The Chain Rule for Differentiation

Table of Contents

Theorem 1 (The Chain Rule): Let

f : A \to B

and

g : B \to \mathbb{R}

. Let

a \in A

. If

$f$

is differentiable on

$A$

and

$g$

is differentiable on

$f(A)$

then

g \circ f : A \to \mathbb{R}

is differentiable on

$A$

and

[g(f(x))]' = g'(f(x)) \cdot f'(x)

Proof: Let $a \in A$ . Since $$f$$ is differentiable at $$a$$ , Carathéodory’s Differentiation Criterion tells us that there exists a continuous function $\varphi$ defined on $$A$$ such that:

(1)

\begin{align} \quad f(x) - f(a) = \varphi(x)(x - a) \quad (*) \end{align}

Similarly, since $$g$$ is differentiable at $$f(a)$$ we have that there exists a continuous function $\psi$ defined on $$B$$ such that:

(2)

\begin{align} \quad g(y) - g(f(a)) = \psi(y)(y - f(a)) \end{align}

(3)

\begin{align} \quad g(f(x)) - g(f(a)) &= \psi(f(x))(f(x) - f(a)) \\ & \overset{(*)} = \psi(f(x))[\varphi(x)(x - a)] \end{align}

By Carathéodory’s Differentiation Criterion we have that $g \circ f$ is differentiable at $$a$$ and that:

(4)

\begin{align} \quad [g \circ f]'(a) &= \psi(f(a)) \cdot \varphi(a) \\ &= g'(f(a)) \cdot f'(a) \end{align}

(5)

\begin{align} \quad [g(f(x))]' = g'(f(x)) \cdot f'(x) \quad \blacksquare \end{align}

The chain rule can be used to prove many nice results.

Theorem 2: Let

f : \mathbb{R} \to \mathbb{R}

be differentiable everywhere.
a) If

$f$

is an even function then

$f'$

is an odd function.
b) If

$f$

is an odd function then

$f'$

is an even function.

Proof of a) Since $$f$$ is an even function we have that $$f(x) = f(-x)$$ for every $x \in \mathbb{R}$ . We differentiable both sides of this equation and apply the chain rule to get:

(6)

\begin{align} \quad f'(x) &= [f(-x)]' \\ &= f'(-x) \cdot [-x]' \\ &= f'(-x) \cdot (-1) \\ &= -f'(-x) \end{align}

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