The Chain Rule for Scalar-Scalar Functions

The Chain Rule is a fundamental theorem in calculus that describes how to differentiate composite functions. It states that if you have two functions, \(f\) and \(g\), and you want to differentiate their composition \(f(g(x))\), you can do so by multiplying the derivative of \(f\) evaluated at \(g(x)\) by the derivative of \(g\) evaluated at \(x\). This is particularly useful when dealing with functions that are composed of other functions, as it allows us to break down the differentiation process into manageable parts.

Theorem (scalar-scalar chain rule)

Let \(f: \mathbb{R} \to \mathbb{R}\) and \(g: \mathbb{R} \to \mathbb{R}\) be differentiable functions. If \(f\) is differentiable at \(u_0 = g(x_0)\) and \(g\) is differentiable at \(x_0\), then the composition \(( f \circ g)(x) = f\bigl(g(x)\bigr)\) is differentiable at \(x_0\), and we have

\[ ( f \circ g)'(x_0) = f'(g(x_0)) \cdot g'(x_0). \]

where \(\circ\) denotes function composition.

Proof. Scalar–Scalar Chain Rule.

Let \(g\) be differentiable at \(x_0\) and \(f\) be differentiable at \(u_0=g(x_0)\).
Define

\[ h(x)=f\bigl(g(x)\bigr). \]

Then

\[ h'(x_0) =\lim_{\Delta x\to0}\frac{h(x_0+\Delta x)-h(x_0)}{\Delta x} =\lim_{\Delta x\to0}\frac{f\bigl(g(x_0+\Delta x)\bigr)-f\bigl(g(x_0)\bigr)}{\Delta x}. \]

Set

\[ \Delta u = g(x_0+\Delta x)-g(x_0), \]

so that \(\Delta u\to0\) and \(\tfrac{\Delta u}{\Delta x}\to g'(x_0)\) by differentiability of \(g\).

We now write

\[ \frac{f\bigl(g(x_0+\Delta x)\bigr)-f\bigl(g(x_0)\bigr)}{\Delta x} \;=\; \frac{f(u_0+\Delta u)-f(u_0)}{\Delta u} \;\times\; \frac{\Delta u}{\Delta x}. \]

By the Mean Value Theorem applied to \(f\) on the interval \([u_0,u_0+\Delta u]\), there exists some \(\xi\) between \(u_0\) and \(u_0+\Delta u\) so that

\[ \frac{f(u_0+\Delta u)-f(u_0)}{\Delta u} = f'(\xi). \]

As \(\Delta x\to0\), we have \(\xi\to u_0\), and hence \(f'(\xi)\to f'(u_0)\).

Therefore

\[ h'(x_0) =\lim_{\Delta x\to0} \underbrace{\frac{f(u_0+\Delta u)-f(u_0)}{\Delta u}}_{\to f'(u_0)} \;\times\; \underbrace{\frac{\Delta u}{\Delta x}}_{\to g'(x_0)} = f'(g(x_0))\,g'(x_0). \]

This completes the proof of the Chain Rule. ◻