Proof of the Chain Rule

Why the Proof Needs to Track the Inner Increment

The Chain Rule says that the derivative of a composition is the product of two rates of change. In the previous tutorial, we stated the rule and used it to differentiate compositions. Here we prove it from the definition of the derivative. The main point is to track the change in the inner function carefully: the outer function receives that change as its input increment, and that increment may be zero even when the original input has changed.

We use the First-Order Approximation Characterization from earlier in this course. It says that differentiability at a point is equivalent to having a linear approximation there with an error that, divided by the input increment, tends to zero. The proof below combines the approximations for the inner and outer functions. A small lemma makes precise why the outer error is still negligible after the inner increment is substituted.

Theorem (Chain Rule): Let $I$ and $J$ be intervals, let $a$ be an interior point of $I$, and let $g:I\to J$ be differentiable at $a$. Suppose $f:J\to\mathbb{R}$ is differentiable at $b=g(a)$, where $b$ is an interior point of $J$. Then $f\circ g$ is differentiable at $a$, and $$ (f\circ g)'(a)=f'(g(a))g'(a). $$

Transferring a Remainder Through a Small Increment

Suppose an error term $r(t)$ is negligible compared with $t$, so that $r(t)/t\to0$ as $t\to0$ through nonzero values. In a composition, the outer error is evaluated not at the original increment $h$, but at the inner increment $k(h)$. To show that the resulting error is negligible compared with $h$, we need $k(h)/h$ to stay bounded. Differentiability of the inner function will give exactly that bound.

Lemma (Remainder Transfer): Suppose $k(h)\to0$ as $h\to0$, and suppose there are constants $M\geq0$ and $\delta_0>0$ such that $|k(h)/h|\leq M$ whenever $0<|h|<\delta_0$. Let $r$ be defined near $0$, with $r(0)=0$, and suppose $r(t)/t\to0$ as $t\to0$ through nonzero values. Then $$ \frac{r(k(h))}{h}\longrightarrow0 \quad\text{as }h\to0. $$

Proof. Let $\varepsilon>0$. Since $r(t)/t\to0$, there is $\eta>0$ such that

0<|t|<\eta \quad\Longrightarrow\quad \left|\frac{r(t)}{t}\right|<\frac{\varepsilon}{M+1}.

Because $k(h)\to0$, we can choose $\delta>0$ so that $0<|h|<\delta$ implies both $|h|<\delta_0$ and $|k(h)|<\eta$. If $k(h)=0$, then $r(k(h))=r(0)=0$. If $k(h)\ne0$, the choices above give

\left|\frac{r(k(h))}{h}\right| = \left|\frac{r(k(h))}{k(h)}\right| \left|\frac{k(h)}{h}\right| < \frac{\varepsilon}{M+1}M < \varepsilon.

Thus, for every sufficiently small nonzero $h$, the quotient $|r(k(h))/h|$ is less than $\varepsilon$, including the case $k(h)=0$. This proves the limit. $\square$

The bound $M$ is important: an error that is small relative to $k(h)$ need not automatically be small relative to $h$ if $k(h)/h$ can grow without bound. In the Chain Rule proof, differentiability of $g$ ensures that the ratio is bounded near $a$.

Proof. Write $b=g(a)$. For $h\ne0$ sufficiently small, $a+h\in I$. Define the inner increment

$$ k(h)=g(a+h)-g(a). $$

Since $g$ is differentiable at $a$, the First-Order Approximation Characterization gives a remainder $r_g(h)$ such that

g(a+h)=g(a)+g'(a)h+r_g(h), \qquad \frac{r_g(h)}{h}\longrightarrow0.

Consequently,

\frac{k(h)}{h}=g'(a)+\frac{r_g(h)}{h} \longrightarrow g'(a).

In particular, $k(h)/h$ is bounded for all sufficiently small nonzero $h$. Also, $k(h)\to0$, because $k(h)/h$ is bounded and $h\to0$. Equivalently, this follows from continuity of $g$ at $a$, which is guaranteed by differentiability at $a$.

Now use differentiability of $f$ at $b$. There is an outer remainder $r_f(t)$, with $r_f(0)=0$, such that for small $t$ with $b+t\in J$,

f(b+t)=f(b)+f'(b)t+r_f(t), \qquad \frac{r_f(t)}{t}\longrightarrow0 \quad (t\to0,\ t\ne0).

For $t=k(h)$, the point $b+t=b+k(h)=g(a+h)$ belongs to $J$, since $g$ maps $I$ into $J$. The Remainder Transfer Lemma applies: $k(h)\to0$, $k(h)/h$ is bounded, and $r_f(t)/t\to0$. It follows that $r_f(k(h))/h\to0$. Substituting the two increments into the approximation for $f$ gives

\begin{aligned} f(g(a+h)) &=f(b+k(h))\\ &=f(b)+f'(b)k(h)+r_f(k(h))\\ &=f(b)+f'(b)\bigl(g'(a)h+r_g(h)\bigr)+r_f(k(h)). \end{aligned}

Subtract $f(b)=f(g(a))$ and divide by $h\ne0$. The resulting difference quotient is

\frac{f(g(a+h))-f(g(a))}{h} = f'(b)g'(a) + f'(b)\frac{r_g(h)}{h} + \frac{r_f(k(h))}{h}.

The second term tends to zero because $r_g(h)/h\to0$, and the third tends to zero by the Remainder Transfer Lemma. Therefore the difference quotient tends to $f'(b)g'(a)$. This proves that $f\circ g$ is differentiable at $a$, with

(f\circ g)'(a)=f'(b)g'(a)=f'(g(a))g'(a).

The proof includes the possibility that $k(h)=0$ for some nonzero $h$. In that case the outer remainder is exactly $r_f(0)=0$; no division by $k(h)$ is made. $\square$

Reading the Proof Through Examples

Worked Example: A Polynomial Composition at a Nonzero Increment Rate

Let $H(x)=(g(x))^2$, where $g(x)=x^2+2x$, and examine the derivative at $a=1$. Here $f(y)=y^2$, so $b=g(1)=1^2+2(1)=3$. The Power Rule gives $f'(y)=2y$, hence $f'(b)=f'(3)=6$. Also, $g'(x)=2x+2$, so $g'(1)=2(1)+2=4$. The Chain Rule predicts $H'(1)=6\cdot4=24$.

The increments verify how the approximations combine. For small $h$,

$$ k(h)=g(1+h)-g(1) =(1+h)^2+2(1+h)-3 =4h+h^2. $$

The outer function is $f(y)=y^2$, and its change at $b=3$ is exactly

$$ f(3+k)-f(3)=(3+k)^2-9=6k+k^2. $$

Substitute $k=4h+h^2$. The composite change is

$$ 6(4h+h^2)+(4h+h^2)^2 =24h+22h^2+8h^3+h^4. $$

After division by $h$, the difference quotient is $24+22h+8h^2+h^3$, which tends to $24$, in agreement with the Chain Rule.

Worked Example: The Inner Increment Has Zero First-Order Term

Let $g(x)=1+(x-2)^2$, $f(y)=y^3$, and $a=2$. Then $g(2)=1$. The derivative $g'(x)=2(x-2)$ gives $g'(2)=0$, while the Power Rule gives $f'(1)=3$. Thus the Chain Rule predicts $(f\circ g)'(2)=3\cdot0=0$.

To check the prediction directly, put $x=2+h$. Then the inner increment from $g(2)=1$ is $k(h)=g(2+h)-1=h^2$. The composite difference quotient is

\frac{f(g(2+h))-f(g(2))}{h} = \frac{(1+h^2)^3-1}{h} = \frac{3h^2+3h^4+h^6}{h} = 3h+3h^3+h^5.

This tends to zero. The inner increment is smaller than a first-order change in $h$: $k(h)/h=h\to0$. In the proof, this is why the linear contribution from the outer derivative also disappears.

Worked Example: A Zero Outer Derivative

Let $g(x)=3x-1$, $f(y)=(y-2)^2$, and $a=1$. The inner value is $g(1)=3(1)-1=2$, and $g'(1)=3$. Since $f'(y)=2(y-2)$, we have $f'(g(1))=f'(2)=0$. The Chain Rule therefore gives $(f\circ g)'(1)=0\cdot3=0$.

Indeed, $g(1+h)=2+3h$, so the inner increment is $k(h)=3h$. The outer change is

$$ f(2+k(h))-f(2)=(k(h))^2=(3h)^2=9h^2. $$

Thus the difference quotient is $9h^2/h=9h$, which tends to zero. Here the inner function has a nonzero derivative, but the first-order contribution from the outer function vanishes because $f'(2)=0$.

Why the Remainder Details Matter

The proof separates the change in the composition into a linear contribution and errors that vanish after division by $h$. The linear contribution is $f'(g(a))g'(a)h$: the inner approximation contributes $g'(a)h$, and the outer approximation multiplies that increment by $f'(g(a))$. The evaluation point $g(a)$ is forced by the fact that the outer function is being approximated near its input $b=g(a)$.

A tempting but incomplete argument is to divide the outer difference quotient by $g(a+h)-g(a)$. That quotient is only defined when this inner increment is nonzero. An inner function may take the same value at $a+h$ and at $a$, even for nonzero $h$. The remainder proof avoids this problem: when the inner increment is zero, the outer change and its remainder are both zero; when it is nonzero, the Remainder Transfer Lemma controls the error.

The proof also shows exactly where differentiability of the inner function is used. Besides providing its linear approximation, it ensures that $k(h)/h$ remains bounded. Without that control, an outer error that is negligible relative to $k(h)$ could fail to be negligible relative to $h$. The Chain Rule’s hypotheses ensure both approximations and the necessary control of their errors.

Check Your Understanding

Use the proof and examples to answer the following.

In the proof, what is the inner increment $k(h)$, and to what does $k(h)/h$ converge?
Why is it necessary to handle separately the case $k(h)=0$ when controlling the outer remainder?
For $g(x)=2+(x-1)^2$ and $f(y)=y^2$, what does the Chain Rule give for $(f\circ g)'(1)$?
Which bound from differentiability of the inner function allows the Remainder Transfer Lemma to be used?
In the Chain Rule formula, at which point is the outer derivative evaluated, and why?

Proof of the Chain Rule

What You'll Learn

Why the Proof Needs to Track the Inner Increment

Transferring a Remainder Through a Small Increment