# Activity6.1Introduction to the Chain Rule¶ permalink

The functions $\fe{f}{t}=\fe{\sin}{t}$ and $\fe{k}{t}=\fe{\sin}{3t}$ are shown in Figure 6.1.1. Since $\fe{\fd{f}}{t}=\fe{\cos}{t}\text{,}$ it is reasonable to speculate that $\fe{\fd{k}}{t}=\fe{\cos}{3t}\text{.}$ But this would imply that $\fe{\fd{k}}{0}=\fe{\fd{f}}{0}=1\text{,}$ and a quick glance of the two functions at $0$ should convince you that this is not true; clearly $\fe{\fd{k}}{0}>\fe{\fd{f}}{0}\text{.}$

The function $k$ moves through three periods for every one period generated by the function $f\text{.}$ Since the amplitudes of the two functions are the same, the only way $k$ can generate periods at a rate of $3:1$ (compared to $f$) is if its rate of change is three times that of $f\text{.}$ In fact, $\fe{\fd{k}}{t}=3\fe{\cos}{3t}\text{;}$ please note that $3$ is the first derivative of $3t\text{.}$ This means that the formula for $\fe{\fd{k}}{t}$ is the product of the rates of change of the outside function ($\lzoo{u}{\fe{\sin}{u}}=\fe{\cos}{u}$) and the inside function ($\lzoo{t}{3t}=3$).

$k$ is an example of a composite function (as illustrated in (6.1.1)). If we define $g$ by the rule $\fe{g}{t}=3t\text{,}$ then $\fe{k}{t}=\fe{f}{\fe{g}{t}}\text{.}$

$$t\xrightarrow{g}3t\xrightarrow{f}\fe{\sin}{3t}\tag{6.1.1}$$ \begin{align*} \fe{g}{t}&=3t\\ \fe{f}{u}&=\fe{\sin}{u}\\ \fe{k}{t}&=\fe{f}{\fe{g}{t}} \end{align*}

Taking the output from $g$ and processing it through a second function, $f\text{,}$ is the action that characterizes $k$ as a composite function.

Note that $\fe{\fd{k}}{0}=\fe{\fd{g}}{0}\fe{\fd{f}}{\fe{g}{0}}\text{.}$ This last equation is an example of what we call the chain rule for differentiation. Loosely, the chain rule tells us that when finding the rate of change for a composite function (at $0$), we need to multiply the rate of change of the outside function, $\fe{\fd{f}}{\fe{g}{0}}\text{,}$ with the rate of change of the inside function, $\fe{\fd{g}}{0}\text{.}$ This is symbolized for general values of $x$ in (6.1.2) where $u$ represents a function of $x$ (e.g. $u=\fe{g}{x}$).

$$\lzoo{x}{\fe{f}{u}}=\fe{\fd{f}}{u}\lzoo{x}{u}\tag{6.1.2}$$

This rule is used to find derivative formulas in Examples 6.1.2 and 6.1.3.

##### Example6.1.2

\begin{align*} \text{Problem}&&&\text{Solution}\\ \text{Find }\lz{y}{x}\text{ if }y&=\fe{\sin}{x^2}\text{.}&\lz{y}{x}&=\fe{\cos}{x^2}\cdot\lzoo{x}{x^2}\\ &&&=\fe{\cos}{x^2}\cdot2x\\ &&&=2x\fe{\cos}{x^2} \end{align*}

The factor of $\lzoo{x}{x^2}$ is called a chain rule factor.

##### Example6.1.3

\begin{align*} \text{Problem}&&&\text{Solution}\\ \text{Find }\fe{\fd{f}}{t}\text{ if }\fe{f}{t}&=\fe{\sec^9}{t}\text{.}&\fe{f}{t}&=\left[\fe{\sec}{t}\right]^9\\ &&\fe{\fd{f}}{t}&=9\left[\fe{\sec}{t}\right]^8\cdot\lzoo{t}{\fe{\sec}{t}}\\ &&&=9\fe{\sec^8}{t}\cdot\fe{\sec}{t}\fe{\tan}{t}\\ &&&=9\fe{\sec^9}{t}\fe{\tan}{t} \end{align*}

The factor of $\lzoo{t}{\fe{\sec}{t}}$ is called a chain rule factor.

##### Example6.1.4

\begin{align*} \text{Problem}&&&\text{Solution}\\ \text{Find }\lz{y}{x}\text{ if }y&=4^x\text{.}&\lz{y}{x}&=\fe{\ln}{4}\cdot4^x \end{align*}

Please note that the chain rule was not applied here because the function being differentiated was not a composite function. Here we just applied the formula for the derivative of an exponential function with base $4\text{.}$

##### Example6.1.5

\begin{align*} \text{Problem}&&&\text{Solution}\\ \text{Find }\lz{y}{x}\text{ if }y&=4^x&y&=e^{\fe{\ln}{4}x}\\ &=(e^{\fe{\ln}{4}})^x&\lz{y}{x}&=e^{\fe{\ln}{4}x}\cdot\lzoo{x}{\fe{\ln}{4}x}\\ &=e^{\fe{\ln}{4}x}\text{.}&&=e^{\fe{\ln}{4}x}\cdot\fe{\ln}{4}\text{.}\\ &&&=\fe{\ln}{4}\cdot4^{x} \end{align*}

While this is the same task as in Example 6.1.4, here we view the function as a composite function, and apply the chain rule.

While you ultimately want to perform the chain rule step in your head, your instructor may want you to illustrate the step while you are first practicing the rule. For this reason, the step will be explicitly shown in every example given in this lab.

# Subsection6.1.1Exercises

Find the first derivative formula for each function. In each case take the derivative with respect to the independent variable as implied by the expression on the right side of the equal sign. Make sure that you use the appropriate name for each derivative (e.g. $\fe{\fd{h}}{t}$).

##### 1

$\fe{h}{t}=\fe{\cos}{\sqrt{t}}$

##### 2

$P=\fe{\sin}{\theta^4}$

##### 3

$\fe{w}{\alpha}=\fe{\cot}{\sqrt[3]{\alpha}}$

##### 4

$z=7\left[\fe{\ln}{t}\right]^3$

##### 5

$\fe{z}{\theta}=\fe{\sin^4}{\theta}$

##### 6

$\fe{P}{\beta}=\fe{\tan^{-1}}{\beta}$

##### 7

$y=\left[\fe{\sin^{-1}}{t}\right]^{17}$

##### 8

$T=2^{\fe{\ln}{x}}$

##### 9

$\fe{y}{x}=\fe{\sec^{-1}}{e^x}$

A function, $f\text{,}$ is shown in Figure 6.1.6. Answer the following questions in reference to this function.

##### 10

Use the graph to rank the following in decreasing order: $\fe{f}{1}\text{,}$ $\fe{\fd{f}}{1}\text{,}$ $\fe{\sd{f}}{1}\text{,}$ $\fe{\fd{f}}{0}\text{,}$ and $\fe{\fd{f}}{-1}\text{.}$

##### 11

The formula for $f$ is $\fe{f}{t}=t\,e^t-e^{t^2}+3\text{.}$ Find the formulas for $\fd{f}$ and $\sd{f}$ and use the formulas to verify your answer to Exercise 6.1.1.10.

##### 12

Find the equation to the tangent line to $f$ at $0\text{.}$

##### 13

Find the equation to the tangent line to $\fd{f}$ at $0\text{.}$

##### 14

There is an antiderivative of $f$ that passes through the point $\point{0}{7}\text{.}$ Find the equation of the tangent line to this antiderivative at $0\text{.}$