The Chain Rule

A function inside a function

$y={\left(3x+1\right)}^5$y = (3x + 1)⁵ is made in two stages. First x goes to u = 3x + 1, the inner function. Then u goes to $y=u^5$y = u⁵, the outer function. Everything inside the bracket travels together as one quantity, u.

Multiplying out ${\left(3x+1\right)}^5$(3x + 1)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, $y=u^5$y = u⁵, and the power rule gives $\frac{dy}{du}=5u^4=5{\left(3x+1\right)}^4$dy/du = 5u⁴ = 5(3x + 1)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 $\frac{dy}{dx}$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 $\frac{du}{dx}=3$du/dx = 3: u changes 3 times as fast as x. A small step $\Delta x$Δx in x moves u by $\Delta u=3\Delta x$Δu = 3Δx, and that step in u moves y by about $5u^4\times \Delta u$5u⁴ × Δu. So y moves by about $5u^4\times 3\times \Delta x$5u⁴ × 3 × Δx.

In general, $\frac{\Delta y}{\Delta x}=\left(\frac{\Delta y}{\Delta u}\right)\times \left(\frac{\Delta u}{\Delta x}\right)$Δy/Δx = (Δy/Δu) × (Δu/Δx). As $\Delta x\to 0$Δx → 0, $\Delta u\to 0$Δu → 0 as well, and the three quotients become the three derivatives: $\frac{dy}{dx}=\frac{dy}{du}\times \frac{du}{dx}$dy/dx = dy/du × 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 $y=u^2$y = u². A step in x moves u twice as far, and a step in u moves y 2u times as far, so $\frac{dy}{dx}=2u\times 2=4u$dy/dx = 2u × 2 = 4u. At x = 1, u = 3 and $\frac{dy}{dx}=12$dy/dx = 12, which is 8x + 4, the derivative of the multiplied-out $4x^2+4x+1$4x² + 4x + 1.

Times the derivative of the inside

So $\frac{dy}{dx}=5{\left(3x+1\right)}^4\times 3=15{\left(3x+1\right)}^4$dy/dx = 5(3x + 1)⁴ × 3 = 15(3x + 1)⁴. At x = 0 the inside is 1, and the gradient is 15. At $x=\frac{1}{3}$x = 1/3 the inside is 2, and the gradient is $15\times 2^4=240$15 × 2⁴ = 240.

Check at x = 0 with a chord. ${\left(1.003\right)}^5=1.0150903$(1.003)⁵ = 1.0150903, so the chord from x = 0 to x = 0.001 has gradient $\frac{0.0150903}{0.001}=15.09$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 $y={\left(x^2+1\right)}^3$y = (x² + 1)³, the inner function is $u=x^2+1$u = x² + 1 and the outer function is $u^3$u³. Then $\frac{dy}{du}=3u^2$dy/du = 3u² and $\frac{du}{dx}=2x$du/dx = 2x, so $\frac{dy}{dx}=3{\left(x^2+1\right)}^2\times 2x=6x{\left(x^2+1\right)}^2$dy/dx = 3(x² + 1)² × 2x = 6x(x² + 1)². At x = 1 that is 6 × 4 = 24. Multiplying out instead, ${\left(x^2+1\right)}^3=x^6+3x^4+3x^2+1$(x² + 1)³ = x⁶ + 3x⁴ + 3x² + 1, which differentiates to $6x^5+12x^3+6x$6x⁵ + 12x³ + 6x, and that is 6 + 12 + 6 = 24 at x = 1.

For $y=\sqrt{x^2+9}$y = √(x² + 9), the inner function is $u=x^2+9$u = x² + 9 and the outer function is $\sqrt{u}=u^{\frac{1}{2}}$√u = u^(1/2). Then $\frac{dy}{du}=\frac{1}{2\sqrt{u}}$dy/du = 1/(2√u) and $\frac{du}{dx}=2x$du/dx = 2x, so $\frac{dy}{dx}=\frac{2x}{2\sqrt{x^2+9}}=\frac{x}{\sqrt{x^2+9}}$dy/dx = 2x/(2√(x² + 9)) = x/√(x² + 9). At x = 4 the root is $\sqrt{25}=5$√25 = 5, and the gradient is $\frac{4}{5}=0.8$4/5 = 0.8. Check: $\sqrt{4.01^2+9}=5.0080036$√(4.01² + 9) = 5.0080036, so the chord from x = 4 to x = 4.01 has gradient 0.8004.

For $y=\frac{1}{x^2+1}={\left(x^2+1\right)}^{-1}$y = 1/(x² + 1) = (x² + 1)⁻¹, the inner function is $x^2+1$x² + 1 and the outer function is $u^{-1}$u⁻¹. The chain rule gives $-{\left(x^2+1\right)}^{-2}\times 2x=-\frac{2x}{{\left(x^2+1\right)}^2}$−(x² + 1)⁻² × 2x = −2x/(x² + 1)², the same as the quotient rule, and −0.5 at x = 1.

The gold curve $y=\sqrt{x^2+9}$y = √(x² + 9), lowest at (0, 3), and its tangent at (4, 5), the gold line with gradient $\frac{x}{\sqrt{x^2+9}}=\frac{4}{5}$x/√(x² + 9) = 4/5.

More than two layers

Each layer adds one factor. $y={\left({\left(2x+1\right)}^2+1\right)}^3$y = ((2x + 1)² + 1)³ has three: the innermost u = 2x + 1, then $w=u^2+1$w = u² + 1, then the outer $y=w^3$y = w³. So $\frac{dy}{dx}=\frac{dy}{dw}\times \frac{dw}{du}\times \frac{du}{dx}=3w^2\times 2u\times 2$dy/dx = dy/dw × dw/du × du/dx = 3w² × 2u × 2.

At x = 0, u = 1 and w = 2, so $\frac{dy}{dx}=3\times 4\times 2\times 2=48$dy/dx = 3 × 4 × 2 × 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$5(3x + 1)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$15(3x + 1)⁵ is 480 at $x=\frac{1}{3}$x = 1/3, not 240.

Giving $\frac{dy}{du}$dy/du as the answer. $5{\left(3x+1\right)}^4$5(3x + 1)⁴ is the rate per unit of u, and a rate per unit of x needs the factor $\frac{du}{dx}$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: $\frac{dV}{dr}$dV/dr, from the power rule, times $\frac{dr}{dx}$dr/dx, the rate at which the radius grows.