Section 2.2: The Derivative

Example 1

Suppose the total cost of producing \(x\) items is given by \(TC(x) = 200+30x-0.1x^2\). Determine the average rate of change of cost when increasing production from 25 units to 100 units.

The cost when producing 25 units is \(TC(25) = 200+30(25)-0.1(25)^2 = \$887.50\)

The cost when producing 100 units is \(TC(100) = 200+30(100)-0.1(100)^2 = \$2200\)

The cost increased by $2200-$887.50 = $1312.50 while production increased by 100 – 25 = 75 items. The average rate of change is:
\[\frac{TC(100) - TC(25)}{100 - 25} = \frac{2200-887.50}{100-25} = \frac{1312.50}{75} = 17.50 \text{dollars per unit}\]

This tells us that on average the cost increases by $17.50 for each unit produced.

The average rate of change is the slope of the secant line between (25, 887.50) and (100, 2200)

Example 2

Suppose the total cost of producing \(x\) items is given by \(TC(x) = 200+30x-0.1x^2\). Estimate the instantaneous rate of change when 25 items are being produced.

We can see in the previous graph that the secant slope on the interval \(25\le x \le 100\) is not a particularly good estimate of the instantaneous rate of change since the cost function seems steeper at \(x = 25\) then over the secant slope. We could improve the estimate by choosing a smaller interval.

Approach 1: Using smaller intervals

Using the interval \(25\le x \le 30\), the average rate of change would be:
\[\frac{TC(30) - TC(25)}{30 - 25} = \frac{1010-887.50}{30-25} = \frac{122.50}{5} = 24.50 \text{dollars per unit}\]

Using an interval on the other side, \(20\le x \le 25\), the average rate of change would be:
\[\frac{TC(25) - TC(20)}{25 - 20} = \frac{887.50-760}{25-20} = \frac{127.5}{5} = 25.50 \text{dollars per unit}\]

Visually, we can see both these secant lines seem to approximate the function pretty well.

We would expect the instantaneous rate of change to be somewhere between these two values. Averaging them, we get an estimate of $25 per unit for the instantaneous rate of change.

Approach 2: Estimating from the graph.

This approach is more commonly used when we only have the graph of a function, and don’t have a formula to evaluate, but we will illustrate it here using the same function.

The instantaneous rate of change is the slope of the tangent line, which is the line that just touches the graph at the point of interest, and has the same rate of change (slope) as the function does at the point. Given a graph, we can sketch in a tangent line by holding a ruler up to the graph, and drawing in a line that touches the graph when \(x = 25\) and has the same slope.

To estimate the instantaneous rate of change, we can now calculate the slope of this drawn-in line. Looking at the graph, the tangent line appears to pass approximately through (30, 1000) and (70, 2000) which would give a slope of \(\frac{2000-1000}{70-30}=\frac{1000}{40}=25\) dollars per unit.

Example 3

Suppose we drop a tomato from the top of a 100 foot building and time its fall.

1. How long did it take for the tomato to drop 100 feet?
2. How far did the tomato fall during the first second?
3. How far did the tomato fall during the last second?
4. How far did the tomato fall between \(t =0.5\) and \(t = 1\)?
5. What was the average velocity of the tomato during its fall?
6. What was the average velocity between \( t=1\) and \(t=2\) seconds?
7. How fast was the tomato falling 1 second after it was dropped?

Some of these questions are pretty easy to answer, while some are more complex.

1. From the table, it took 2.5 seconds for the tomato to drop 100 feet
2. During the first second, the tomato fell 100 – 84 = 16 feet
3. During the last second, the tomato fell 64 – 0 = 64 feet
4. Between \(t =0.5\) and \(t = 1\) the tomato fell 96 – 84 = 12 feet
5. Velocity is similar to speed, and is a rate of change. We can calculate average velocity the same way we did average rate of change earlier.
   \[\text{Average velocity}=\frac{\text{distance fallen}}{\text{total time}}=\frac{\Delta\text{position}}{\Delta\text{time}}=\frac{-100 \text{ ft}}{2.5 \text{ s}}=-40 \text{ ft/s}\]
6. Between \( t=1\) and \(t=2\) seconds, \[\text{Average velocity}=\frac{\Delta\text{position}}{\Delta\text{time}}=\frac{36\text{ ft}- 84\text{ ft}}{2\text{ s} - 1\text{ s}}=\frac{-48 \text{ ft}}{1 \text{ s}}=-48 \text{ ft/s}\]

7. This question is significantly different from the previous two questions about average velocity. Here we want the instantaneous velocity, the velocity at an instant in time. Unfortunately the tomato is not equipped with a speedometer so we will have to give an approximate answer. Like in our Approach 1 in the previous example, we will estimate it using secant slopes.

   One crude approximation of the instantaneous velocity after 1 second is simply the average velocity during the entire fall, -40 ft/s . But the tomato fell slowly at the beginning and rapidly near the end so the "-40 ft/s" estimate may or may not be a good answer.

   We can get a better approximation of the instantaneous velocity at \(t=1\) by calculating the average velocities over a short time interval near \(t = 1\). The average velocity between \(t = 0.5\) and \(t = 1\) is \(\dfrac{-12\text{ feet}}{0.5\text{ s}} = -24\text{ ft/s}\), and the average velocity between \(t = 1\) and \(t = 1.5\) is \(\dfrac{-20\text{ feet}}{0.5\text{ s}} = -40\text{ ft/s}\) so we can be reasonably sure that the instantaneous velocity is between -24 ft/s and -40 ft/s. The average, –32 ft/s, would be a good estimate for the instantaneous velocity.

In general, the shorter the time interval over which we calculate the average velocity, the better the average velocity will approximate the instantaneous velocity. The average velocity over a time interval is \( \dfrac{\Delta\text{position}}{\Delta\text{time}} \), which is the slope of the secant line through two points on the graph of height versus time. The instantaneous velocity at a particular time and height is the slope of the tangent line to the graph at the point given by that time and height.

Example 4

Suppose we set up a machine to count the number of bacteria growing on a Petri plate. At first there are few bacteria so the population grows slowly. Then there are more bacteria to divide so the population grows more quickly. Later, there are more bacteria and less room and nutrients available for the expanding population, so the population grows slowly again. Finally, the bacteria have used up most of the nutrients, and the population declines as bacteria die.

Use the population graph to estimate the answer to the questions below.

1. What is the bacteria population at time \(t = 3\) days?
2. What is the population increment from \(t = 3\) to \(t =10\) days?
3. What is the rate of population growth from \(t = 3\) to \(t = 10\) days?
4. What is the rate of population growth on the third day, at \(t = 3\) ?

1. From the graph, at \(t = 3\), the population is about 0.5 thousand, or 500 bacteria.
2. At \(t = 10\), the population is about 4.5 thousand, so the increment is about 4000 bacteria.

3. The rate of growth from \(t = 3\) to \(t = 10\) is the average rate of change in population during that time: \[ \begin{align*} \text{average change in population } & = \frac{\text{change in population}}{\text{change in time}}\\ & = \frac{\Delta\text{population}}{\Delta\text{time}} \\ & = \frac{4000\text{ bacteria}}{7\text{ days}} \\ & \approx 570\text{ bacteria/day}. \end{align*} \]

   This is the slope of the secant line through the two points (3, 500) and (10, 4500).

4. This question is asking for the instantaneous rate of population change, the slope of the line which is tangent to the population curve at (3, 500). If we sketch a line approximately tangent to the curve at (3, 500) and pick two points near the ends of the tangent line segment , we can estimate that instantaneous rate of population growth is approximately 320 bacteria/day .

   

Example 5

Find the slope of the line \(L\) in the graph below which is tangent to \(f(x) = x^2\) at the point (2,4).

We could estimate the slope of \(L\) from the graph, but we won't. Instead, we will use the idea that secant lines over tiny intervals approximate the tangent line.

We can see that the line through (2,4) and (3,9) on the graph of \(f\) is an approximation of the slope of the tangent line, and we can calculate that slope exactly: \(m = \frac{\Delta y}{\Delta x} = \frac{9-4}{3-2} = 5\). But \(m = 5\) is only an estimate of the slope of the tangent line and not a very good estimate. It's too big. We can get a better estimate by picking a second point on the graph of \(f\) which is closer to (2,4) – the point (2,4) is fixed and it must be one of the points we use.

From the second figure, we can see that the slope of the line through the points (2,4) and (2.5,6.25) is a better approximation of the slope of the tangent line at (2,4): \(m = \frac{\Delta y}{\Delta x} = \frac{6.25 - 4}{2.5 - 2} = \frac{2.25}{0.5} = 4.5 \), a better estimate, but still an approximation. We can continue picking points closer and closer to (2,4) on the graph of \(f\), and then calculating the slopes of the lines through each of these points and the point (2,4):

The only thing special about the x–values we picked is that they are numbers which are close, and very close, to \(x = 2\). Someone else might have picked other nearby values for \(x\). As the points we pick get closer and closer to the point (2,4) on the graph of \( y = x^2\), the slopes of the lines through the points and (2,4) are better approximations of the slope of the tangent line, and these slopes are getting closer and closer to 4.

We can bypass much of the calculating by not picking the points one at a time: let's look at a general point near (2,4). Define \( x = 2 + h\) so \(h\) is the increment from 2 to \(x\). If \(h\) is small, then \(x = 2 + h\) is close to 2 and the point \((2+h, f(2+h) ) = \left(2+h, (2+h)^2\right) \) is close to (2,4). The slope \(m\) of the line through the points (2,4) and \(\left(2+h, (2+h)^2\right)\) is a good approximation of the slope of the tangent line at the point (2,4):

The value \( m = 4 + h \) is the slope of the secant line through the two points (2,4) and \(\left( 2+h, (2+h)^2 \right)\). As \(h\) gets smaller and smaller, this slope approaches the slope of the tangent line to the graph of \(f\) at (2,4).

More formally, we could write: \[\text{Slope of the tangent line} = \dfrac{\Delta y}{\Delta x} = \lim\limits_{h\to 0} (4+h). \]

We can easily evaluate this limit using direct substitution, finding that as the interval \(h\) shrinks towards 0, the secant slope approaches the tangent slope, 4.

Example 6

Find the slope of the tangent line to \( f(x)=\frac{1}{x} \) at \(x = 3\).

The slope of the tangent line is the value of the derivative \(f'(3)\). \( f(3)=\frac{1}{3}\) and \( f(3+h)=\frac{1}{3+h} \), so, using the formal limit definition of the derivative, \[ f'(3)=\lim\limits_{h\to 0}\frac{f(3+h)-f(3)}{h}=\lim\limits_{h\to 0}\frac{\frac{1}{3+h}-\frac{1}{3}}{h}. \]

We can simplify by giving the fractions a common denominator: \[ \begin{align*} \lim\limits_{h\to 0}\frac{\frac{1}{3+h}-\frac{1}{3}}{h} & = \lim\limits_{h\to 0}\frac{\frac{1}{3+h}\cdot\frac{3}{3}-\frac{1}{3}\cdot\frac{3+h}{3+h}}{h} \\ & = \lim\limits_{h\to 0}\frac{\frac{3}{9+3h}-\frac{3+h}{9+3h}}{h} \\ & = \lim\limits_{h\to 0}\frac{\frac{3-(3+h)}{9+3h}}{h} \\ & = \lim\limits_{h\to 0}\frac{\frac{3-3-h}{9+3h}}{h} \\ & = \lim\limits_{h\to 0}\frac{\frac{-h}{9+3h}}{h} \\ & = \lim\limits_{h\to 0}\frac{-h}{9+3h}\cdot\frac{1}{h} \\ & = \lim\limits_{h\to 0}\frac{-1}{9+3h} \\ \end{align*} \] and the evaluate using direct substitution: \[\lim\limits_{h\to 0}\frac{-1}{9+3h}=\frac{-1}{9+3(0)}=-\frac{1}{9}.\]

Thus, the slope of the tangent line to \( f(x)=\frac{1}{x} \) at \(x = 3\) is \( -\frac{1}{9} \).

Example 7

Below is the graph of a function \( y=f(x) \). We can use the information in the graph to fill in a table showing values of \( f'(x): \)

At various values of \(x\), draw your best guess at the tangent line and measure its slope. You might have to extend your lines so you can read some points. In general, your estimate of the slope will be better if you choose points that are easy to read and far away from each other. Here are estimates for a few values of \(x\) (parts of the tangent lines used are shown above in the graph):

We can estimate the values of \(f'(x)\) at some non-integer values of \(x\), too: \(f'(0.5) \approx 0.5\) and \(f'(1.3) \approx -0.3\).

We can even think about entire intervals. For example, if \(0 \lt x \lt 1\), then \(f(x)\) is increasing, all the slopes are positive, and so \(f'(x)\) is positive.

The values of \(f'(x)\) definitely depend on the values of \(x\), and \(f'(x)\) is a function of \(x\). We can use the results in the table to help sketch the graph of \(f'(x)\).

Example 8

Shown is the graph of the height \(h(t)\) of a rocket at time \(t\).

Sketch the graph of the velocity of the rocket at time \(t\). (Velocity is the derivative of the height function, so it is the slope of the tangent to the graph of position or height.)

We can estimate the slope of the function at several points. The lower graph below shows the velocity of the rocket. This is \(v(t) = h'(t)\).

Example 9

Below is the graph of \(y = g(x)\). At what values of \(x\) does the graph of \(g(x)\) have horizontal tangent lines?

The tangent lines to the graph of \(g(x)\) are horizontal (slope = 0) when \(x\approx -1, 1, 2.5, \text{ and } 5\).

Example 10

Find \( \frac{d}{dx}\left( 2x^2-4x-1 \right) \).

Setting up the derivative using a limit, \[ f'(x)=\lim\limits_{h\to 0}\frac{f(x+h)-f(x)}{h}.\]

We will start by simplifying \( f(x+h) \) by expanding: \[ \begin{align*} f(x+h) & = 2(x+h)^2-4(x+h)-1 \\ & = 2(x^2+2xh+h^2)-4(x+h)-1 \\ & = 2x^2+4xh+2h^2-4x-4h-1 \end{align*} \]

Now finding the limit: \[ \begin{align*} f'(x) & = \lim\limits_{h\to 0}\frac{f(x+h)-f(x)}{h} \\ & = \lim\limits_{h\to 0} \frac{(2x^2+4xh+2h^2-4x-4h-1)-(2x^2-4x-1)}{h} \\ & = \lim\limits_{h\to 0} \frac{2x^2+4xh+2h^2-4x-4h-1-2x^2+4x+1}{h} \qquad \text{(Substitute in the formulas.)} \\ & = \lim\limits_{h\to 0} \frac{4xh+2h^2-4h}{h} \qquad \text{(Now simplify.)}\\ & = \lim\limits_{h\to 0} \frac{h(4x+2h-4)}{h} \qquad \text{(Factor out the \( h \), then cancel.)} \\ & = \lim\limits_{h\to 0} (4x+2h-4) \end{align*} \] We can find the limit of this expression by direct substitution: \[ f'(x)=\lim\limits_{h\to 0} (4x+2h-4)=4x-4\]

Notice that the derivative depends on \(x\), and that this formula will tell us the slope of the tangent line to \(f\) at any value \(x\). For example, if we wanted to know the tangent slope of \(f\) at \(x = 3\), we would simply evaluate: \( f'(3)=4(3)-4=8 \).

Example 7

Suppose the demand curve for widgets was given by \( D(p)=\frac{1}{p} \), where \(D\) is the quantity of widgets, in thousands, at a price of \(p\) dollars. Interpret the derivative of \(D\) at \(p = \)$3.

Note that we calculated \( D'(3) \) earlier to be \( D'(3)=-\frac{1}{9}\approx -0.111 \).

Since \(D\) has units thousands of widgets and the units for \(p\) is dollars of price, the units for \(D'\) will be \( \frac{\text{thousands of widgets}}{\text{dollar of price}} \). In other words, it shows how the demand will change as the price increases.

Specifically, \( D'(3)\approx -0.111 \) tells us that when the price is $3, the demand will decrease by about 0.111 thousand items for every dollar the price increases.