The chain rule for the derivative of a composition of functions

Summary

This lesson proves the chain rule: if $g$ is differentiable at $a$ and $f$ is differentiable at $g(a)$ , the composition $f \circ g$ is differentiable at $a$ , and its derivative is the product $f'(g(a)) \, g'(a)$ . The proof uses the Carathéodory characterization, which divides by nothing; the lesson explains why the argument that divides by $g(x) - g(a)$ is not enough and in which case it does work. With the rule, compositions of powers, roots, rational functions and trigonometric functions are differentiated in a few lines, which with the earlier rules required long expansions or were out of reach. The lesson also attends to the domain of the composition, which the formula for the derivative does not show, and presents the rule in Leibniz notation, where it looks like a cancellation of fractions without being one.

Learning objectives

Prove the chain rule with the Carathéodory characterization, and explain why the proof that divides by $g(x) - g(a)$ fails when $g'(a) = 0$ .
Differentiate compositions of two or more known functions, evaluating each derivative at the point that corresponds to it.
Determine the domain of a composition before differentiating it, and recognize that the hypotheses of the rule are sufficient, but not necessary.
Write and read the chain rule in Leibniz notation.

The derivative of a composition

With the rules from the lesson on differentiation rules and the formulas from the lesson on the derivatives of the elementary functions, one can differentiate $\left(2x^2 + 1\right)^{12}$ , but only by expanding the power into a polynomial of degree $24$ , or by applying the product rule over and over. And one cannot yet differentiate $\sin(\cos x)$ or $\sqrt{1 + x^2}$ : no rule says what to do with a function applied to another. Both expressions are compositions: $\left(2x^2 + 1\right)^{12}$ is the power $t^{12}$ evaluated at $t = 2x^2 + 1$ , and $\sin(\cos x)$ is the sine evaluated at $t = \cos x$ .

Two particular cases are already known. Solved problem 2 of the lesson on the elementary functions showed that $f(cx + b)$ has at $a$ the derivative $c f'(ca + b)$ : the inner function is linear, with slope $c$ , and the derivative of the outer function is evaluated at the point $ca + b$ , not at $a$ . And proposed problem 4 of the lesson on differentiation rules obtained, by induction, that $\left(f(x)\right)^n$ has at $a$ the derivative $n \left(f(a)\right)^{n - 1} f'(a)$ . In both cases the derivative is a product: the derivative of the outer function, evaluated at the value of the inner one, times the derivative of the inner one. The idea is that of rates of change: if $t = g(x)$ changes $g'(a)$ times as fast as $x$ , and $f(t)$ changes $f'(g(a))$ times as fast as $t$ , then $f(g(x))$ changes $f'(g(a)) \, g'(a)$ times as fast as $x$ .

TheoremChain rule

Let $g$ be a function defined on $A$ , differentiable at a point $a$ of $A$ that is an accumulation point of $A$ , and $f$ a function defined on a set $B$ that contains the values of $g$ , differentiable at $b = g(a)$ . Then $f \circ g$ is differentiable at $a$ , and

(f \circ g)'(a) = f'(g(a)) \, g'(a)

The two increments are factored with the Carathéodory characterization and then combined: the increment of $f$ is evaluated at $t = g(x)$ , and that of $g$ replaces $t - b$ . The figure shows, with $g(x) = x^2$ , $f(t) = \sin t$ and $a = 1$ , the secant and the tangent of $g$ , those of $f$ at $b = 1$ and, at the end, those of $\sin\left(x^2\right)$ , whose slope at $1$ is the product $\cos 1 \cdot 2$ .

Proof

$g(x) - g(a) = \dato{psi}{\psi(x)(x - a)}, \quad x \in A, \qquad f(t) - f(b) = \dato{fi}{\varphi(t)(t - b)}, \quad t \in B$
By the theorem on the characterizations of the derivative, from the lesson on the derivative as a limit, there exist $\psi \colon A \to \mathbb{R}$ , continuous at $a$ with $\psi(a) = g'(a)$ , and $\varphi \colon B \to \mathbb{R}$ , continuous at $b$ with $\varphi(b) = f'(b)$ , such that:
$f(g(x)) - f(g(a)) = \varphi(g(x)) \left(g(x) - g(a)\right) = \resaltar{\varphi(g(x)) \, \psi(x)}(x - a)$
Let $x$ be in $A$ . The number $t = g(x)$ is in $B$ , and the second equality holds for it, whether or not it equals $b$ ; then $t - b = g(x) - g(a)$ is replaced by the first factorization.
$\gamma(a) = \varphi(g(a)) \, \psi(a) = \varphi(b) \, \psi(a) = \resaltar{f'(g(a)) \, g'(a)}$
Since $g$ is differentiable at $a$ , it is continuous at $a$ , by the theorem that differentiable implies continuous; and $\varphi$ is continuous at $b = g(a)$ . By the theorem on the composition of continuous functions, from the lesson on limit and continuity, $\varphi \circ g$ is continuous at $a$ , and the factor $\gamma = (\varphi \circ g) \, \psi$ is continuous there too, by the product rule for operations with continuous functions. Since $a$ is an accumulation point of $A$ , which is the domain of $f \circ g$ , the Carathéodory characterization gives the derivative.

If $g$ is differentiable on all of $A$ and $f$ on all of $B$ , the rule holds at every accumulation point of $A$ , and it is written as an equality between functions: $(f \circ g)' = (f' \circ g) \, g'$ . The derivative of the outer function is evaluated at $g(x)$ , not at $x$ ; this is the most frequent error in applying the rule. The advantage of the proof lies in its second move: the factorization of $f$ also holds at $t = b$ , where both of its sides are zero, so it does not matter whether $g$ repeats its value $g(a)$ near $a$ .

In the rest of this lesson

The domain of a composition
Why dividing is not enough
Compositions of known functions
The chain rule in Leibniz notation
Solved problems
Horizontal tangents of a triple composition
Differentiating from a table of values
Sufficient, not necessary hypotheses
A bounded function with an unbounded derivative
A function composed with itself
Proposed problems
The derivative of an even function
Differentiating an identity
The absolute value away from the zeros
A join with the chain rule
An outer function that is not differentiable