The multivariable chain rule and the total differential

A path through the plane

When both inputs of $f(x, y)$ depend on a third variable $t$, the output depends on $t$ too. Think of $t$ as time and of $(x(t), y(t))$ as a point moving across the plane; $z = f(x(t), y(t))$ is the height of the surface above it.

Take a rectangle whose width is $x(t) = 3 + 2t$ cm and height $y(t) = 2 - t$ cm, $t$ seconds from now. Its area is $A = xy$. At $t = 0$ it is 3 by 2, its width grows at 2 cm per second, and its height shrinks at 1 cm per second.

Take the two changes one at a time, with the partial derivatives of lesson 4.1.2. With the height held at 2, the growing width adds area at $A_x \cdot x' = y \cdot x' = 2 \times 2 = 4$ cm² per second. With the width held at 3, the shrinking height changes it at $A_y \cdot y' = x \cdot y' = 3 \times (-1) = -3$, so together they give 1 cm² per second.

Expanding confirms it. $A(t) = (3 + 2t)(2 - t) = 6 + t - 2t^2$ has slope $1 - 4t$, which is 1 at $t = 0$. Chord slopes from $t = 0$ give 0.8, 0.98 and 0.998 for steps 0.1, 0.01 and 0.001.

The rates $(x'(t), y'(t))$ form the velocity $\mathbf{r}'(t)$ of the moving point, so the rule is a dot product: $\frac{dz}{dt} = \nabla f \cdot \mathbf{r}'(t)$. At speed 1 this is the directional derivative of lesson 4.2.1; at speed 2 the height changes twice as fast.

A second example

Follow the tilted bowl $f(x, y) = x^2 + xy + y^2 - 3x$ of lesson 4.2.1 along the path $x = t^2$, $y = 1 - t$. At $t = 1$ the point is $(1, 0)$, where $f_x = 2x + y - 3 = -1$ and $f_y = x + 2y = 1$. The rates are $x' = 2t = 2$ and $y' = -1$:

Substituting first gives the same answer. $z(t) = t^4 - t^3 - t^2 - 2t + 1$, so $z'(t) = 4t^3 - 3t^2 - 2t - 2$, and $z'(1) = 4 - 3 - 2 - 2 = -3$. Central differences of $z(t)$ at 1 agree:

Pulling together, pulling apart, or not at all

Each term carries its own sign, so the inputs can add to the change or work against each other.

• Pulling apart. In the rectangle at $t = 0$, the width adds 4 and the height takes away 3. By $t = 1$ the rectangle is 5 by 1, so the terms are $1 \times 2 = 2$ and $5 \times (-1) = -5$: the area now shrinks at 3 cm² per second.
• Along a level curve. On the circle $x = \cos t$, $y = \sin t$, the function $f = x^2 + y^2$ is always 1. The two terms, $2x \cdot (-\sin t)$ and $2y \cdot \cos t$, cancel exactly: at $t = \frac{\pi}{6}$ they are $-0.866025$ and $0.866025$.
• At a flat point. Where $\nabla f = \mathbf{0}$, $\frac{dz}{dt} = 0$ whatever the velocity.

The circle is one case of a general fact. A path along a level curve has its velocity at right angles to $\nabla f$ (lesson 4.2.1), so $\nabla f \cdot \mathbf{r}' = 0$. In the figure, that happens at $t = 1.38$.

When time is an input too

Sometimes $t$ is one of the inputs itself: $z = f(t, y(t))$. Then the rule has a term for $t$ directly and a term for $t$ acting through $y$.

Take $f(t, y) = ty$ with $y = t^2$, so $z = t^3$ and $\frac{dz}{dt} = 3t^2 = 3$ at $t = 1$. The partial $f_t = y = 1$ is only part of it. The rest, $f_y \cdot y' = t \cdot 2t = 2$, is the effect of $y$ moving.

Case study 4.2.4 meets both kinds: an option's value changes with time directly, and through the stock price and volatility, which move as time passes.

The total differential

Multiply the chain rule by a short time $dt$. The point moves by small steps $dx$ and $dy$, and the height by about $f_x\,dx + f_y\,dy$. The same estimate holds for any small steps from a point, whatever path they come from.

In words: $f(a + dx,\, b + dy) \approx f(a, b) + f_x\,dx + f_y\,dy$, with the partials taken at $(a, b)$. The right side is the tangent plane: the flat surface that touches $f$ at $(a, b)$ with the same slopes. It is the degree-1 copy of lesson 2.7.1, in two variables.

Example 1: the rectangle. Grow the 3 by 2 rectangle by $dx = 0.1$ and $dy = 0.05$. Then $dA = y\,dx + x\,dy = 0.2 + 0.15 = 0.35$, while the true change is $3.1 \times 2.05 - 6 = 0.355$. The estimate misses only the small corner, $dx \cdot dy = 0.005$.

Halve both steps and the corner, which is the error, falls to a quarter: 0.00125. Halve them again and it is 0.0003125.

Example 2: a distance. Take $f(x, y) = \sqrt{x^2 + y^2}$ at $(3, 4)$, where $f = 5$ and $\nabla f = (0.6, 0.8)$ (lesson 4.2.1). For steps $dx = 0.1$ and $dy = -0.2$, $df = 0.06 - 0.16 = -0.1$. So $\sqrt{3.1^2 + 3.8^2} \approx 4.9$; the true value is 4.90408.

Each halving divides the error by about 4: the ratios are 4.04, 4.02 and 4.01.

Three variables, and where it fails

With three inputs the estimate has three terms. A box 2 by 3 by 4 cm has volume $V = xyz = 24$ cm³. Lengthen each side by 0.1 cm: $dV = yz\,dx + xz\,dy + xy\,dz = 12(0.1) + 8(0.1) + 6(0.1) = 2.6$, against a true change of 2.691.

The estimate fails in two ways. Large steps: from $(3, 4)$, the steps $(2, -4)$ reach $(5, 0)$, again at distance 5, so the true change is 0; the estimate says $-2$. Corners: the cone at the origin has no partial derivatives (lesson 4.2.1), so it has no tangent plane and no $df$ there.

Two inner variables

Now let $x$ and $y$ each depend on two variables, $s$ and $t$. Then $z$ has two partial derivatives, and each follows the chain rule with one of $s$, $t$ moving and the other held fixed.

Example: polar coordinates. A point at distance $r$ from the origin, at angle $\theta$, has $x = r\cos\theta$ and $y = r\sin\theta$, as in the polar form of lesson 1.7.3. Take $f = xy$, so $f_x = y$ and $f_y = x$. The inner partials are $x_r = \cos\theta$, $y_r = \sin\theta$, $x_\theta = -r\sin\theta$ and $y_\theta = r\cos\theta$.

At $r = 2$, $\theta = \frac{\pi}{6}$, the point is $(\sqrt{3}, 1) \approx (1.732051, 1)$, with $\cos\theta \approx 0.866025$ and $\sin\theta = 0.5$. Then

In numbers, $z_r = 0.866025 + 0.866025 \approx 1.732051$ and $z_\theta = -1 + 3 = 2$.

Substituting first agrees: $z = r^2\cos\theta\sin\theta$, so $z_r = 2r\cos\theta\sin\theta = \sqrt{3}$ and $z_\theta = r^2(\cos^2\theta - \sin^2\theta) = 4 \times 0.5 = 2$.

In matrix form, the row of answers is the gradient, written as a row, times the matrix of inner partials (lesson 3.2.1):

That matrix is called the Jacobian matrix of the map from $(r, \theta)$ to $(x, y)$; lesson 4.7.2 uses it to change variables in integrals.

Lesson 4.3.1 adds second-order terms to the total differential. Case study 4.2.4 uses the first-order estimate to split a day's change in an option's value into its causes.

Common mistakes

1. One term per input. For the rectangle at $t = 0$, $\frac{dA}{dt} = 4 - 3 = 1$; leaving out the height gives 4.
2. Take the partials at the moving point. At $t = 1$ the rectangle is 5 by 1, so the terms are $1 \times 2$ and $5 \times (-1)$, not the values at $t = 0$.
3. $\partial$ and $d$ are different. For $f = ty$ with $y = t^2$, the partial $f_t$ is 1 but $\frac{dz}{dt}$ is 3.
4. The total differential is for small steps. Its error grows like the square of the steps: the step $(2, -4)$ on the distance gives $-2$ for a true change of 0.
5. Steps in the inputs' own units. $dx$ must be in the units $f_x$ is measured per; mixing centimetres and metres, or percent and decimals, gives nonsense.
6. No tangent plane at a corner. Where a partial does not exist, neither does $df$.