The chain rule

A function inside a function

$(3x + 1)^5$ is not a power of $x$. It is a power of the expression $3x + 1$, so the power rule of lesson 2.3.1 does not apply to it directly. It is a composition (lesson 1.1.3): an inner function $g(x) = 3x + 1$ fed into an outer function $f(u) = u^5$.

The letter $u$ names the inner function's output, so $u = 3x + 1$ and $y = u^5$.

To spot the two parts, evaluate the function at a number. At $x = 2$, you first work out $3 \times 2 + 1 = 7$ (the inner function), then $7^5$ (the outer one). The last step is the outer function.

You could expand $(3x + 1)^5$ into six terms and use lesson 2.3.1, but it is long. $\sqrt{x^2 + 1}$ cannot be expanded at all. The chain rule handles both in one line.

Slopes multiply

Rates that feed into each other multiply. A car that uses 0.06 litres of fuel per km, driving at 80 km per hour, uses $0.06 \times 80 = 4.8$ litres per hour. The units cancel like fractions: litres per km times km per hour is litres per hour.

A composition works the same way. When $x$ moves, $u$ moves $\frac{du}{dx}$ times as much, and then $y$ moves $\frac{dy}{du}$ times as much as $u$. So $y$ moves $\frac{dy}{du} \cdot \frac{du}{dx}$ times as much as $x$.

Take $y = (2x + 1)^2$ at $x = 1$. The inner function $u = 2x + 1$ has slope 2, and at $x = 1$ it gives $u = 3$. The outer function $y = u^2$ has slope $2u$, which is 6 at $u = 3$.

As the step shrinks, each stretch settles on a slope, and the total settles on their product.

Algebra agrees. Expanding, $(2x + 1)^2 = 4x^2 + 4x + 1$, with derivative $8x + 4$, which is 12 at $x = 1$.

That last point matters. For $(2x + 1)^2$ at $x = 1$, the outer slope $2u$ is taken at $u = 3$, giving 6. Taking it at $x = 1$ would give an outer slope of 2 and a total of $2 \times 2 = 4$, which is wrong.

Worked examples

A linear inside. For $(3x + 1)^5$, the outer $u^5$ gives $5u^4$ and the inner $3x + 1$ gives 3. So the derivative is $5(3x + 1)^4 \cdot 3 = 15(3x + 1)^4$, which is 15 at $x = 0$. An inside $ax + b$ always adds the factor $a$.

A falling inside. For $(5 - 2x)^3$ the inner slope is $-2$. The derivative is $3(5 - 2x)^2 \cdot (-2) = -6(5 - 2x)^2$, so the slope is $-54$ at $x = 1$. The inner function falls, and its negative factor carries through.

A root. Write $\sqrt{x^2 + 1}$ as $(x^2 + 1)^{1/2}$. The outer power gives $\frac{1}{2}(x^2 + 1)^{-1/2}$, and the inside gives $2x$:

At $x = 1$ that is $\frac{1}{\sqrt{2}} \approx 0.707107$. Central differences agree.

A negative power. Write $\frac{1}{x^2 + 1}$ as $(x^2 + 1)^{-1}$. The chain rule gives $-(x^2 + 1)^{-2} \cdot 2x = -\frac{2x}{(x^2 + 1)^2}$, the same answer as the quotient rule in lesson 2.3.2.

A flat inside. $(x^2 + 1)^3$ has derivative $3(x^2 + 1)^2 \cdot 2x = 6x(x^2 + 1)^2$. At $x = 0$ the inner slope is 0, so the whole slope is 0. At $x = 1$ it is 24.

The chain rule also agrees with the power rule. $(x^2)^3 = x^6$, and the chain rule gives $3(x^2)^2 \cdot 2x = 6x^5$, as the power rule does.

Chains of three

A function can have more than two layers. Take $\sqrt{16 + (x^2 + 2)^2}$ and peel it from the outside in.

Each layer adds one stretch factor: $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dv} \cdot \frac{dv}{dx}$. At $x = 1$ that is $0.1 \times 6 \times 2 = 1.2$. Central differences give 1.201943, 1.200019 and 1.2 for gaps 0.1, 0.01 and 0.001.

As a formula, the three factors are

Work from the outside in, one layer at a time, and write one factor for each. Skipping a layer is the usual slip.

Where the chain rule breaks

The chain rule needs both slopes: the inner one at $x$, and the outer one at $u = g(x)$. If either is missing, so is the slope of the composition.

$\sqrt{x^2}$ is $|x|$, the corner of lesson 2.2.2. Its outer function $\sqrt{u}$ has no slope at $u = 0$ (lesson 2.3.1), and its inside $x^2$ is 0 at $x = 0$. The rule gives $\frac{x}{\sqrt{x^2}} = \frac{x}{|x|}$: 1 for every $x > 0$, $-1$ for every $x < 0$, and nothing at 0.

Compare $\sqrt{x^2 + 1}$. Its inside is never below 1, so the outer slope always exists, and the curve has a slope everywhere.

Which rule?

Evaluate the function at a number in your head, and look at the last step. That step decides the first rule; the pieces it leaves may need others.

Product, then chain. For $x(2x + 1)^3$, the product rule gives $1 \cdot (2x + 1)^3 + x \cdot \frac{d}{dx}(2x + 1)^3$. The chain rule gives $\frac{d}{dx}(2x + 1)^3 = 3(2x + 1)^2 \cdot 2$. Together:

At $x = 1$ the slope is $3^2 \times 9 = 81$.

Quotient, then chain. For $\frac{x}{\sqrt{x^2 + 1}}$, the slope of the bottom comes from the chain rule: $\frac{x}{\sqrt{x^2 + 1}}$, as worked out above. The quotient rule then gives the first line below. Multiplying its top and bottom by $\sqrt{x^2 + 1}$ clears the small fraction:

At $x = 0$ the slope is 1.

Chain, then quotient. For $\left(\frac{x + 1}{x - 1}\right)^2$, the outer square gives $2 \cdot \frac{x + 1}{x - 1}$. The inside, by the quotient rule, has slope $-\frac{2}{(x - 1)^2}$. Multiplied, the derivative is $-\frac{4(x + 1)}{(x - 1)^3}$, which is $-12$ at $x = 2$.

Lessons 2.3.4 and 2.3.5 use the chain rule with $e$, $\ln$, sine and cosine as outer functions. Module 2.4 relies on it to find peaks, and lesson 2.6.1 runs it backwards as integration by substitution. Case study 2.3.7 uses it on every discounted payment of a bond, $(1 + y)^{-t}$.

Common mistakes

1. Forgetting the inner derivative. $(3x + 1)^5$ gives $15(3x + 1)^4$, not $5(3x + 1)^4$.
2. Taking the outer slope at $x$. It is taken at $u = g(x)$: for $(2x + 1)^2$ at 1, the outer slope is 6, not 2.
3. Putting the inner derivative inside. $(x^2 + 1)^3$ gives $3(x^2 + 1)^2 \cdot 2x$, not $3(2x)^2$.
4. Mixing up inner and outer. The outer function is the last step of the evaluation. In $\sqrt{x^2 + 1}$ it is the root.
5. Skipping a layer. A chain of three functions gives three factors.
6. No slope where a link has none. $\sqrt{x^2} = |x|$ has no slope at 0, because $\sqrt{u}$ has none at $u = 0$.