The chain rule: a careful proof and two applications

Composite functions and the chain rule

If $f$ maps $x$ to an intermediate value and $g$ acts on that value, their composition is $g(f(x))$ . For example, $f(x)=x^2+1$ and $g(u)=u^2$ give $g(f(x))=(x^2+1)^2$ .

If $f$ is differentiable at $x$ and $g$ is differentiable at $f(x)$ , then

\boxed{\frac{d}{dx}g(f(x))=g'(f(x))f'(x).}

A useful reminder is differentiate the outside, keep the inside, then multiply by the derivative of the inside. The derivative of the outside must be evaluated at $f(x)$ , not at $x$ .

For the example,

\frac{d}{dx}(x^2+1)^2=2(x^2+1)\cdot2x=4x(x^2+1).

Expanding the polynomial first gives the same result and provides a simple check.

A proof that also handles zero inner increments

Fix a point $x$ and write $b=f(x)$ . Differentiability of $g$ at $b$ means that the function

R(u)=\begin{cases}\dfrac{g(b+u)-g(b)}u,&u\ne0,\\g'(b),&u=0\end{cases}

is continuous at $u=0$ .

Let $\Delta f=f(x+h)-f(x)$ . For every sufficiently small nonzero $h$ ,

\frac{g(f(x+h))-g(f(x))}{h}=R(\Delta f)\frac{\Delta f}{h}.

This identity remains valid when $\Delta f=0$ , because both numerators are then zero. Since differentiability of $f$ implies continuity, $\Delta f\to0$ . Taking limits gives

\lim_{h\to0}R(\Delta f)\frac{\Delta f}{h}=g'(f(x))f'(x).

The familiar proof that divides by $f(x+h)-f(x)$ is a useful heuristic, but this extra definition fills the gap when that increment vanishes, possibly for infinitely many nearby values of $h$ .

Problem 1: a logistic function and a convergent iteration

Let

f(x)=\frac1{1+e^{-x}},\qquad x\in\mathbb R.

Find the maximum of $f'(x)$ .
Prove that $f(x)=x$ has exactly one real solution.
For any real starting value $a_1$ , prove that $a_{n+1}=f(a_n)$ converges to that solution.

Part 1: differentiate from the outside inward

The outside reciprocal function has derivative $-u^{-2}$ , while the inside has derivative $-e^{-x}$ . Thus

f'(x)=\frac{e^{-x}}{(1+e^{-x})^2}=\frac1{e^x+e^{-x}+2}.

For $t=e^x>0$ , the inequality $(\sqrt t-1/\sqrt t)^2\ge0$ gives $t+1/t\ge2$ . Therefore

0<f'(x)\le\frac14,

with equality precisely at $e^x=1$ , or $x=0$ . The maximum is $\boxed{1/4}$ .

Part 2: a unique fixed point

Put $g(x)=f(x)-x$ . Then $g'(x)=f'(x)-1\le-3/4<0$ , so $g$ is strictly decreasing. Since $0<f(x)<1$ ,

\lim_{x\to-\infty}g(x)=+\infty,\qquad\lim_{x\to+\infty}g(x)=-\infty.

Continuity gives at least one zero, and strict decrease gives at most one. Call it $\alpha$ . As $\alpha=f(\alpha)$ , it lies in $(0,1)$ .

Part 3: a contraction estimate

Subtract $\alpha=f(\alpha)$ from the recurrence:

a_{n+1}-\alpha=f(a_n)-f(\alpha).

If $a_n\ne\alpha$ , the mean value theorem gives an intermediate point $c_n$ such that

f(a_n)-f(\alpha)=f'(c_n)(a_n-\alpha).

The derivative bound yields

|a_{n+1}-\alpha|\le\frac14|a_n-\alpha|.

If $a_n=\alpha$ , both sides are zero, so the inequality holds in that case too. Iteration gives

0\le|a_n-\alpha|\le\left(\frac14\right)^{n-1}|a_1-\alpha|\longrightarrow0.

Hence $\boxed{a_n\to\alpha}$ for every real initial value. The bound is an inequality, not generally an equality; a maximum derivative provides an upper bound on the contraction factor.

Problem 2: compare an exponential with a cosine

Problem. Prove that for $0<x<\pi/2$ ,

e^{-x^2/2}>\cos x.

Instead of differentiating their difference repeatedly, multiply by the positive factor $e^{x^2/2}$ . It suffices to show

F(x)=e^{x^2/2}\cos x<1.

The chain and product rules give

F'(x)=e^{x^2/2}(x\cos x-\sin x).

Let $G(x)=x\cos x-\sin x$ . Then

G'(x)=-x\sin x<0\qquad(0<x<\pi/2).

Since $G(0)=0$ , we have $G(x)<0$ , so $F'(x)<0$ . Consequently $F(x)<F(0)=1$ , proving the required strict inequality.

The important rearrangement is to turn a comparison into a product whose derivative contains a common positive exponential factor. The sign question then reduces to an elementary trigonometric expression.

A practical checking routine

Identify every layer before differentiating. In a reciprocal containing an exponential, for example, both the reciprocal and the exponential need the chain rule. Check the sign of each derivative, and distinguish a bound on a derivative from an equality. When a proof involves a difference quotient, do not divide by a quantity that may be zero without handling that case.