The Chain Rule

A function inside a function

$y = \left(3x + 1\right)^{5}$ is made in two stages. First x goes to u = 3x + 1, the inner function. Then u goes to $y = u^{5}$, the outer function. Everything inside the bracket travels together as one quantity, u.

Multiplying out $\left(3x + 1\right)^{5}$ would give six terms to differentiate one at a time. The chain rule differentiates each stage once instead.

The outer function first

As a function of u, $y = u^{5}$, and the power rule gives $\frac{dy}{du} = 5u^{4} = 5\left(3x + 1\right)^{4}$. The power comes down to the front and drops by one, while the inside is left as it stands.

This is the gradient of y with respect to u. It is not yet $\frac{dy}{dx}$, because u does not change at the same rate as x.

Why the rates multiply

The inner function is u = 3x + 1, so $\frac{du}{dx} = 3$: u changes 3 times as fast as x. A small step $\Delta x$ in x moves u by $\Delta u = 3\Delta x$, and that step in u moves y by about $5u^{4} \times \Delta u$. So y moves by about $5u^{4} \times 3 \times \Delta x$.

In general, $\frac{\Delta y}{\Delta x} = \left(\frac{\Delta y}{\Delta u}\right) \times \left(\frac{\Delta u}{\Delta x}\right)$. As $\Delta x \to 0$, $\Delta u \to 0$ as well, and the three quotients become the three derivatives: $\frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx}$. The gradient in u is scaled once on the way from x to u and again on the way from u to y.

Three number lines: x, then u = 2x + 1, then $y = u^{2}$. A step in x moves u twice as far, and a step in u moves y 2u times as far, so $\frac{dy}{dx} = 2u \times 2 = 4u$. At x = 1, u = 3 and $\frac{dy}{dx} = 12$, which is 8x + 4, the derivative of the multiplied-out $4x^{2} + 4x + 1$.

Times the derivative of the inside

So $\frac{dy}{dx} = 5\left(3x + 1\right)^{4} \times 3 = 15\left(3x + 1\right)^{4}$. At x = 0 the inside is 1, and the gradient is 15. At $x = \frac{1}{3}$ the inside is 2, and the gradient is $15 \times 2^{4} = 240$.

Check at x = 0 with a chord. $\left(1.003\right)^{5} = 1.0150903$, so the chord from x = 0 to x = 0.001 has gradient $\frac{0.0150903}{0.001} = 15.09$.

A linear inside, such as 3x + 1, contributes just its coefficient. An inside that is not linear contributes its own derivative, which depends on x.

Insides that are not linear

For $y = \left(x^{2} + 1\right)^{3}$, the inner function is $u = x^{2} + 1$ and the outer function is $u^{3}$. Then $\frac{dy}{du} = 3u^{2}$ and $\frac{du}{dx} = 2x$, so $\frac{dy}{dx} = 3\left(x^{2} + 1\right)^{2} \times 2x = 6x\left(x^{2} + 1\right)^{2}$. At x = 1 that is 6 × 4 = 24. Multiplying out instead, $\left(x^{2} + 1\right)^{3} = x^{6} + 3x^{4} + 3x^{2} + 1$, which differentiates to $6x^{5} + 12x^{3} + 6x$, and that is 6 + 12 + 6 = 24 at x = 1.

For $y = \sqrt{x^{2} + 9}$, the inner function is $u = x^{2} + 9$ and the outer function is $\sqrt{u} = u^{\frac{1}{2}}$. Then $\frac{dy}{du} = \frac{1}{{2\sqrt{u}}}$ and $\frac{du}{dx} = 2x$, so $\frac{dy}{dx} = \frac{2x}{{2\sqrt{x^{2} + 9}}} = \frac{x}{\sqrt{x^{2} + 9}}$. At x = 4 the root is $\sqrt{25} = 5$, and the gradient is $\frac{4}{5} = 0.8$. Check: $\sqrt{4.01^{2} + 9} = 5.0080036$, so the chord from x = 4 to x = 4.01 has gradient 0.8004.

For $y = \frac{1}{{x^{2} + 1}} = \left(x^{2} + 1\right)^{-1}$, the inner function is $x^{2} + 1$ and the outer function is $u^{-1}$. The chain rule gives $-\left(x^{2} + 1\right)^{-2} \times 2x = - \frac{2x}{\left(x^{2} + 1\right)^{2}}$, the same as the quotient rule, and −0.5 at x = 1.

The gold curve $y = \sqrt{x^{2} + 9}$, lowest at (0, 3), and its tangent at (4, 5), the gold line with gradient $\frac{x}{\sqrt{x^{2} + 9}} = \frac{4}{5}$.

More than two layers

Each layer adds one factor. $y = \left(\left(2x + 1\right)^{2} + 1\right)^{3}$ has three: the innermost u = 2x + 1, then $w = u^{2} + 1$, then the outer $y = w^{3}$. So $\frac{dy}{dx} = \frac{dy}{dw} \times \frac{dw}{du} \times \frac{du}{dx} = 3w^{2} \times 2u \times 2$.

At x = 0, u = 1 and w = 2, so $\frac{dy}{dx} = 3 \times 4 \times 2 \times 2 = 48$. The chord from x = 0 to x = 0.001 has gradient 48.14, and with a step of 0.000001 the chord gradient is 48.0001.

The usual mistakes

Stopping after the outer function. $5\left(3x + 1\right)^{4}$ leaves out the factor 3, so every gradient comes out 3 times too small: 5 at x = 0 instead of 15.

Multiplying by the inner derivative without lowering the power. $15\left(3x + 1\right)^{5}$ is 480 at $x = \frac{1}{3}$, not 240.

Giving $\frac{dy}{du}$ as the answer. $5\left(3x + 1\right)^{4}$ is the rate per unit of u, and a rate per unit of x needs the factor $\frac{du}{dx}$.

A balloon being inflated

In the application below, the volume of a balloon depends on its radius, and the radius depends on time. The chain rule multiplies the two rates: $\frac{dV}{dr}$, from the power rule, times $\frac{dr}{dx}$, the rate at which the radius grows.