The Multivariable Chain Rule, multivariable and vector calculus lesson

A drone flies through a room where the temperature varies from place to place. As the drone moves, each of its coordinates changes, and each change affects the recorded temperature at its own rate. The chain rule adds these contributions up. In matrix form it says something short and memorable, the derivative of a composition is the product of the derivatives.

Key terms

Composition of multivariable functions

Definition Let $U \subseteq \mathbb{R}^n$ and $V \subseteq \mathbb{R}^m$ be open, let $g \colon U \to \mathbb{R}^m$ with $g(U) \subseteq V$ , and let $f \colon V \to \mathbb{R}^p$ . The composition $f \circ g \colon U \to \mathbb{R}^p$ is

(f \circ g)(\mathbf{x}) = f\big(g(\mathbf{x})\big).

The condition $g(U) \subseteq V$ makes sure that every output of $g$ is an allowed input of $f$ .

In plain words Feed the inputs into $g$ , then feed what comes out into $f$ . The middle variables are the outputs of $g$ .

Example Let $f(x, y) = x^2 y$ and let $g(t) = \langle \cos t, \sin t \rangle$ , a point moving around the unit circle. Then

(f \circ g)(t) = \cos^2 t \, \sin t.

At $t = \frac{\pi}{4}$ the point is $\left( \frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2} \right)$ and the value is $\frac{\sqrt{2}}{4}$ .

Jacobian multiplication

Definition With $U$ , $V$ , $f$ and $g$ as above, suppose $g$ is differentiable at $\mathbf{a}$ in $U$ and $f$ is differentiable at the point $g(\mathbf{a})$ . Then $f \circ g$ is differentiable at $\mathbf{a}$ and

D(f \circ g)(\mathbf{a}) = Df\big(g(\mathbf{a})\big)\, Dg(\mathbf{a}).

In matrices, the $p \times n$ Jacobian of $f \circ g$ at $\mathbf{a}$ is the $p \times m$ Jacobian of $f$ evaluated at $g(\mathbf{a})$ times the $m \times n$ Jacobian of $g$ evaluated at $\mathbf{a}$ . The hypothesis on $f$ is differentiability, not mere existence of partial derivatives. For $f(x, y) = \frac{xy}{x^2 + y^2}$ with $f(0, 0) = 0$ and $g(t) = \langle t, t \rangle$ , both partial derivatives of $f$ at the origin are $0$ , so a formal product of partial derivatives would predict $(f \circ g)'(0) = 0$ . Yet $f(g(t)) = \frac{1}{2}$ for $t \ne 0$ and $f(g(0)) = 0$ , so the composition is not even continuous at $0$ . The rule fails because $f$ is not differentiable at the origin.

In plain words Each derivative is a matrix, and composing functions corresponds to multiplying those matrices. Evaluate the outer derivative where the inner function lands, not at the original input.

Example Let $g(u, v) = \langle uv, \ u + v \rangle$ and $f(x, y) = \langle x^2 - y, \ xy \rangle$ , both polynomial and so differentiable everywhere. At $(u, v) = (1, 2)$ the inner point is $g(1, 2) = (2, 3)$ . The two Jacobians are

\begin{aligned} Jg(1, 2) &= \begin{bmatrix} v & u \\ 1 & 1 \end{bmatrix}_{(1,2)} \\[0.4em] &= \begin{bmatrix} 2 & 1 \\ 1 & 1 \end{bmatrix} \end{aligned}

and

\begin{aligned} Jf(2, 3) &= \begin{bmatrix} 2x & -1 \\ y & x \end{bmatrix}_{(2,3)} \\[0.4em] &= \begin{bmatrix} 4 & -1 \\ 3 & 2 \end{bmatrix}. \end{aligned}

Their product is

\begin{aligned} J(f \circ g)(1, 2) &= \begin{bmatrix} 4 & -1 \\ 3 & 2 \end{bmatrix} \begin{bmatrix} 2 & 1 \\ 1 & 1 \end{bmatrix} \\[0.4em] &= \begin{bmatrix} 7 & 3 \\ 8 & 5 \end{bmatrix}. \end{aligned}

As a check, the composition is $\langle u^2 v^2 - u - v, \ u^2 v + u v^2 \rangle$ , whose partial derivatives at $(1, 2)$ are $7$ , $3$ , $8$ and $5$ .

One independent parameter

Definition Let $f \colon V \to \mathbb{R}$ with $V \subseteq \mathbb{R}^2$ open, and let $x(t)$ and $y(t)$ be functions on an open interval $I$ with $(x(t), y(t))$ in $V$ . If $x$ and $y$ are differentiable at $t$ and $f$ is differentiable at $(x(t), y(t))$ , then $z(t) = f(x(t), y(t))$ is differentiable at $t$ and

\frac{dz}{dt} = \frac{\partial f}{\partial x} \frac{dx}{dt} + \frac{\partial f}{\partial y} \frac{dy}{dt},

where both partial derivatives are evaluated at $(x(t), y(t))$ . This is the Jacobian product of a $1 \times 2$ row and a $2 \times 1$ column. With $n$ middle variables there are $n$ terms.

In plain words As $t$ moves, $x$ and $y$ move, and each one pushes $z$ at its own rate. Add the two pushes.

Example For $f(x, y) = x^2 y$ on the circle $x = \cos t$ , $y = \sin t$ at $t = \frac{\pi}{4}$ , the point is $\left( \frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2} \right)$ . There $f_x = 2xy = 1$ and $f_y = x^2 = \frac{1}{2}$ , while $x' = -\frac{\sqrt{2}}{2}$ and $y' = \frac{\sqrt{2}}{2}$ . So

\begin{aligned} \frac{dz}{dt} &= 1 \cdot \left( -\frac{\sqrt{2}}{2} \right) + \frac{1}{2} \cdot \frac{\sqrt{2}}{2} \\[0.4em] &= -\frac{\sqrt{2}}{4}. \end{aligned}

Differentiating $\cos^2 t \sin t$ directly gives $-2\cos t \sin^2 t + \cos^3 t$ , which is also $-\frac{\sqrt{2}}{4}$ at $t = \frac{\pi}{4}$ .

Several independent parameters

Definition Let $z = f(x, y)$ with $f$ defined on an open set $V \subseteq \mathbb{R}^2$ , and let $x = x(s, t)$ and $y = y(s, t)$ be defined on an open set $W$ of the $st$ plane with $(x(s, t), y(s, t))$ in $V$ for all $(s, t)$ in $W$ . If $x$ and $y$ are differentiable at $(s, t)$ and $f$ is differentiable at $(x(s, t), y(s, t))$ , then

\begin{aligned} \frac{\partial z}{\partial s} &= \frac{\partial f}{\partial x} \frac{\partial x}{\partial s} + \frac{\partial f}{\partial y} \frac{\partial y}{\partial s}, \\[0.4em] \frac{\partial z}{\partial t} &= \frac{\partial f}{\partial x} \frac{\partial x}{\partial t} + \frac{\partial f}{\partial y} \frac{\partial y}{\partial t}, \end{aligned}

with the partial derivatives of $f$ evaluated at $(x(s, t), y(s, t))$ . Each formula is one entry of the product of a $1 \times 2$ Jacobian and a $2 \times 2$ Jacobian.

In plain words Hold one parameter fixed and apply the one parameter rule in the other. Do it once for each parameter.

Example Let $z = x^2 + xy$ in polar coordinates $x = r\cos\theta$ , $y = r\sin\theta$ , at $(r, \theta) = \left(2, \frac{\pi}{4}\right)$ . The point is $(\sqrt{2}, \sqrt{2})$ , where $f_x = 2x + y = 3\sqrt{2}$ and $f_y = x = \sqrt{2}$ . Also $x_r = y_r = \frac{\sqrt{2}}{2}$ , $x_\theta = -r\sin\theta = -\sqrt{2}$ and $y_\theta = r\cos\theta = \sqrt{2}$ . So

\begin{aligned} \frac{\partial z}{\partial r} &= 3\sqrt{2} \cdot \tfrac{\sqrt{2}}{2} + \sqrt{2} \cdot \tfrac{\sqrt{2}}{2} = 4, \\[0.4em] \frac{\partial z}{\partial \theta} &= 3\sqrt{2}(-\sqrt{2}) + \sqrt{2} \cdot \sqrt{2} = -4. \end{aligned}

Dependency diagrams

Definition A dependency diagram is a tree with the final variable at the top, the middle variables below it, and the independent variables at the bottom. Each edge is labeled with the partial derivative of the upper variable with respect to the lower one. Under the hypotheses of the chain rule, the partial derivative of the top variable with respect to an independent variable is the sum, over all routes from the top down to that variable, of the product of the labels along the route. Every label is evaluated at the actual point, the middle variables at their values there.

In plain words Draw who depends on whom. Each route from the top to your variable gives one product, and you add the products. Routes that never reach the variable contribute nothing.

Example Let $w = x e^y$ with $x = s^2 + t$ and $y = st$ . Here $x$ depends on both $s$ and $t$ , and so does $y$ , so there are two routes to $s$ and two to $t$ . At $(s, t) = (1, 1)$ we have $x = 2$ and $y = 1$ , so $w_x = e^y = e$ and $w_y = x e^y = 2e$ , while $x_s = 2$ , $y_s = t = 1$ , $x_t = 1$ and $y_t = s = 1$ . Adding the products along the routes gives

\begin{aligned} \frac{\partial w}{\partial s} &= e \cdot 2 + 2e \cdot 1 = 4e, \\[0.4em] \frac{\partial w}{\partial t} &= e \cdot 1 + 2e \cdot 1 = 3e. \end{aligned}

A common slip is to evaluate $w_x$ and $w_y$ at $(s, t) = (1, 1)$ instead of at $(x, y) = (2, 1)$ .

Total versus partial derivative

Definition Let $T(x, y, t)$ be differentiable on an open set of $\mathbb{R}^3$ , and let $x(t)$ and $y(t)$ be differentiable on an open interval with $(x(t), y(t), t)$ in that set. The partial derivative $\frac{\partial T}{\partial t}$ is the rate of change of $T$ in $t$ with $x$ and $y$ held fixed, here at $x = x(t)$ and $y = y(t)$ . The total derivative of $T$ along the path is the derivative of the composition $t \mapsto T(x(t), y(t), t)$ ,

\frac{dT}{dt} = \frac{\partial T}{\partial t} + \frac{\partial T}{\partial x} \frac{dx}{dt} + \frac{\partial T}{\partial y} \frac{dy}{dt},

with all partial derivatives of $T$ evaluated at $(x(t), y(t), t)$ . The two derivatives are different numbers in general, so the notation must say which is meant.

In plain words The partial derivative is what a thermometer fixed to the floor sees. The total derivative is what a thermometer riding along the path sees, which includes the effect of moving.

Example Let $T(x, y, t) = 20 + x^2 y - 2t$ and let the path be $x = 1 + t$ , $y = 3 - t$ . At $t = 1$ the point is $(2, 2)$ , where $T_x = 2xy = 8$ , $T_y = x^2 = 4$ and $T_t = -2$ . Also $x' = 1$ and $y' = -1$ . So

\frac{dT}{dt} = -2 + 8 \cdot 1 + 4 \cdot (-1) = 2,

while $\frac{\partial T}{\partial t} = -2$ . The fixed thermometer sees cooling and the moving one sees warming.

Check yourself

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

Show the answer

At $t = 2$ the point is $(3, 4)$ , where $z_x = 2xy = 24$ and $z_y = x^2 = 9$ . With $x' = 1$ and $y' = 2t = 4$ ,

\frac{dz}{dt} = 24 \cdot 1 + 9 \cdot 4 = 60.

Let $g(t) = \langle \cos t, \sin t \rangle$ and $F(x, y) = \langle x^2 - y^2, \ 2xy \rangle$ . Find $(F \circ g)'(0)$ by multiplying Jacobians.

Show the answer

Both maps are differentiable everywhere. The inner point is $g(0) = (1, 0)$ and $g'(0) = \langle 0, 1 \rangle$ . The outer Jacobian at $(1, 0)$ is

\begin{aligned} JF(1, 0) &= \begin{bmatrix} 2x & -2y \\ 2y & 2x \end{bmatrix}_{(1,0)} \\[0.4em] &= \begin{bmatrix} 2 & 0 \\ 0 & 2 \end{bmatrix}, \end{aligned}

and multiplying it by the column $g'(0)$ gives $(F \circ g)'(0) = \langle 0, 2 \rangle$ . This agrees with $F(g(t)) = \langle \cos 2t, \sin 2t \rangle$ .

Let $f(x, t) = xt$ and $w(t) = f(t^2, t)$ . Find $\frac{\partial f}{\partial t}$ and $\frac{dw}{dt}$ at $t = 1$ .

Show the answer

At $t = 1$ the point is $(x, t) = (1, 1)$ . The partial derivative is $f_t = x = 1$ . The total derivative adds the effect of $x = t^2$ moving,

\frac{dw}{dt} = f_t + f_x \cdot 2t = 1 + 1 \cdot 2 = 3,

which matches $w(t) = t^3$ .

The Multivariable Chain Rule

Key terms

Composition of multivariable functions

Jacobian multiplication

One independent parameter

Several independent parameters

Dependency diagrams

Total versus partial derivative

Check yourself

How this lesson connects