carathéodory’s theorem, the continuous function \Phi := (\varphi \circ f) \cdot \psi (since f is continuous at c) with \Phi(c) = g'(f(c))f'(c) establishes the result.

Here’s a simplest application of the chain rule.

Theorem 2. Given that \sin' = \cos, we have \cos = -{\sin}.

Proof. Recall from trigonometry that \cos(x) = \sin(x+\pi/2) and \sin(x+\pi) = -{\sin (x)}. By the chain rule,

\begin{aligned} \frac{\mathrm d}{\mathrm dx} \cos(x) &= \frac{\mathrm d}{\mathrm dx} \sin(x+\pi/2) \\ &= \cos(x + \pi/2) \cdot 1\\ &= \sin((x+\pi/2) + \pi/2) \\ &= \sin(x+\pi) = -{\sin(x)}. \end{aligned}

With the chain rule, we obtain the product rule as a corollary. We even obtain the derivatives of n-th roots (coupled with the inverse function theorem to establish existence). But we will elaborate on both of them in a future post.

—Joel Kindiak, 20 Oct 24, 1504H

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