Deriving the Chain Rule from the Difference Quotient

Why a Composite Function Needs a Derivative Rule

A composite function changes in two stages. A small change in the input first changes the value of an inner function; that change then changes the value of an outer function. The derivative of the composite must account for both stages.

The chain rule expresses this relationship. Here we derive it from the definition of derivative, rather than treating it as a formula to memorize. The key step is to track the change in the inner function and relate it to the original input increment.

For functions \(g\) and \(f\), their composition is the function \(f\circ g\) defined by \((f\circ g)(x)=f(g(x))\), wherever \(g(x)\) is in the domain of \(f\).

The Chain Rule

The derivative \(f'(g(a))\) measures the outer function’s rate of change at the value reached by the inner function. The factor \(g'(a)\) measures how quickly the inner function’s value changes as the input moves from \(a\). Both factors are needed.

Factoring Through the Inner Change

Write \(b=g(a)\). For a nonzero input increment \(h\), define the change in the inner function by \(\Delta g=g(a+h)-g(a)\). When \(\Delta g\ne 0\), the composite difference quotient can be factored as

This factorization suggests the proof. As \(h\to 0\), differentiability of \(g\) implies \(g(a+h)\to g(a)\), so \(\Delta g\to 0\). The first factor then approaches \(f'(g(a))\), and the second approaches \(g'(a)\). Their product approaches the claimed chain-rule value.

There is an important detail: \(\Delta g\) may equal zero even when \(h\ne 0\). In that case, the first factor in the displayed product divides by zero and is not defined. A proof must therefore avoid assuming that the inner change is always nonzero.

A Complete Proof That Includes Zero Inner Changes

Differentiability of \(f\) at \(b\) means that the difference between \(f(b+k)\) and its linear change \(f'(b)k\), divided by \(k\), tends to zero as \(k\to 0\). We encode this remainder in a function \(\rho\): for \(y\ne b\), set

By the definition of \(f'(b)\), \(\rho(y)\to 0\) as \(y\to b\). For every \(y\) sufficiently close to \(b\), including \(y=b\), its definition gives

For \(y=b\), both sides are zero, so this identity does not require division by \(y-b\). Now take \(y=g(a+h)\). For all sufficiently small nonzero \(h\), this value lies in the interval on which \(f\) is defined. Substitution and division by \(h\) give

As \(h\to 0\), the first difference quotient on the right tends to \(g'(a)\). In particular, it remains bounded for all sufficiently small nonzero \(h\). Also, differentiability of \(g\) at \(a\) implies continuity there, so \(g(a+h)\to b\). Consequently \(\rho(g(a+h))\to 0\). The second term therefore tends to zero: it is the product of a quantity tending to zero and a bounded quantity. Taking limits yields

This proves the chain rule under the stated assumptions. Because the identity used in the proof remains valid when \(g(a+h)=g(a)\), the argument also covers zero inner changes.

Worked Examples

A Common Mistake: Omitting the Inner Rate of Change

For \(f(t)=t^3\) and \(g(x)=x^2+1\), one might differentiate the outer function and write \(3(x^2+1)^2\), stopping there. That accounts for how the outer function changes with its own input, but not how quickly \(x^2+1\) changes with \(x\). The missing factor is \(g'(x)=2x\). The correction is

At \(x=1\), the incomplete calculation gives \(3(2)^2=12\), while the corrected derivative gives \(6(1)(2)^2=24\). The difference is precisely the inner rate of change.

When the Rule Applies

The chain rule is a pointwise result: it applies at \(a\) when \(g\) is differentiable at \(a\), and \(f\) is differentiable at \(g(a)\). The composite must be defined near \(a\); the theorem’s open-interval assumptions ensure this. If either required derivative does not exist, this theorem alone does not establish that the composite is differentiable.

The difference-quotient derivation also explains the structure of the answer. A small input increment \(h\) produces an inner change approximately equal to \(g'(a)h\). The outer function responds at a rate approximately \(f'(g(a))\). Combining these rates gives the product \(f'(g(a))g'(a)\). The proof makes that approximation exact by taking limits, and its remainder argument avoids dividing by an inner change that might be zero.