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The Multivariable Chain Rule

This section use the ideas of partial derivatives and provides an algebraic description for the chain rule (which is used in ).

The Multivariable Chain Rule

This section use the ideas of partial derivatives and provides an algebraic description for the chain rule (which is used in ). Some instructors will do a cursory coverage this material, but we relate the concept and development of the chain rule to our tools of linearization from . We have relegated the use of tree diagrams to a couple of exercises. Most motivation and execution of the chain rule comes from our description of change as a linear combination of the change through each independent variable.

Introduction

In single-variable calculus, we encountered situations in which some quantity was related through a composition of related quantities. For instance, suppose \(P\) is the price of a product and \(P\) depends on the wages \(x\) of people making the product. The wages of people making the product will also change over time because of seasonal availability and other factors. So a change in the value of time \(t\) produces a change in the value of \(x\), which produces a change in the value of \(P\). This means that we can write \(P\) as a function of \(x\) and \(x\) as a function of \(t\); thus we can express \(P\)'s dependence on \(t\) by a composition of functions: \((P\circ x)(t) = P(x(t))\).

Applying the chain rule from single-variable calculus to find the rate of change of the price \(P\) with respect to time \(t\) likely leads you to write \[\begin{aligned}\end{aligned}\]. However, this form of the chain rule is only useful when we have algebraic expressions for \(P(x)\) and \(x(t)\). It may be that we know about the rate of change of \(P\) with respect to \(x\) at a specific value \(x_0\) and the rate of change of \(x\) with respect to \(t\) at a specific value \(t_0\) for which we have \(x(t_0)=x_0\). Using the language of differentials and local linearity we can express these changes as follows. For a small change in \(x\), the change in the price of the product is approximately proportional to the change in \(x\): \[\begin{aligned}\end{aligned}\], where \(\frac{dP}{dx}\restrict{x=x_0}{}\) denotes the instantaneous rate of change in the product price per unit of labor cost at \(x=x_0\). Similarly, a small change in the value of \(t\) will also cause a proportional change in the value of \(x\): \[\begin{aligned}\end{aligned}\].

We can combine these relationships to express how a change in \(t\) produces a change in \(P\): \[\begin{aligned}\end{aligned}\] We should interpret this as telling us that the change in the price of the product is approximately the current instantaneous rate of change of the price per step in \(x\) times the current rate of change in the price of labor times the step in time.

As we look at functions of two (or more) variables in this text, we will encounter a situation where \(z = f(x,y)\) and both \(x\) and \(y\) depend on another variable \(t\). A change in \(t\) then produces changes in both \(x\) and \(y\), which then causes \(z\) to change. In this section, we will see how to describe the change in \(z\) that is caused by a change in \(t\), leading us to multivariable versions of the chain rule involving both regular and partial derivatives.

Condensed — the full section is in Boelkins et al., Active Calculus Multivariable.

The Chain Rule

As Preview Activity suggests, the following version of the chain rule holds in general.

If \(z = f(x,y)\) is a differentiable function of \(x\) and \(y\) and \(x\) and \(y\) are each differentiable functions of \(t\), then \[\begin{aligned}\end{aligned}\].

It is important to note the differences among the derivatives in . Since \(z\) is a function of the two variables \(x\) and \(y\), the derivatives in the chain rule for \(z\) with respect to \(x\) and \(y\) are partial derivatives. However, since \(x = x(t)\) and \(y = y(t)\) are functions of the single variable \(t\), their derivatives are the standard derivatives of functions of one variable. When we compose \(z\) with \(x(t)\) and \(y(t)\), \(z\) is represented as a function of the single variable \(t\). Therefore, the derivative of \(z\) with respect to \(t\) is a derivative from single-variable calculus as well.

To understand why this chain rule works in general, suppose that \(z\) depends on \(x\) and \(y\) so that we can express the change in \(z\) in terms of the differential as \[\begin{aligned}\end{aligned}\]. Further suppose that \(x\) and \(y\) each depend on \(t\), so that \[\begin{aligned}\end{aligned}\].

Combining equations and , we find that \[\begin{aligned}\end{aligned}\] which is the chain rule in this particular context, as expressed in equation.

This approach to understanding the change in \(z\) using the differential or linearization also separates how much of the change in \(z\) is coming through each of the intermediate variables \(x\) and \(y\). In the context of the Preview Activity where \(P\) is the amount of the large particulate matter in the air as function of location \((x,y)\) and location is changing according to time along the path \(\vr(t)=\langle x(t),y(t)\rangle\), the chain rule expresses the rate of change in \(P\) as a function of time as \[\begin{aligned}\end{aligned}\].

Because the amount of large particulate matter in terms of the \(x\)-coordinate is governed by tree pollen, the term \(\frac{\partial P}{\partial x} \frac{dx}{dt}\) describes the rate of change in the large particulate matter per unit time along our drive attributable to tree pollen. Similarly, the amount of large particulate matter in terms of the \(y\)-coordinate is governed by industrial pollution. Hence, the second term \(\frac{\partial P}{\partial y} \frac{dy}{dt}\) describes the rate of change in the large particulate matter per unit time along our drive attributable to industrial pollution. Because this function is locally linear, the total rate of change in large particulate matter is the sum of these individual rates of change.

Condensed — the full section is in Boelkins et al., Active Calculus Multivariable.

Practice (1)

Try each one on paper first. Reveal the answer to check; verified ones can be opened in the solver for every step.

  1. There are several proposed formulas to approximate the surface area of the human body. One modelDuBois D, DuBois DF. A formula to estimate the approximate surface area if height and weight be known. Arch Int Med 1916;17:863-71. uses the formula \[\begin{aligned}\end{aligned}\] where \(A\) is the surface area in square meters, \(h\) is the height in centimeters, and \(w\) is the weight in kilograms.

    Since a person's height \(h\) and weight \(w\) change over time \(h\) and \(w\) are functions of time \(t\). Let us think about what is happening to a child whose height is \(60\) centimeters and weight is \(9\) kilograms. Suppose, furthermore, that \(h\) is increasing at an instantaneous rate of 20 centimeters per year and \(w\) is increasing at an instantaneous rate of \(5\) kg per year.

    Determine the instantaneous rate at which the child's surface area is changing at this point in time.

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    We want to find \(\frac{dA}{dt}\mid_{(60,9)}\) given that \(\frac{dh}{dt} = 20\) and \(\frac{dw}{dt} = 5\). The Chain Rule tells us that \[\begin{aligned}\frac{dA}{dt} \amp = \frac{\partial A}{\partial h} \frac{dh}{dt} + \frac{\partial A}{\partial w} \frac{dw}{dt} \\ \amp = 0.00522h^{-0.275}w^{0.425}(20) + 0.00306h^{0.725}w^{-0.575}(5)\end{aligned}\].

    So \[\begin{aligned}\end{aligned}\].

Symbols used here

\frac{\partial f}{\partial x}
partial derivative
Derivative with respect to x, holding the other variables fixed.
P(A),\ P(A \mid B)
probability, conditional probability
Chance of A; chance of A given that B happened.
f'(x),\ \frac{dy}{dx}
derivative
Instantaneous rate of change; slope of the graph.
\nabla f
gradient (nabla, del)
Vector of partial derivatives; points uphill.
\int f(x)\,dx,\ \int_a^b
integral
Antiderivative (indefinite) or signed area from a to b (definite).
\iint,\ \oint
double / contour integral
Integral over a region of the plane; integral around a closed curve.
\mathbf{v},\ \vec{v}
vector
A quantity with magnitude and direction; a column of numbers.
\mathbf{u} \cdot \mathbf{v},\ \|\mathbf{v}\|
dot product, norm
Σ u_i v_i; the length of v, √(v·v).

How to: The Multivariable Chain Rule

  1. How can we take derivatives involving compositions of multivariable functions?
  2. For locally linear functions, how can a partial derivative of a composition of multivariable functions separate into dependence on the input variables?

Questions people ask

What is a partial derivative?

The ordinary derivative with respect to one variable while every other variable is frozen — the slope of the surface in one coordinate direction.

What does the gradient point at?

Uphill: the direction of steepest increase, with length equal to that steepest slope. It is perpendicular to the level curves.

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Parts of this page are adapted from Boelkins et al., Active Calculus Multivariable (CC BY-SA 4.0). Condensed and re-explained here; errors are ours.

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