Proof of the chain rule

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Proof Builder

Reading the chain rule proof line by line is easier in the Proof Builder because each structural move — defining $k$ , extending $Q$ , applying the product of limits — stays visible as a separate step.

What the chain rule says

If $g$ is differentiable at $x$ and $f$ is differentiable at $g (x)$ , then the composition $h (x) = f (g (x))$ is differentiable at $x$ and $h^{'} (x) = f^{'} (g (x)) \cdot g^{'} (x)$ .

Informally: the rate of change of the outer function evaluated at the inner output, times the rate of change of the inner function. The proof shows this informal picture is exact.

Why the naive approach almost works

The most direct attempt is to write $\frac{f ( g ( x + h )) - f ( g ( x ))}{h}$ and multiply and divide by $k = g (x + h) - g (x)$ , giving $\frac{f ( g ( x ) + k ) - f ( g ( x ))}{k} \cdot \frac{k}{h}$ .

This works when $k \neq = 0$ for all small $h \neq = 0$ . But it fails when $g$ is locally flat — when $k = 0$ even though $h \neq = 0$ — because the factor $1/ k$ is undefined.

The fix is to define a function $Q$ that equals the difference quotient of $f$ when $k \neq = 0$ and equals $f^{'} (g (x))$ when $k = 0$ . Because $f$ is differentiable at $g (x)$ , this $Q$ is continuous at $0$ , so the limiting argument goes through in all cases.

Step-by-step proof strategy

Work through these moves in order. Each step is one logical unit.

1Write the limit definition of $(f \circ g)^{'} (x)$ using the standard difference quotient.
2Define $k = g (x + h) - g (x)$ . This is the increment in the inner function as the input shifts by $h$ .
3Note that $k \to 0$ as $h \to 0$ because differentiability implies continuity.
4Define $Q (t) = \frac{f ( g ( x ) + t ) - f ( g ( x ))}{t}$ for $t \neq = 0$ and $Q (0) = f^{'} (g (x))$ . Check that $Q$ is continuous at $0$ .
5Rewrite the difference quotient of $f \circ g$ as $Q (k) \cdot \frac{k}{h}$ . This identity holds whether $k = 0$ or not.
6Take the limit as $h \to 0$ : the first factor converges to $Q (0) = f^{'} (g (x))$ and the second factor converges to $g^{'} (x)$ .
7Conclude by the product of limits: $(f \circ g)^{'} (x) = f^{'} (g (x)) \cdot g^{'} (x)$ .

Connecting to concrete examples

Consider $sin (x^{2})$ . Here $g (x) = x^{2}$ and $f (u) = sin (u)$ . The chain rule gives $\frac{d}{d x} sin (x^{2}) = cos (x^{2}) \cdot 2 x$ .

The two-factor structure always appears: derivative of the outer function evaluated at the inner value, times derivative of the inner function. The proof makes precise why those two contributions multiply.

Common Pitfall

The chain rule is not just 'bring down the power.' It applies to any differentiable composition, including cases like $ln (sin (x))$ or $e^{x^{2}}$ , wherever there is an inner and outer function.

Try a Variation

Try differentiating $ln (cos (x))$ using the chain rule. Identify which function is $f$ and which is $g$ , then apply the rule and simplify.

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