The Chain Rule

Main idea

When $x$ and $y$ both depend on $t$ , $z = f(x, y)$ changes for two reasons at once: because $x$ is moving and because $y$ is moving. Work out each effect (how sensitive $z$ is to that input, times how fast the input moves) and add them.

\frac{dz}{dt} = {\htmlData{sym=0}{\frac{\partial z}{\partial x}\frac{dx}{dt}}} + \frac{\partial z}{\partial y}\frac{dy}{dt}

What the symbols mean

$\frac{\partial z}{\partial x}\frac{dx}{dt}$
Path through x
How much z changes because x is changing.

Implicit differentiation

F(x, y) = 0 \implies \frac{dy}{dx} = -\frac{F_x}{F_y}

Explanation

The chain rule is simply the rule for how fast something changes when it depends on things that are themselves changing. It's called a "chain" because the effect passes along links: time moves your position, and your position changes the temperature you feel. The formula $\frac{dz}{dt} = \frac{\partial z}{\partial x}\frac{dx}{dt} + \frac{\partial z}{\partial y}\frac{dy}{dt}$ looks long, but it's two copies of one idea added together: $\frac{\partial z}{\partial x}\frac{dx}{dt}$ (how much $z$ cares about $x$ , times how fast $x$ is moving) plus the same thing for $y$ . Two effects, added up. Bear with the notation; the idea is really that simple.

In one variable, the chain rule said that if $z$ depends on $x$ and $x$ depends on $t$ , then $\frac{dz}{dt} = \frac{dz}{dx}\frac{dx}{dt}$ : rates multiply along the chain. Now $z$ depends on $t$ through two routes. Where does the plus sign come from? It's the differential from the last lesson. In a short time $dt$ , $x$ moves by $\frac{dx}{dt}dt$ and $y$ moves by $\frac{dy}{dt}dt$ , and $dz = f_x\,dx + f_y\,dy$ adds up their effects. Divide by $dt$ and you have the formula.

Tree diagrams keep the bookkeeping straight. Draw $z$ at the top, branches down to $x$ and $y$ , and from each of those down to $t$ . Each route from $z$ down to $t$ contributes the product of the derivatives along it, and you add the routes. This scales to any number of variables, which is why it's worth the habit.

Several parameters. If $x = g(s, t)$ and $y = h(s, t)$ , then $\frac{\partial z}{\partial s} = z_x x_s + z_y y_s$ , and similarly for $t$ . It's the same tree; you just follow the routes that end at $s$ .

Implicit differentiation, made easy. Remember implicit differentiation from Calculus I, where you differentiated both sides of something like $x^2 + y^2 = 25$ ? Here's a shortcut for it. A curve $F(x, y) = 0$ has $y$ secretly depending on $x$ . Differentiating $F(x, y(x)) = 0$ with the chain rule gives $F_x + F_y\frac{dy}{dx} = 0$ , and solving for $\frac{dy}{dx}$ gives the formula above.

See it

A drone flies a circle of radius 3 through air whose temperature is T(x, y) = 20 + 0.5x² − 0.3y² + 0.4x. The reading it records rises and falls as it goes around, from a high of about 25.7 °C at (3, 0) to a low of about 17.3 °C near the top and bottom of the circle. The label shows dT/dt, which the chain rule builds from how T varies with position and how fast the drone moves.

Worked examples

Example 1

$z = x^2y$ , with $x = \cos t$ and $y = \sin t$ . Find $\frac{dz}{dt}$ .

1
Gather the four pieces: $z_x = 2xy$ and $z_y = x^2$ , and $x' = -\sin t$ and $y' = \cos t$ .

Practice

Type your answer and check it. Solve them all to complete the lesson.

Problem 1+5 XP

$z = x^2 + y^2$ with $x = t$ , $y = t^2$ . Find $\frac{dz}{dt}$ at $t = 1$ .

Problem 2+5 XP

$z = xy$ with $x = e^t$ , $y = \cos t$ . Find $\frac{dz}{dt}$ at $t = 0$ .

Problem 3+5 XP

Use $\frac{dy}{dx} = -\frac{F_x}{F_y}$ on $x^2 + y^2 = 25$ at $(3, 4)$ .

Problem 4+5 XP

At some point $f_x = 2$ and $f_y = -1$ . If $x = 3s$ and $y = s^2$ , find $\frac{\partial z}{\partial s}$ there at $s = 1$ .

Problem 5+5 XP

A rectangle's length is 10 cm and growing at 2 cm/s; its width is 6 cm and shrinking at 1 cm/s. How fast is the area changing?

Problem 6+5 XP

A weather drone flies with velocity $\langle 3, -2\rangle$ m/s through air where $\frac{\partial T}{\partial x} = 0.4$ and $\frac{\partial T}{\partial y} = 0.1$ °C per meter. How fast is the temperature it measures changing?

Challenge problem+25 XP

The hardest one here. Get it, and you've got this lesson.

Let $z = f(x, y)$ with $x = r\cos\theta$ and $y = r\sin\theta$ . At the point where $r = 2$ and $\theta = \frac\pi3$ , you know $f_x = 3$ and $f_y = 4$ . Find $\frac{\partial z}{\partial\theta}$ there.

Where it's used

Backpropagation in neural networks is this chain rule on a huge tree: each weight's effect on the loss is summed over every path through the network.

Gradient DescentCalc III · Computer Science Temperature Felt by a DroneCalc III · Robotics

Differentials Directional Derivatives and the Gradient