The Chain Rule for Differentiation

# The Chain Rule for Differentiation

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:

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

- And $f'(a) = \varphi(a)$.

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

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

- And $g'(f(a)) = \psi(a)$.

- Set $y = f(x)$. Then:

\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:

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

- Since this holds true for all $a \in A$ we have that:

\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:

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

- So $f'$ is an odd function. $\blacksquare$

**Proof of b)**Analogous to (a).