Math Core

Quantities are often linked in chains. The temperature you feel depends on your position, and your position depends on time. The volume of a balloon depends on its radius and height, which both change as it inflates. The multivariable chain rule tells you how the final quantity changes by adding up the contribution of every link in the chain.

Case 1: one parameter

Suppose $z = f(x, y)$ and both $x$ and $y$ are functions of $t$ . Then $z$ is ultimately a function of $t$ alone. How fast does it change?

Over a short time $\Delta t$ , the inputs change by $\Delta x$ and $\Delta y$ . From the linear approximation in the previous lesson, the output changes by about

\Delta z \approx f_x\,\Delta x + f_y\,\Delta y.

Dividing by $\Delta t$ and letting $\Delta t \to 0$ gives the first version of the chain rule.

Chain rule, one parameter

If $z = f(x, y)$ is differentiable and $x = x(t)$ , $y = y(t)$ are differentiable, then

\frac{dz}{dt} = \frac{\partial z}{\partial x}\frac{dx}{dt} + \frac{\partial z}{\partial y}\frac{dy}{dt}.

Each term is "how sensitive $z$ is to one input" times "how fast that input is changing." For three intermediate variables $x$ , $y$ , $z$ feeding into $w$ , there are three such terms.

Worked example: Along a circle

Let $z = x^2y + y^3$ , where $x = \cos t$ and $y = \sin t$ . Find $\dfrac{dz}{dt}$ at $t = 0$ .

At $t = 0$ : $x = 1$ , $y = 0$ , $\dfrac{dx}{dt} = -\sin 0 = 0$ and $\dfrac{dy}{dt} = \cos 0 = 1$ .

The partials are $z_x = 2xy$ and $z_y = x^2 + 3y^2$ , which equal $0$ and $1$ at $(1, 0)$ . So

\frac{dz}{dt} = (0)(0) + (1)(1) = 1.

Worked example: A changing cylinder

The radius of a cylinder is increasing at $0.2$ cm/s while its height is decreasing at $0.5$ cm/s. How fast is the volume changing when $r = 3$ cm and $h = 8$ cm?

$V = \pi r^2 h$ , so

\frac{dV}{dt} = V_r\frac{dr}{dt} + V_h\frac{dh}{dt} = 2\pi r h\,(0.2) + \pi r^2\,(-0.5).

At $r = 3$ , $h = 8$ : $\dfrac{dV}{dt} = 2\pi(24)(0.2) - \pi(9)(0.5) = 9.6\pi - 4.5\pi = 5.1\pi \approx 16.0$ cm³/s. The volume is still increasing, because growth in the radius wins.

Case 2: two parameters

Now suppose $x$ and $y$ each depend on two variables, $x = x(s, t)$ and $y = y(s, t)$ . Then $z$ is a function of $s$ and $t$ , and it has two partial derivatives. Hold $t$ fixed and apply Case 1 to get $\partial z/\partial s$ ; hold $s$ fixed to get $\partial z/\partial t$ .

Chain rule, two parameters

\frac{\partial z}{\partial s} = \frac{\partial z}{\partial x}\frac{\partial x}{\partial s} + \frac{\partial z}{\partial y}\frac{\partial y}{\partial s}, \qquad \frac{\partial z}{\partial t} = \frac{\partial z}{\partial x}\frac{\partial x}{\partial t} + \frac{\partial z}{\partial y}\frac{\partial y}{\partial t}.

Tree diagrams

A tree diagram keeps the bookkeeping straight. Put $z$ at the top, draw a branch to each intermediate variable ( $x$ and $y$ ), and from each of those draw branches to the independent variables ( $s$ and $t$ ). Label every branch with the corresponding partial derivative. To find $\partial z/\partial s$ , follow every path from $z$ down to $s$ , multiply along each path, and add the results. This rule works for any number of variables and any number of levels.

Worked example: Two parameters

Let $z = e^x\sin y$ with $x = st^2$ and $y = s^2t$ . Find $\dfrac{\partial z}{\partial s}$ .

The pieces are $z_x = e^x\sin y$ , $z_y = e^x\cos y$ , $x_s = t^2$ and $y_s = 2st$ . So

\frac{\partial z}{\partial s} = e^x\sin y\cdot t^2 + e^x\cos y\cdot 2st = e^{st^2}\left(t^2\sin(s^2t) + 2st\cos(s^2t)\right).

Common mistake

Evaluate every factor at the same point. If you are asked for $\partial z/\partial s$ at $(s, t) = (2, 1)$ , first compute the corresponding $x$ and $y$ , then evaluate $z_x$ and $z_y$ at that $(x, y)$ and evaluate $x_s$ and $y_s$ at $(s, t) = (2, 1)$ . Plugging $s$ and $t$ directly into $z_x$ as if they were $x$ and $y$ is the most common error.

Implicit differentiation

The chain rule also explains implicit differentiation. Suppose the equation $F(x, y) = 0$ defines $y$ as a function of $x$ . Differentiate both sides with respect to $x$ , treating $x$ as $x$ and $y$ as $y(x)$ :

F_x\frac{dx}{dx} + F_y\frac{dy}{dx} = 0 \quad\Longrightarrow\quad \frac{dy}{dx} = -\frac{F_x}{F_y}, \quad\text{provided } F_y \ne 0.

The same argument works one dimension up. If $F(x, y, z) = 0$ defines $z$ implicitly as a function of $x$ and $y$ , then

\frac{\partial z}{\partial x} = -\frac{F_x}{F_z}, \qquad \frac{\partial z}{\partial y} = -\frac{F_y}{F_z}, \qquad F_z \ne 0.

Worked example: Implicit derivatives

(a) Find $\dfrac{dy}{dx}$ if $x^3 + y^3 = 6xy$ .

Move everything to one side: $F(x, y) = x^3 + y^3 - 6xy$ . Then $F_x = 3x^2 - 6y$ and $F_y = 3y^2 - 6x$ , so

\frac{dy}{dx} = -\frac{3x^2 - 6y}{3y^2 - 6x} = \frac{2y - x^2}{y^2 - 2x}.

(b) Find $\dfrac{\partial z}{\partial x}$ if $x^2 + 2y^2 + 3z^2 = 12$ .

With $F = x^2 + 2y^2 + 3z^2 - 12$ , $F_x = 2x$ and $F_z = 6z$ , so $\dfrac{\partial z}{\partial x} = -\dfrac{2x}{6z} = -\dfrac{x}{3z}$ .

Tip

Mind the minus sign in $-F_x/F_y$ . A quick sanity check: on the circle $x^2 + y^2 = 25$ , the formula gives $\dfrac{dy}{dx} = -\dfrac{x}{y}$ , which is negative in the first quadrant, just as the picture of the circle says it should be.

Practice

Practice 1

Let $z = x^2 + xy$ with $x = t^2$ and $y = 3t$ . Use the chain rule to find $\dfrac{dz}{dt}$ at $t = 1$ .