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 $[g(f(x))]' = g'(f(x)) \cdot f'(x)$.
(1)
\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:
(2)
\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:
(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}
• Since this holds true for all $a \in A$ we have that:
(5)
 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:
• So $f'$ is an odd function. $\blacksquare$