Differentials

A basketball suspended in the air beside an outdoor hoop — Photo by Filipe Cantador on Unsplash.

A moving object has a velocity at every instant, even though no time passes during a single instant. How can we measure change at one particular input when change normally requires comparing two different values?

In this lesson, we'll use average rates over increasingly small intervals and the local linearity of a graph to make sense of this apparent contradiction. This will lead us to differentials and the ratio $\displaystyle \dv{y}{x}$ , which can describe instantaneous velocity, sensitivity to change, and other kinds of local response.

By the end of this lesson, you should be able to:

Estimate an instantaneous rate of change using average rates over increasingly small intervals.
Interpret differentials and the ratio $\displaystyle \dv{y}{x}$ using signs, units, and the context of a model.
Recognize instantaneous velocity and sensitivity to change as different interpretations of the same mathematical idea.

Imagine that you throw a basketball straight up into the air. It rises quickly, slows down until it briefly hangs in the air, and then falls back to the ground.

At any moment during the flight, it feels natural to ask:

How fast is the ball moving right now?

But there's a paradox hidden in that question. Velocity compares a change in position with the time over which that change occurs. Yet during a single instant, no time passes... so how can we measure velocity at that instant?

Suppose the height $$ H $$ of the basketball, in feet, can be modeled as a function of the time $$ t $$ , in seconds, after it leaves your hands:

$H(t)=6+40t-16t^2\text.$

The ball begins at a height of $$ H(0)= $$ feet. After one second, its height is $$ H(1)= $$ feet, so during that first second its change in height is $\Delta H=$ feet.

We can start to follow the ball's height throughout its flight:

$$ t $$ (seconds)	$$ H(t) $$ (feet)
$$ 0 $$	$$ 6 $$
$$ 0.5 $$
$$ 1 $$
$$ 1.25 $$
$$ 1.5 $$
$$ 2 $$
$$ 2.5 $$

The ball reaches its greatest height of feet at $$ t= $$ seconds. Notice that it has the same height at $$ t=1 $$ and $$ t=1.5 $$ , even though it's moving in opposite directions at those two times — in other words, knowing the height alone doesn't tell us how the ball is moving.

Loading the graph…If it does not appear, check the Desmos API configuration.

Let's try to determine how the ball is moving at exactly $$ t=1 $$ second. We already know that from $$ t=0 $$ to $$ t=1 $$ , the ball rises $$ 24 $$ feet in $$ 1 $$ second. Its average velocity during that interval is therefore

$\frac{\Delta H}{\Delta t}=\frac{30-6}{1-0}=24\text{ feet per second.}$

But notice that the graph is curved, so the ball's velocity is gradually changing throughout that entire interval; the average velocity doesn't necessarily tell us its instantaneous velocity at the time $$ t=1 $$ .

To focus more closely on that time, consider what happens just after $$ t=1 $$ . From $$ t=1 $$ to $$ t=1.1 $$ , the change in time is $\Delta t=0.1$ second and $$ H(1.1)=30.64 $$ feet, so

$\begin{aligned}\Delta H&=30.64-30\\ &=0.64\text{ feet}.\end{aligned}$

The average velocity over this shorter interval is therefore $\displaystyle \frac{\Delta H}{\Delta t}=$ feet per second.

We can repeat this calculation over shorter and shorter intervals beginning at $$ t=1 $$ :

$\Delta t$	$\displaystyle \frac{\Delta H}{\Delta t}$ (feet per second)
$$ 1 $$	$$ -8 $$
$$ 0.1 $$	$$ 6.4 $$
$$ 0.01 $$
$$ 0.001 $$

The change in sign may be surprising: during the full second after $$ t=1 $$ , the ball ends lower than it began (a lot can happen in one whole second!), so its average velocity is negative. But over intervals that stay close to $$ t=1 $$ , the average velocities appear to approach $$ 8 $$ feet per second.

We should also look at what happens just before $$ t=1 $$ . A negative value of $\Delta t$ means that we compare $$ t=1 $$ with an earlier time:

$\Delta t$	$\displaystyle \frac{\Delta H}{\Delta t}$ (feet per second)
$$ -0.1 $$	$$ 9.6 $$
$$ -0.01 $$
$$ -0.001 $$

From both sides, as $\Delta t$ gets smaller and smaller (that is, closer to zero), the average velocity seems to be getting closer to feet per second.

This process also has a geometric interpretation: each average velocity is the slope of the dotted line connecting the two relevant points on the height graph. To focus on what happens at $$ t=1 $$ , we can zoom in on shorter and shorter intervals of the form $[1,1+\Delta t]$ .

Loading the graph…If it does not appear, check the Desmos API configuration.

Move the slider to zoom in on the interval from

\( t=1 \)

t=1+\Delta t

. The dotted line connects the endpoints of the interval.

As you zoom in, the interval becomes shorter, the dotted line changes to connect its new endpoints, and the graph itself appears more and more like .

Now let's take a bold step: what if we could zoom in "infinitely close" to the curve at $$ t=1 $$ ?

At that scale, we might imagine that the graph is indistinguishable from a perfectly straight line. The ordinary changes $\Delta t$ and $\Delta H$ would become infinitesimal changes — changes smaller than any finite measurement, but which still preserve their ratio.

We call these infinitely small changes differentials. We write $\dd{t}$ for an infinitesimal change in time and $\dd{H}$ for the corresponding infinitesimal change in height. Based on our evidence from smaller and smaller intervals, we might conclude that the ratio between these two infinitesimal changes would be

$\dv{H}{t}=8\text{ feet per second.}$

The notation $\displaystyle \dv{H}{t}$ is read " $\dd{H}$ over $\dd{t}$ ." Here it represents the rate at which height is changing with respect to time at one instant.

So we now have a convincing answer to our original question: at $$ t=1 $$ , the ball has an instantaneous velocity of $$ 8 $$ feet per second upward.

Differentials and instantaneous rates

For a nonlinear function, the ratio $\Delta y/\Delta x$ generally depends on the interval we choose. When we choose two endpoints and compute $\Delta y/\Delta x$ over that interval, what we get is called the average rate of change of $$ y $$ with respect to $$ x $$ over that interval.

Example

In the previous lesson, we modeled the revenue $$ R $$ from a concert, in dollars, as a function of the ticket price $$ p $$ , in dollars:

$R(p)=-80p^2+960p\text.$

Compare the average rate of change from $$ p=3 $$ to $$ p=4 $$ with the average rate of change from $$ p=4 $$ to $$ p=5 $$ .

Solution

Over the first interval,

$\begin{aligned}\frac{\Delta R}{\Delta p}&=\frac{R(4)-R(3)}{4-3}\\ &=\frac{2560-2160}{1}\\ &=400\text{ dollars of revenue per dollar of ticket price.}\end{aligned}$

Over the second interval,

$\begin{aligned}\frac{\Delta R}{\Delta p}&=\frac{R(5)-R(4)}{5-4}\\ &=\frac{2800-2560}{1}\\ &=240\text{ dollars of revenue per dollar of ticket price.}\end{aligned}$

Note that the average rates are different even though both intervals have the same width. Since the graph is curved, the relationship between changes in price and changes in revenue depends on which part of the graph we examine.

For a linear function, the rate of change is constant, so we can say that every extra unit of input produces a fixed change in the output. For a nonlinear function, however, this interpretation doesn't hold over the entire graph: the output responds differently depending on the input where we start.

To describe what happens at one particular input $$ x=a $$ , we can examine average rates over smaller positive and negative changes beginning at $$ a $$ and see whether these rates seem to approach a fixed value. As the intervals become smaller, we're also zooming in closer to the graph around the point $$ (a,f(a)) $$ .

The straight line that locally matches the curve gives us a way to describe change at a single input: we imagine zooming in "infinitely far," so that the finite changes $\Delta x$ and $\Delta y$ become infinitely small, and the curve becomes indistinguishable from a line.

Loading the graph…If it does not appear, check the Desmos API configuration.

Drag the point to examine the zoomed-in graph at different points.

Loading the graph…If it does not appear, check the Desmos API configuration.

An infinitely magnified view near the selected point.

The ratio of these differentials represents how much $$ y $$ changes per infinitesimal change in $$ x $$ . This is the rate of change at one particular input:

If the idea of an "infinitely small" change feels a little strange, you're in good company! Infinitesimals have been a controversial topic throughout the history of mathematics. After all, infinity is notoriously difficult to get a handle on — just how small is "infinitely small," anyway?

For now, we're going to use infinitesimals as an intuitive way to describe what we see when we zoom in closer and closer to a curve. These intuitive-but-somewhat-vague ideas are remarkably useful, and later we'll develop a more precise way to justify them. This will allow us to use them with confidence to describe change in a variety of mathematical and real-world contexts.

Interpreting instantaneous rates of change

The meaning of an instantaneous rate of change comes from the quantities being compared. Two especially important interpretations are sensitivity to change and instantaneous velocity.

When the independent variable is a quantity that can increase or decrease, an instantaneous rate often describes the sensitivity of the output: how strongly the output responds to a small change in the input.

Example

Let $$ Y $$ be the yield of a rice crop, measured in bushels per acre, and let $$ F $$ be the fertilizer applied, measured in pounds per acre. Suppose

$Y(F)=250+0.8F-0.002F^2\text.$

Suppose that when $$ F=100 $$ , we have $\displaystyle \dv{Y}{F}=0.4$ . Interpret this sensitivity, and use it to estimate the change in yield if the fertilizer is decreased by $$ 5 $$ pounds per acre.

Solution

The sensitivity $\displaystyle \dv{Y}{F}=0.4$ means that near $$ F=100 $$ , an infinitesimal change in fertilizer produces $$ 0.4 $$ times as much change in yield. We can manipulate the differentials as a ratio:

$\begin{aligned}\dv{Y}{F}&=0.4\\ \dd{Y}&=0.4\,\dd{F}\text.\end{aligned}$

Decreasing the fertilizer by $$ 5 $$ pounds per acre gives $\Delta F=-5$ . We use the differential relationship to approximate this finite change:

$\begin{aligned}\Delta Y&\approx 0.4\Delta F\\ &=0.4(-5)\\ &=-2\text{ bushels per acre.}\end{aligned}$

Thus we estimate that the yield will decrease by about $$ 2 $$ bushels per acre. The signs matter: because the sensitivity is positive, a negative change in fertilizer produces a negative change in yield.

We can compare this estimate with the actual change predicted by the model:

$\begin{aligned}\Delta Y&=Y(95)-Y(100)\\ &=307.95-310\\ &=-2.05\text{ bushels per acre.}\end{aligned}$

The differential estimate $$ -2 $$ is close, but it isn't exact because $\Delta F=-5$ is a finite change and the graph is curved. In many situations, once the ratio of differentials is known, manipulating it is faster than repeatedly evaluating the original function and makes the direction, size, and units of the response easier to interpret.

When position is a function of time, its instantaneous rate of change is called instantaneous velocity. If $$ s(t) $$ is position, then

$\dv{s}{t}=\frac{\text{infinitesimal change in position}}{\text{infinitesimal change in time}}\text.$

Velocity is related to speed, but it's not quite the same — speed is the magnitude (absolute value) of velocity, $\displaystyle \abs{\dv{s}{t}}$ , so speed is never negative.

Example

Suppose the height $$ H $$ of a basketball, in feet, is modeled as a function of time $$ t $$ , in seconds, by

$H(t)=6+40t-16t^2\text.$

Estimate the instantaneous velocity of the basketball at $$ t=2 $$ .

Solution

We calculate average velocities over smaller and smaller time intervals beginning at $$ t=2 $$ . Positive values of $\Delta t$ look forward in time, while negative values look backward. For example, when $\Delta t=0.1$ , we have

$\begin{aligned}\frac{\Delta H}{\Delta t}&=\frac{H(2.1)-H(2)}{2.1-2}\\ &=\frac{19.44-22}{0.1}\\ &=-25.6\text{ feet per second.}\end{aligned}$

Repeating this process over smaller intervals from both directions gives us the following values:

$\Delta t$	$\displaystyle \frac{\Delta H}{\Delta t}$ (feet per second)
$$ -0.1 $$	$$ -22.4 $$
$$ -0.01 $$	$$ -23.84 $$
$$ -0.001 $$	$$ -23.984 $$
$$ 0.001 $$	$$ -24.016 $$
$$ 0.01 $$	$$ -24.16 $$
$$ 0.1 $$	$$ -25.6 $$

As $\Delta t$ approaches zero from either direction, the average velocities approach $$ -24 $$ feet per second. We therefore estimate

$\eval{\dv{H}{t}}_{t=2}=-24\text{ feet per second.}$

The negative sign means that the basketball is moving downward; its speed is $$ 24 $$ feet per second.

Interpreting signs

Because $\displaystyle \dv{y}{x}$ is a ratio of differentials, its sign tells us how the signs of $\dd{x}$ and $\dd{y}$ are related. If $\dv{y}{x}=m\text,$ then we can manipulate the differentials to write $\dd{y}=m\dd{x}\text{.}$ Therefore, the sign of $\displaystyle \dv{y}{x}$ determines whether an infinitesimal nudge in $$ x $$ produces a nudge in $$ y $$ in the same or opposite direction:

If $\displaystyle \dv{y}{x}>0$ , then $\dd{x}$ and $\dd{y}$ have the same sign. Nudging $$ x $$ upward nudges $$ y $$ upward, while nudging $$ x $$ downward nudges $$ y $$ downward.
If $\displaystyle \dv{y}{x}<0$ , then $\dd{x}$ and $\dd{y}$ have opposite signs. Nudging $$ x $$ upward nudges $$ y $$ downward, while nudging $$ x $$ downward nudges $$ y $$ upward.
If $\displaystyle \dv{y}{x}=0$ , then an infinitesimal nudge in either direction produces almost no change in $$ y $$ . Geometrically, the line that locally matches the graph is horizontal at that point.

A horizontal locally matching line often appears at the top or bottom of a graph, so a value of $\displaystyle \dv{y}{x}=0$ can be a clue that the output has reached a maximum or minimum. However, the value zero only tells us what the graph is doing at that particular point; by itself, it doesn't tell us what happens on either side. We'll address how to more carefully find the maximum or minimum of a function in Chapter 2.

If $\displaystyle \dv{y}{x}$ represents a sensitivity, then its sign tells us whether the two quantities respond in the same or opposite directions. For example, continuing with the crop yield example above:

At $$ F=100 $$ , we have $\displaystyle \eval{\dv{Y}{F}}_{F=100}=0.4$ , which is positive; this means that increasing fertilizer slightly increases yield, while decreasing fertilizer slightly decreases yield.
At $$ F=200 $$ , we have $\displaystyle \eval{\dv{Y}{F}}_{F=200}=0$ , so a small nudge in either direction has almost no effect on yield.
At $$ F=400 $$ , we have $\displaystyle \eval{\dv{Y}{F}}_{F=400}=-0.8$ , which is negative; this means that increasing fertilizer slightly decreases yield, while decreasing fertilizer slightly increases yield.

(You should of course verify these ratios yourself by computing $\displaystyle \frac{\Delta Y}{\Delta F}$ over smaller and smaller intervals.)

In a motion problem, we usually interpret time as flowing forward, so we almost always focus on a positive nudge in $$ t $$ .

A positive value of $\displaystyle \dv{s}{t}$ means that the object is moving in the positive direction — usually right or up — at that moment.
A negative value of $\displaystyle \dv{s}{t}$ means that the object is moving in the negative direction — usually left or down — at that moment.

In the basketball example, $\displaystyle \eval{\dv{H}{t}}_{t=2}=-24$ feet per second, so the ball is moving downward at that moment.

Whether $\displaystyle \dv{y}{x}$ represents a velocity, a sensitivity, or something else, the underlying idea is the same: it compares infinitesimal changes in two related quantities. We can estimate this ratio using average rates over smaller and smaller intervals, visualize it as the slope of the line that locally matches the graph, and interpret its sign and units using the context of the model.

However, you might have rightfully noticed that repeatedly computing ratios over increasingly small intervals can get rather tedious. Don't worry — we'll soon develop systematic techniques for finding these instantaneous rates much more efficiently.

Let $$ f(x)=${f} $$ . Estimate the instantaneous rate of change at $$ x=${a} $$ .
1. Complete the table. Each row uses an interval from $$ x=${a} $$ to $x=${a}+\Delta x$ .
  
  $\Delta x$ $\displaystyle \dfrac{\Delta f}{\Delta x}$
  
  $$ -0.1 $$
  
  $$ -0.01 $$
  
  $$ 0.01 $$
  
  $$ 0.1 $$
  
  Show at least one calculation of $\displaystyle \frac{\Delta f}{\Delta x}$ . Be sure to identify $\Delta x$ and $\Delta f$ before finding their ratio.
  
  Solution
  
  The completed quotient column, from top to bottom, is
  
  $${dq1},\quad ${dq2},\quad ${dq3},\quad ${dq4}\text.$
  
  For the first row, $$ x $$ changes from $$ ${a} $$ to $${dleft_x}$ . Therefore,
  
  $\begin{aligned}\Delta x&=${dleft_x}-${a}=-0.1,\\ \Delta f&=${df_left_value}-${parenthesize_negative(f_a)}=${ddf_left},\\ \frac{\Delta f}{\Delta x}&=\frac{${ddf_left}}{-0.1}=${dq1}.\end{aligned}$
2. What instantaneous rate of change do the quotients appear to approach?
  
  $\displaystyle \left.\dv{f}{x}\right|_{x=${a}}=$
  
  Solution
  
  The quotients on both sides approach $$ ${rate} $$ , so
  
  $\left.\dv{f}{x}\right|_{x=${a}}=\boxed{${rate}}\text.$
In this exercise, you'll investigate instantaneous rates for three familiar families of functions.
1. Suppose $$ y=${constant} $$ . Compute $\displaystyle \frac{\Delta y}{\Delta x}$ over any interval you choose. What does this suggest about $\displaystyle \dv{y}{x}$ at every input? Explain using the graph of the function.
  
  $\displaystyle \dv{y}{x}=$
  
  Solution
  The output never changes, so $\Delta y=0$ over every interval and every ratio $\displaystyle \frac{\Delta y}{\Delta x}$ is zero. The graph is horizontal, so its locally matching line also has slope zero. Therefore $\displaystyle \dv{y}{x}=0$ at every input.
2. Suppose $$ y=${line} $$ . Compute the average rate of change from $x=${line_a}$ to $x=${line_a+0.01}$ .
  
  $\dfrac{\Delta y}{\Delta x}=$
  
  What can you conclude about the instantaneous rate of change of $$ y=mx+b $$ at any input?
  
  Solution
  The average rate of change of a linear function is its slope, which in this case is $$ ${m} $$ . A linear function is already its own locally matching line, so $\displaystyle \dv{y}{x}=m$ at every input.
3. For $$ y=${quad} $$ , use $\Delta x=0.01$ to estimate the instantaneous rate at $$ x=0 $$ , $$ x=1 $$ , $$ x=2 $$ , and $$ x=3 $$ .
  
  $$ x=0: $$
  
  $$ x=1: $$
  
  $$ x=2: $$
  
  $$ x=3: $$
  
  Then conjecture a formula for the instantaneous rate at $$ x=a $$ .
  
  $\displaystyle \left.\dv{y}{x}\right|_{x=a}=$
  
  Solution
  
  The ratios $\displaystyle \frac{\Delta y}{\Delta x}$ are $$ ${dq0} $$ , $$ ${dq1} $$ , $$ ${dq2} $$ , and $$ ${dq3} $$ , respectively. Using still smaller changes would give rates approaching $$ 0 $$ , $$ ${2*k} $$ , $$ ${4*k} $$ , and $$ ${6*k} $$ . The pattern suggests
  
  $\left.\dv{y}{x}\right|_{x=a}=${quadratic_formula}\text.$
Each ratio below represents the sensitivity of one real-world quantity to another. Decide whether you would expect each sensitivity to be positive, negative, or zero. Explain your reasoning, and give the units. Think about what should happen to the dependent variable if the independent variable is increased slightly.
1. $\displaystyle \dv{B}{P}$ , where $$ B $$ is the brightness of a lamp, measured in lumens, and $$ P $$ is the electrical power supplied to it, measured in watts.
  
  Solution
  We expect a positive sensitivity: supplying slightly more power should make the lamp brighter. The units are lumens per watt.
2. $\displaystyle \dv{S}{D}$ , where $$ S $$ is the strength of a wireless signal, measured in bars, and $$ D $$ is the distance from the signal source, measured in meters.
  
  Solution
  We expect a negative sensitivity: moving slightly farther from the source should weaken the signal. The units are bars per meter.
3. $\displaystyle \dv{T}{t}$ , where $$ T $$ is the temperature of a potato in degrees Fahrenheit and $$ t $$ is the number of minutes it has been baking.
  
  Solution
  We expect a positive sensitivity while the potato is baking: leaving it in the oven slightly longer should raise its temperature. The units are degrees Fahrenheit per minute.
4. $\displaystyle \dv{t}{T}$ , where $$ t $$ is the time required to bake a potato fully, measured in minutes, and $$ T $$ is the oven temperature, measured in degrees Fahrenheit.
  
  Solution
  We expect a negative sensitivity: increasing the oven temperature slightly should decrease the required baking time. The units are minutes per degree Fahrenheit.
5. $\displaystyle \dv{C}{h}$ , where $$ C $$ is the number of calories in a candy bar and $$ h $$ is its elevation above sea level, measured in feet.
  
  Solution
  We expect the sensitivity to be zero: the candy bar's calories aren't sensitive at all to a change in elevation. The units are calories per foot.
A car is driven along a fixed $$ 100 $$ -mile route. The table shows the total energy $$ E $$ used for the trip as a function of the car's speed $$ v $$ . Speed is measured in miles per hour and energy in kilowatt-hours.

$$ v $$ (mph) $$ ${v1} $$ $$ ${v2} $$ $$ ${v3} $$ $$ ${v4} $$ $$ ${v5} $$

$$ E $$ (kWh) $$ ${de1} $$ $$ ${de2} $$ $$ ${de3} $$ $$ ${de4} $$ $$ ${de5} $$
1. Use the data at $$ v=${v2} $$ and $$ v=${v3} $$ to estimate $\displaystyle \left.\dv{E}{v}\right|_{v=${v3-3}}$ . Include units and interpret its sign.
  
  $\displaystyle \left.\dv{E}{v}\right|_{v=${v3-3}}\approx$
  
  Solution
  
  $\frac{\Delta E}{\Delta v}=\frac{${de3}-${de2}}{${v3}-${v2}}=${drate}\text{ kilowatt-hours per mile per hour.}$
  
  Near $$ ${v3-3} $$ mph, increasing speed by one mile per hour decreases the trip's energy use by approximately $${rate_size}$ kilowatt-hours.
2. Suppose the energy use is minimized at $$ v=${v3} $$ mph. What would you expect $\displaystyle \dv{E}{v}$ to equal there? Explain your reasoning.
  
  Solution
  At the minimum, we expect $\displaystyle \dv{E}{v}=0$ . Near the minimum, the graph's locally matching line is horizontal, so a small change in speed has almost no effect on energy use.
The graph shows $$ P(x) $$ , the profit earned by a company in thousands of dollars, as a function of $$ x $$ , the number of units produced in hundreds.

Loading the graph…If it does not appear, check the Desmos API configuration.

Profit as a function of production.
1. For approximately which production levels is $$ P(x) $$ positive? What does that mean for the company?
  
  Solution
  The graph is above the horizontal axis from $x=${left_root}$ to $x=${right_root}$ . Since $$ x $$ is measured in hundreds, the company earns a positive profit when it produces between $${left_units}$ and $${right_units}$ units.
2. For approximately which values of $$ x $$ is $\displaystyle \dv{P}{x}$ positive? Negative? Explain how you can tell from the graph and what each sign means for the company.
  
  Solution
  The graph rises for $$ 0<x<${vertex} $$ , so $\displaystyle \dv{P}{x}>0$ there: producing slightly more units increases profit. It falls for $$ ${vertex}<x<30 $$ , so $\displaystyle \dv{P}{x}<0$ there: producing slightly more units decreases profit.
3. About how many units should the company produce to maximize profit? What is $\displaystyle \dv{P}{x}$ at that production level, and what are its units?
  
  Solution
  The maximum occurs at $$ x=${vertex} $$ , or $${vertex_units}$ units, where the profit is $$ ${height} $$ thousand dollars. At the top, the graph's locally matching line is horizontal, so $\displaystyle \dv{P}{x}=0$ . Its units are thousands of dollars per hundred units produced.

$\Delta x$	$\displaystyle \dfrac{\Delta f}{\Delta x}$
$\( -0.1 \)$
$\( -0.01 \)$
$\( 0.01 \)$
$\( 0.1 \)$

$\( v \)$ (mph)	$\( ${v1} \)$	$\( ${v2} \)$	$\( ${v3} \)$	$\( ${v4} \)$	$\( ${v5} \)$
$\( E \)$ (kWh)	$\( ${de1} \)$	$\( ${de2} \)$	$\( ${de3} \)$	$\( ${de4} \)$	$\( ${de5} \)$

Extra practice

For each function, calculate $\displaystyle \frac{\Delta y}{\Delta x}$ over intervals with smaller and smaller nonzero values of $\Delta x$ , using values from both directions. Then use those ratios to estimate $\displaystyle \dv{y}{x}$ at the two given inputs.

$$ y=${cubic} $$ , at $x=${cubic_x1}$ and $x=${cubic_x2}$

$\displaystyle \left.\dv{y}{x}\right|_{x=${cubic_x1}}=$

$\displaystyle \left.\dv{y}{x}\right|_{x=${cubic_x2}}=$

Solution

At $x=${cubic_x1}$ :

$\Delta x$	$\displaystyle \frac{\Delta y}{\Delta x}$
$$ -0.1 $$	$${fixed(cubic_average(cubic_x1,h1),4)}$
$$ -0.01 $$	$${fixed(cubic_average(cubic_x1,h2),4)}$
$$ 0.01 $$	$${fixed(cubic_average(cubic_x1,h3),4)}$
$$ 0.1 $$	$${fixed(cubic_average(cubic_x1,h4),4)}$

At $x=${cubic_x2}$ :

$\Delta x$	$\displaystyle \frac{\Delta y}{\Delta x}$
$$ -0.1 $$	$${fixed(cubic_average(cubic_x2,h1),4)}$
$$ -0.01 $$	$${fixed(cubic_average(cubic_x2,h2),4)}$
$$ 0.01 $$	$${fixed(cubic_average(cubic_x2,h3),4)}$
$$ 0.1 $$	$${fixed(cubic_average(cubic_x2,h4),4)}$

The ratios approach $${cubic_rate1}$ and $${cubic_rate2}$ , so

$\left.\dv{y}{x}\right|_{x=${cubic_x1}}=${cubic_rate1}\qquad\text{and}\qquad \left.\dv{y}{x}\right|_{x=${cubic_x2}}=${cubic_rate2}\text.$

$y=${root_function}$ , at $x=${root_x1}$ and $x=${root_x2}$

$\displaystyle \left.\dv{y}{x}\right|_{x=${root_x1}}=$

$\displaystyle \left.\dv{y}{x}\right|_{x=${root_x2}}=$

Solution

At $x=${root_x1}$ :

$\Delta x$	$\displaystyle \frac{\Delta y}{\Delta x}$
$$ -0.1 $$	$${fixed(root_average(root_x1,h1),4)}$
$$ -0.01 $$	$${fixed(root_average(root_x1,h2),4)}$
$$ 0.01 $$	$${fixed(root_average(root_x1,h3),4)}$
$$ 0.1 $$	$${fixed(root_average(root_x1,h4),4)}$

At $x=${root_x2}$ :

$\Delta x$	$\displaystyle \frac{\Delta y}{\Delta x}$
$$ -0.1 $$	$${fixed(root_average(root_x2,h1),4)}$
$$ -0.01 $$	$${fixed(root_average(root_x2,h2),4)}$
$$ 0.01 $$	$${fixed(root_average(root_x2,h3),4)}$
$$ 0.1 $$	$${fixed(root_average(root_x2,h4),4)}$

The ratios approach $${root_rate1}$ and $${root_rate2}$ , so

$\left.\dv{y}{x}\right|_{x=${root_x1}}=${root_rate1}\qquad\text{and}\qquad \left.\dv{y}{x}\right|_{x=${root_x2}}=${root_rate2}\text.$

$$ y=${reciprocal} $$ , at $x=${reciprocal_x1}$ and $x=${reciprocal_x2}$

$\displaystyle \left.\dv{y}{x}\right|_{x=${reciprocal_x1}}=$

$\displaystyle \left.\dv{y}{x}\right|_{x=${reciprocal_x2}}=$

Solution

At $x=${reciprocal_x1}$ :

$\Delta x$	$\displaystyle \frac{\Delta y}{\Delta x}$
$$ -0.1 $$	$${fixed(reciprocal_average(reciprocal_x1,h1),4)}$
$$ -0.01 $$	$${fixed(reciprocal_average(reciprocal_x1,h2),4)}$
$$ 0.01 $$	$${fixed(reciprocal_average(reciprocal_x1,h3),4)}$
$$ 0.1 $$	$${fixed(reciprocal_average(reciprocal_x1,h4),4)}$

At $x=${reciprocal_x2}$ :

$\Delta x$	$\displaystyle \frac{\Delta y}{\Delta x}$
$$ -0.1 $$	$${fixed(reciprocal_average(reciprocal_x2,h1),4)}$
$$ -0.01 $$	$${fixed(reciprocal_average(reciprocal_x2,h2),4)}$
$$ 0.01 $$	$${fixed(reciprocal_average(reciprocal_x2,h3),4)}$
$$ 0.1 $$	$${fixed(reciprocal_average(reciprocal_x2,h4),4)}$

The ratios approach $${reciprocal_rate1}$ and $${reciprocal_rate2}$ , so

$\left.\dv{y}{x}\right|_{x=${reciprocal_x1}}=${reciprocal_rate1}\qquad\text{and}\qquad \left.\dv{y}{x}\right|_{x=${reciprocal_x2}}=${reciprocal_rate2}\text.$

$$ y=${quadratic} $$ , at $x=${quad_x1}$ and $x=${quad_x2}$

$\displaystyle \left.\dv{y}{x}\right|_{x=${quad_x1}}=$

$\displaystyle \left.\dv{y}{x}\right|_{x=${quad_x2}}=$

Solution

At $x=${quad_x1}$ :

$\Delta x$	$\displaystyle \frac{\Delta y}{\Delta x}$
$$ -0.1 $$	$${fixed(quadratic_average(quad_x1,h1),4)}$
$$ -0.01 $$	$${fixed(quadratic_average(quad_x1,h2),4)}$
$$ 0.01 $$	$${fixed(quadratic_average(quad_x1,h3),4)}$
$$ 0.1 $$	$${fixed(quadratic_average(quad_x1,h4),4)}$

At $x=${quad_x2}$ :

$\Delta x$	$\displaystyle \frac{\Delta y}{\Delta x}$
$$ -0.1 $$	$${fixed(quadratic_average(quad_x2,h1),4)}$
$$ -0.01 $$	$${fixed(quadratic_average(quad_x2,h2),4)}$
$$ 0.01 $$	$${fixed(quadratic_average(quad_x2,h3),4)}$
$$ 0.1 $$	$${fixed(quadratic_average(quad_x2,h4),4)}$

The ratios approach $${quad_rate1}$ and $${quad_rate2}$ , so

$\left.\dv{y}{x}\right|_{x=${quad_x1}}=${quad_rate1}\qquad\text{and}\qquad \left.\dv{y}{x}\right|_{x=${quad_x2}}=${quad_rate2}\text.$

$\( t \)$ (seconds)	$\( H(t) \)$ (feet)
$\( 0 \)$	$\( 6 \)$
$\( 0.5 \)$
$\( 1 \)$
$\( 1.25 \)$
$\( 1.5 \)$
$\( 2 \)$
$\( 2.5 \)$

$\Delta t$	$\displaystyle \frac{\Delta H}{\Delta t}$ (feet per second)
$\( -0.1 \)$	$\( -22.4 \)$
$\( -0.01 \)$	$\( -23.84 \)$
$\( -0.001 \)$	$\( -23.984 \)$
$\( 0.001 \)$	$\( -24.016 \)$
$\( 0.01 \)$	$\( -24.16 \)$
$\( 0.1 \)$	$\( -25.6 \)$

$\Delta t$	$\displaystyle \frac{\Delta H}{\Delta t}$ (feet per second)
$\( 1 \)$	$\( -8 \)$
$\( 0.1 \)$	$\( 6.4 \)$
$\( 0.01 \)$
$\( 0.001 \)$