Chapter 1 Review

Answer each question to the best of your ability before reading the solution.

Define a function. What are the different ways a function can be represented?
Solution
A function is a rule or relationship that assigns exactly one output to each allowed input. A function can be represented:
- verbally, by describing how inputs lead to outputs;
- numerically, by providing input-output data;
- graphically, by drawing the graph of all input-output points; or
- algebraically, by giving a formula that turns inputs into outputs.
Explain what the average rate of change of $$ y $$ with respect to $$ x $$ is. What units is it measured in?

Solution

If $$ x $$ changes from $x\tsub{initial}$ to $x\tsub{final}$ and $$ y $$ changes from $y\tsub{initial}$ to $y\tsub{final}$ , then

$\Delta x=x\tsub{final}-x\tsub{initial}\qand\Delta y=y\tsub{final}-y\tsub{initial}\text{.}$

Provided that $\Delta x\ne0$ , the average rate of change is $\displaystyle \frac{\Delta y}{\Delta x}$ . It measures the change in $$ y $$ per unit change in $$ x $$ across the interval, so its units are the units of $$ y $$ per unit of $$ x $$ .
Explain the difference between $\Delta x$ and $\dd{x}$ .

Solution
A differential is an infinitesimal change in a quantity. The symbol $\Delta x$ represents a finite change between two values of $$ x $$ , while $\dd{x}$ represents an infinitesimal change in $$ x $$ .
Explain how the average rate of change $\displaystyle \frac{\Delta y}{\Delta x}$ leads to the instantaneous rate of change $\displaystyle \dv{y}{x}$ .

Solution
We can calculate $\displaystyle \frac{\Delta y}{\Delta x}$ over increasingly small intervals around a point. As the finite changes $\Delta x$ and $\Delta y$ shrink toward the infinitesimal changes $\dd{x}$ and $\dd{y}$ , the average rates approach the instantaneous ratio $\displaystyle \dv{y}{x}$ .
What is the derivative of a function, and what notation can be used for it?

Solution
If $$ y=f(x) $$ , then $\displaystyle \dv{y}{x}=f'(x)$ is the derivative function: it gives the instantaneous rate of change of $$ f $$ (or the slope of the graph of $$ f $$ ) at any value of $$ x $$ .
What does the sensitivity of an output $$ y $$ to an input $$ x $$ describe? Explain how the equation $\displaystyle \dv{y}{x}=f'(x)$ expresses this sensitivity.

Solution
Sensitivity describes how strongly an output responds to a small change in an input. Starting with the equation $\displaystyle \dv{y}{x}=f'(x)$ and multiplying by $\dd{x}$ gives $\dd{y}=f'(x)\dd{x}$ , which shows that an input change $\dd{x}$ produces an output change scaled by $$ f'(x) $$ . The magnitude of $$ f'(x) $$ describes the size of the response per unit of input, while its sign indicates whether the output responds in the same or opposite direction as the input.
Suppose $$ h(t) $$ describes the height of a ball in feet, where $$ t $$ is measured in seconds. What are the units of $\displaystyle \dv{h}{t}$ ? If the ball is thrown upward, what would you expect to happen to the sign of $\displaystyle \dv{h}{t}$ as the ball rises, reaches its greatest height, and then falls?

Solution
The units are feet per second (units of $$ h $$ per unit of $$ t $$ ). While the ball is rising, its height is increasing, so $\displaystyle \dv{h}{t}>0$ . At its greatest height, the ball is momentarily neither rising nor falling, so $\displaystyle \dv{h}{t}=0$ . As the ball falls, its height is decreasing, so $\displaystyle \dv{h}{t}<0$ .
What does the symbol $\displaystyle \dv{}{x}$ tell you to do?

Solution
The symbol $\displaystyle \dv{x}$ is an operator that instructs you to differentiate the expression that follows with respect to $$ x $$ , such as $\displaystyle \dv{x}(x^2)=2x$ . The notation itself can be interpreted as taking the differential first, such as $\displaystyle {\dd(x^2)=2x\dd{x}}$ , and then dividing through by $\dd{x}$ .
State the Power Rule. Explain how rewriting reciprocals and roots as powers allows the rule to be applied to expressions such as $\displaystyle \frac1x$ and $\sqrt{x}$ .

Solution
The Power Rule states that $\displaystyle \dv{x}(x^n)=nx^{n-1}$ . A reciprocal can be written as $\displaystyle \frac1x=x^{-1}$ , so its derivative is $-x^{-2}=-1/x^2$ . A square root can be written as $\sqrt{x}=x^{1/2}$ , so its derivative is $\displaystyle \frac12x^{-1/2}=\frac{1}{2\sqrt{x}}$ .
Explain how the various derivative rules can be used to find the derivative of a polynomial. Use $\displaystyle p(x)=4x^5-3x^2+7x-9$ as an example.

Solution

The Sum and Difference Rules allow us to differentiate term by term, the Constant Multiple Rule preserves each coefficient, and the Power Rule differentiates each power of $$ x $$ . The derivative of the constant term (the one without an $$ x $$ ) is zero. Thus, for our given example

$\begin{aligned}p'(x)&=4(5x^4)-3(2x)+7(1)-0\\ &=20x^4-6x+7\text{.}\end{aligned}$

The same process works for any polynomial: differentiate each term, preserve its coefficient, and combine the resulting terms.
State the Product Rule. Briefly explain where its two terms come from.

Solution

The Product Rule has the particularly simple differential form

$\dd(fg)=g\dd{f}+f\dd{g}\text{.}$

When both factors change, the product receives one contribution from the change in $$ f $$ while $$ g $$ has its current value, and another from the change in $$ g $$ while $$ f $$ has its current value.

If $$ f $$ and $$ g $$ are functions of $$ x $$ , then $\dd{f}=f'(x)\dd{x}$ and $\dd{g}=g'(x)\dd{x}$ . Substituting and dividing by $\dd{x}$ gives the derivative form

$\dv{x}\big(f(x)g(x)\big)=g(x)f'(x)+f(x)g'(x)\text{.}$
State the Quotient Rule. Where does the squared factor in its denominator come from?

Solution

Where $g\ne0$ , the Quotient Rule in differential form is

$\dd(\frac fg)=\frac{g\dd{f}-f\dd{g}}{g^2}\text{.}$

To see where the $$ g^2 $$ comes from, rewrite the quotient as $fg^{-1}$ . The Product and Power Rules give

$\begin{aligned}\dd(fg^{-1})&=g^{-1}\dd{f}+f\dd(g^{-1})\\ &=g^{-1}\dd{f}-fg^{-2}\dd{g}\\ &=\frac{g\dd{f}-f\dd{g}}{g^2}\text{.}\end{aligned}$

Notice that the factor $g^{-2}$ produced by differentiating $g^{-1}$ is the source of the squared denominator.

If $$ f $$ and $$ g $$ are functions of $$ x $$ , we can substitute $\dd{f}=f'(x)\dd{x}$ and $\dd{g}=g'(x)\dd{x}$ , then divide by $\dd{x}$ to obtain

$\dv{x}\left(\frac{f(x)}{g(x)}\right)=\frac{g(x)f'(x)-f(x)g'(x)}{(g(x))^2}\text{.}$
How can you recognize when the Chain Rule is needed, and how do you use it?

Solution

The Chain Rule is needed when one function is composed with, or nested inside, another. If $$ y=f(u) $$ and $$ u=g(x) $$ , the differential form follows the change through the two layers:

$\dd{y}=f'(u)\dd{u}\qand\dd{u}=g'(x)\dd{x}\text{.}$

Substituting the second differential into the first gives

$\dd{y}=f'(g(x))g'(x)\dd{x}\text{.}$

Dividing by $\dd{x}$ produces the derivative form

$\dv{y}{x}=f'(g(x))g'(x)\text{.}$

Equivalently we can also write the Chain Rule as $\displaystyle \dv{y}{x}=\dv{y}{u}\dv{u}{x}$ .

In practice, you can differentiate the outer function while leaving its input in place, then multiply by the derivative of the inner function.
Why is implicit differentiation useful, and how does the method work?

Solution
Implicit differentiation finds how variables change within a relationship without first solving explicitly for one variable. It's generally useful when isolating a variable would be difficult or inconvenient. To do implicit differentiation, we first take the differential of both sides, applying the differentiation rules to every term. Then we divide by the differential of the chosen independent variable and solve algebraically for the derivative we want.
Explain how to solve a related rates problem. What common pitfalls should you watch for?
Solution
First identify the changing quantities, draw a picture if possible, and write an equation that relates these quantities. Treating these quantities as functions of time, we then take the differential of the relationship, and divide by $\dd{t}$ to produce an equation involving their rates. Substitute the measurements and known rates from the specified moment, solve for the unknown rate, and finally interpret its sign and units.

Common pitfalls to watch out for include:
- Substituting measurements before differentiating.
- Confusing the value of a quantity with its rate of change.
- Treating a changing quantity as a constant.
- Losing the sign of a rate.
- Reporting an answer without meaningful units or interpretation.