The Chain Rule

⚡ Key Takeaways

The Chain Rule is the differentiation rule for composite functions: if $y$ depends on $u$ , and $u$ depends on $x$ , then $\frac{dy}{dx} = \frac{dy}{du}\cdot\frac{du}{dx}$ . In words: the rate of change of a composition is the product of the rates of change of each link. Where it applies: any nested or "function-inside-a-function" expression in calculus, physics, economics, and machine learning. Who/when: rooted in the work of Leibniz (1670s) and given rigorous form by Cauchy and later analysts in the 19th century.

💡 The Core Insight

A composite function is a relay race of dependence. If a small nudge in $x$ changes $u$ at rate $\frac{du}{dx}$ , and a small nudge in $u$ changes $y$ at rate $\frac{dy}{du}$ , then the nudge in $x$ ripples through $u$ to $y$ at the product of those rates. Sensitivities compose by multiplication.

DefinitionComposite Function

Given two functions $g$ and $f$ , the composite function $f \circ g$ is defined by

1

\quad (f \circ g)(x) = f\big(g(x)\big)

Derivation hint: substitute the output of $g$ as the input of $f$ ; the domain is all $x$ where $g(x)$ lies in the domain of $f$ . Here $g$ is the inner function, $f$ is the outer function. If we write $u = g(x)$ , then $y = f(u)$ , and $y$ depends on $x$ through the intermediate variable $u$ .

TheoremThe Chain Rule

Let $g$ be differentiable at $x$ and let $f$ be differentiable at $u = g(x)$ . Then the composite $f \circ g$ is differentiable at $x$ , and

2

\quad (f \circ g)'(x) = f'\big(g(x)\big)\cdot g'(x)

Derivation hint: differentiate the outer function with the inner left intact, then multiply by the derivative of the inner. In Leibniz notation, with $y = f(u)$ and $u = g(x)$ :

3

\quad \frac{dy}{dx} = \frac{dy}{du}\cdot\frac{du}{dx}

Derivation hint: read each factor as a local rate of change; the $du$ 's "cancel" symbolically, though they are not genuine fractions.

Symbols: $x$ is the independent variable; $u = g(x)$ is the intermediate quantity; $y = f(u)$ is the final output; $f'$ and $g'$ denote derivatives; $\frac{dy}{du}$ is evaluated at $u = g(x)$ , while $\frac{du}{dx}$ is evaluated at $x$ .

⚠️ Common mistake

The phrase " $du$ cancels" is a mnemonic, not a proof. The differentials $du$ in $\frac{dy}{du}$ and $\frac{du}{dx}$ are not ordinary numbers being divided. A correct proof must handle the case where $g'(x) \neq 0$ separately from where the difference quotient's denominator could vanish.

ProofSketch via Difference Quotients

We want $\frac{dy}{dx}$ where $y = f(g(x))$ . Start from the limit definition.

Step 1. Write the difference quotient for the composite at $x$ :

4

\quad \frac{f(g(x+h)) - f(g(x))}{h}

Derivation hint: this is the increment in $y$ divided by the increment in $x$ , before taking $h \to 0$ .

Step 2. Let $\Delta u = g(x+h) - g(x)$ , the change in the inner function. Multiply and divide by $\Delta u$ (assuming $\Delta u \neq 0$ ):

5

\quad \frac{f(g(x)+\Delta u) - f(g(x))}{\Delta u}\cdot\frac{\Delta u}{h}

Derivation hint: insert $\frac{\Delta u}{\Delta u}=1$ and regroup so each factor is a recognizable difference quotient.

Step 3. As $h \to 0$ , continuity of $g$ forces $\Delta u \to 0$ . The first factor tends to $f'(g(x))$ and the second to $g'(x)$ :

6

\quad \lim_{h\to 0}\frac{\Delta y}{h} = f'(g(x))\cdot g'(x)

Derivation hint: the limit of a product is the product of the limits when both exist.

Step 4 (the gap). The division by $\Delta u$ fails if $\Delta u = 0$ for $h$ near $0$ . A rigorous proof replaces the first factor by a patched function $E(\Delta u)$ that equals the difference quotient when $\Delta u \neq 0$ and equals $f'(g(x))$ when $\Delta u = 0$ , restoring continuity. This is Carathéodory's clean approach. The conclusion $(f\circ g)'(x) = f'(g(x))g'(x)$ then holds in all cases. $\blacksquare$

∎

mathematician · Gottfried Wilhelm Leibniz | 1646–1716

Leibniz co-invented calculus and, crucially, designed the notation $\frac{dy}{dx}$ that makes the Chain Rule look almost self-evident. His differentials $dy$ , $du$ , $dx$ were conceived as "infinitely small" increments, and within that framework $\frac{dy}{du}\cdot\frac{du}{dx}$ reads as honest fraction cancellation. Though his infinitesimals were later replaced by limits, the notation survived precisely because it encodes the Chain Rule so transparently. The form in equation [3] is sometimes called the Leibniz form of the rule.

🧠 Intuition

Imagine a chain of gears. Turn the input gear (variable $x$ ); it spins a middle gear ( $u$ ) which is, say, 3 times faster; that middle gear spins the output gear ( $y$ ) which is 2 times faster than the middle one. How fast does $y$ spin compared to $x$ ? $2 \times 3 = 6$ times. Each link multiplies its gear ratio. Equivalently, think of unit conversions. If you travel 60 miles per hour and there are 1.6 kilometres per mile, you travel $60 \times 1.6 = 96$ km/h. The $\frac{\text{miles}}{\text{hour}}$ and $\frac{\text{km}}{\text{mile}}$ multiply, with "miles" cancelling — exactly the Chain Rule pattern.

Figure

A composite function as a relay: a nudge in x ripples through u to y, with rates multiplying.

ExampleDifferentiating a Power of a Polynomial

Given: $y = (3x^2 + 5)^4$ . Find $\frac{dy}{dx}$ .

Step 1 — Identify the links. Inner: $u = g(x) = 3x^2 + 5$ . Outer: $y = f(u) = u^4$ .

Step 2 — Differentiate each.

7

\quad \frac{dy}{du} = 4u^3, \qquad \frac{du}{dx} = 6x

Derivation hint: apply the power rule to $u^4$ and to $3x^2+5$ term-by-term.

Step 3 — Multiply (Chain Rule).

8

\quad \frac{dy}{dx} = 4u^3 \cdot 6x = 24x(3x^2+5)^3

Derivation hint: substitute $u = 3x^2+5$ back in and combine constants.

Check at $x = 1$ : $u = 8$ , so $\frac{dy}{dx} = 24(1)(8)^3 = 24 \times 512 = 12288$ .

\boxed{\dfrac{dy}{dx} = 24x\,(3x^2+5)^3 \quad(\text{dimensionless})}

ExampleA Physical Rate: Expanding Balloon

Given: A spherical balloon's radius grows at $\frac{dr}{dt} = 2\ \text{cm/s}$ . How fast is its volume growing when $r = 10\ \text{cm}$ ? Volume $V = \frac{4}{3}\pi r^3$ .

Step 1 — Identify the chain. $V$ depends on $r$ ; $r$ depends on $t$ . We want $\frac{dV}{dt}$ .

Step 2 — Differentiate the outer link.

9

\quad \frac{dV}{dr} = 4\pi r^2

Derivation hint: power rule on $\frac{4}{3}\pi r^3$ .

Step 3 — Chain Rule.

10

\quad \frac{dV}{dt} = \frac{dV}{dr}\cdot\frac{dr}{dt} = 4\pi r^2 \cdot \frac{dr}{dt}

Derivation hint: this is "related rates" — multiply the geometric rate by the time rate.

Step 4 — Substitute numbers. $r = 10\ \text{cm}$ , $\frac{dr}{dt} = 2\ \text{cm/s}$ :

11

\quad \frac{dV}{dt} = 4\pi (10)^2 (2) = 800\pi \approx 2513\ \text{cm}^3/\text{s}

Derivation hint: square the radius, multiply through, keep $\pi$ exact then approximate.

\boxed{\dfrac{dV}{dt} = 800\pi \approx 2513\ \text{cm}^3/\text{s}}

⚠️ Common mistake

The most frequent error is forgetting the inner derivative. Students differentiate $\sin(x^2)$ as $\cos(x^2)$ and stop. The correct answer is $\cos(x^2)\cdot 2x$ . The factor $g'(x) = 2x$ — the " $\times$ the derivative of the inside" — is exactly what the Chain Rule supplies. If the inside were just $x$ , that factor would be $1$ and you'd never notice it was missing.

CounterexampleThe Rule Is Not Mere Substitution of Derivatives

One might guess that $(f\circ g)'(x) = f'(g(x))$ alone, as if differentiating the outer function were enough. Take $f(u) = u^2$ and $g(x) = 5x$ , so $f(g(x)) = 25x^2$ . The true derivative is $50x$ . But $f'(g(x)) = 2(5x) = 10x$ — wrong by a factor of 5. The missing factor is exactly $g'(x) = 5$ , and $10x \times 5 = 50x$ ✓. This shows the inner derivative is indispensable, not optional.

⚠️ Common mistake

Do not confuse the Chain Rule with the Product Rule. The Chain Rule is for compositions $f(g(x))$ — one function fed into another. The Product Rule is for products $f(x)\cdot g(x)$ — two functions multiplied. The expression $x^2\sin x$ needs the Product Rule; $\sin(x^2)$ needs the Chain Rule. Ask: "Is one function inside the other, or are they side by side?"

connection

The Chain Rule sits at the heart of differentiation's machinery and links to many neighbours. It is essential for Implicit Differentiation, where we treat $y$ as a hidden function of $x$ and differentiate $y^n$ as $ny^{n-1}\frac{dy}{dx}$ . It powers Related Rates problems, as in the balloon example, by chaining a geometric derivative to a time derivative. It underlies the Derivative of Inverse Functions via $\frac{dx}{dy} = 1/\frac{dy}{dx}$ . Combined with the Fundamental Theorem of Calculus, it produces the technique of Integration by Substitution — substitution is the Chain Rule run backwards. Its multivariable cousin generalizes equation [3] to sums of partial-derivative products.

🔬 Where this appears

⏳ History

- 1676 — Leibniz, in private manuscripts, manipulates differentials

dy

,

du

,

dx

in ways that effectively use the Chain Rule, though without a formal statement.

- 1684 — Leibniz publishes Nova Methodus, the first calculus paper, establishing the

\frac{dy}{dx}

notation that makes the rule transparent.

- 1690s — The Bernoulli brothers (Jacob and Johann) and L'Hôpital apply chained differentials freely to curves and mechanics problems.

- 1734 — Berkeley's The Analyst attacks the logical vagueness of infinitesimals, including the loose "cancellation" justifying the rule, spurring demand for rigor.

- 1755 — Euler systematizes differential calculus in Institutiones Calculi Differentialis, treating composite differentiation as routine.

- 1821 — Cauchy's Cours d'analyse recasts derivatives via limits, providing the conceptual basis for a rigorous Chain Rule proof.

- 1860s–1870s — Weierstrass's epsilon–delta formalism allows fully rigorous statements and proofs of the rule.

- 1880s — Carathéodory's "increment" formulation of the derivative yields the clean, gap-free proof now common in textbooks.

- 1986 — Rumelhart, Hinton, and Williams popularize backpropagation, recasting the Chain Rule as the computational core of modern machine learning.

open-problem

The Chain Rule as stated requires both functions to be differentiable at the relevant points. Subtleties arise at the edges:

Non-differentiable inner functions. If $g$ has a corner (e.g. $g(x) = |x|$ at $x=0$ ), the composite may still be differentiable, but the Chain Rule formula does not directly apply — case-by-case analysis is needed.
The "fake fraction" problem. Equation [3] looks like cancellation but isn't; in multivariable calculus the naïve cancellation breaks entirely, replaced by $\frac{dz}{dt} = \frac{\partial z}{\partial x}\frac{dx}{dt} + \frac{\partial z}{\partial y}\frac{dy}{dt}$ — a sum of products.
Higher derivatives. The second derivative of a composite is not simply $f''(g)\,g'^2$ ; it equals $f''(g)(g')^2 + f'(g)g''$ , and the general $n$ -th derivative is given by the intricate Faà di Bruno's formula, a still-studied combinatorial object.
Non-standard analysis. Whether Leibniz's infinitesimals can be rigorously revived (Robinson, 1960s) so that the "cancellation" is literally valid remains a live area of mathematical logic.

Formal Statement

Why It Holds / Derivation

Building Intuition

Worked Examples

Common Misconceptions

Connections

Real-World Applications

Historical Development

Subtleties & Limits

See Also