Section 4.5 - Chain Rules

Section Objectives

1. State and use the chain rules for any number of independent and intermediate variables.
2. Use implicit differentiation.

Chain rule from Calculus I

The chain rule from Calculus I says

In this context, we think of $x$ as the independent variable and $u$ as an intermediate variable: $y$ depends on the independent variable $x$ through the action of $u$. Using the new language that we will soon develop, we will say this is the chain rule for one independent and one intermediate variable (Chain Rule 1-1).

Chain rules

Suppose we have a function that depends on the single variable $t$, but it does so through the action of two intermediate varibles, $x$ and $y$. For example,

In such a case, we might want to substitute in order to write $z$ in terms of the single variable: $z=\cos^2 3t + \sin^2 5t$. However, this is not always convenient or appropriate. The chain rule tells us how to differentiate $z$ through the intermediate variables.

Chain Rule 1-2:

Suppose $z=f(x,y)$ is a differentiable function of $x$ and $y$ and $x=x(t)$ and $y=y(t)$ are differentiable functions of $t$. Then $z=f(x(t),y(t))$ is a differentiable function of $t$ and

where the ordinary derivatives are evaluated at $t$ and the partial derivatives are evaluated at $(x,y)$.

Examples

• $z=4x^2+3y^2$, $x=\sin t$, and $y=\cos t$. Compute $dz/dt$.

• $z=\ln(x+y)$, $x=e^{2t}$, and $y=e^{5t}$. Compute $dz/dt$ when $t=0$.

What if we have two independent variables and two intermediate variables? Well, there's a chain rule for that.

Chain Rule 2-2:

Suppose $z=f(x,y)$ is a differentiable function of $x$ and $y$ and $x=x(s,t)$ and $y=y(s,t)$ are differentiable functions of $s$ and $t$. Then $z=f(x(s,t),y(s,t))$ is a differentiable function of $s$ and $t$ and

and

Notice that all derivatives in Chain Rule 2-2 are partial derivatives. Does that make sense?

Examples

• $z=3x^2-2xy+y^2$, $x=3u+2v$, and $y=4u-v$. Compute $\partial z/\partial u$ and $\partial z/\partial v$.

By now, you've probably figured out the chain rule patterns. The last version of the chain rule we will explicitly formulate is Chain Rule 2-3.

Chain Rule 2-3:

Suppose $w=f(x,y,z)$ is a differentiable function of $x$, $y$ and $z$ and $x=x(s,t)$, $y=y(s,t)$, and $z=z(s,t)$ are differentiable functions of $s$ and $t$. Then $w=f(x(s,t),y(s,t),z(s,t))$ is a differentiable function of $s$ and $t$ and

and

Examples

• $w=3x^2-2xy+4z^2$, $x=e^s \sin t$, $y=e^s \cos t$, and $z=e^s$. Compute $\partial w/\partial y$.

Implicit differentiation

One very nice application of the chain rule provides a new approach to implicit differentiation. Suppose that the equation $f(x,y)=0$ implicitly defines $y$ as a function of $x$. Now let $z=f(x,y)$ and think about $z$ as a function of the single independent variable $x$ through the action of the intermediate variables $x$ and $y$. Notice that $x$ is taking the role of both the independent variable and an intermediate variable. Let's use Chain Rule 1-2 to determine $dz/dx$.

Now, since $dx/dx=1$, it follows that

or

Examples

• Find $dy/dx$ if $x^3-4xy+y^3=5x$. Do this problem two ways: the new approach and the Calculus I approach.

• The idea generalizes! Suppose $f(x,y,z)=0$ implicitly defines $z$ as a function of $x$ and $y$. Then

Find $\partial z/\partial x$ and $\partial z/\partial y$ if $3x^2z-x^2y^2+2z^3+3yz-5=0$.