Section 2.1 started with an example of using the position of an object (in this case, a falling amusement-park rider) to find the object's velocity. This type of example is often used when introducing the derivative because we tend to readily recognize that velocity is the instantaneous rate of change in position. In general, if \(f\) is a function of \(x\text{,}\) then \(\fp(x)\) measures the instantaneous rate of change of \(f\) with respect to \(x\text{.}\) Put another way, the derivative answers “When \(x\) changes, at what rate does \(f\) change?” Thinking back to the amusement-park ride, we asked “When time changed, at what rate did the height change?” and found the answer to be “By \(-64\) feet per second.”

Now imagine driving a car and looking at the speedometer, which reads “60 mph.” Five minutes later, you wonder how far you have traveled. Certainly, lots of things could have happened in those \(5\) minutes; you could have intentionally sped up significantly, you might have come to a complete stop, you might have slowed to 20 mph as you passed through construction. But suppose that you know, as the driver, none of these things happened. You know you maintained a fairly consistent speed over those \(5\) minutes. What is a good approximation of the distance traveled?

One could argue the only good approximation, given the information provided, would be based on “\(\text{distance} = \text{rate}\times\text{time.}\)” In this case, we assume a constant rate of 60 mph with a time of \(5\) minutes or \(5/60\) of an hour. Hence we would approximate the distance traveled as \(5\) miles.

Referring back to the falling amusement-park ride, knowing that at \(t=2\) the velocity was \(-64\) ft/s, we could reasonably approximate that \(1\) second later the riders' height would have dropped by about \(64\) feet. Knowing that the riders were accelerating as they fell would inform us that this is an under-approximation. If all we knew was that \(f(2) = 86\) and \(\fp(2) = -64\text{,}\) we'd know that we'd have to stop the riders quickly otherwise they would hit the ground!

In both of these cases, we are using the instantaneous rate of change to predict future values of the output.

It is useful to recognize the units of the derivative function. If \(y\) is a function of \(x\text{,}\) i.e., \(y=f(x)\) for some function \(f\text{,}\) and \(y\) is measured in feet and \(x\) in seconds, then the units of \(y' = \fp\) are “feet per second,” commonly written as “ft/s.” In general, if \(y\) is measured in units \(P\) and \(x\) is measured in units \(Q\text{,}\) then \(y'\) will be measured in units “\(P\) per \(Q\)”, or “\(P/Q\text{.}\)” Here we see the fraction-like behavior of the derivative in the notation: the units of \(\frac{dy}{dx}\)are \(\frac{\text{units of }y}{\text{units of }x}\text{.}\)

The previous examples made use of an important approximation tool that we first used in our previous “driving a car at 60 mph” example at the beginning of this section. Five minutes after looking at the speedometer, our best approximation for distance traveled assumed the rate of change was constant. In Examples 2.2.1 and 2.2.2 we made similar approximations. We were given rate of change information which we used to approximate total change. Notationally, we would say that \begin{equation*} f(c+h) \approx f(c) + \fp(c)\cdot h. \end{equation*}

This approximation is best when \(h\) is “small.” Small is a relative term; when dealing with the world population, \(h=22\text{ days} = 28{,}800\text{ minutes}\) is small in comparison to years. When manufacturing widgets, \(100\) widgets is small when one plans to manufacture thousands.

One of the most fundamental applications of the derivative is the study of motion. Let \(s(t)\) be a position function, where \(t\) is time and \(s(t)\) is distance. For instance, \(s\) could measure the height of a projectile or the distance an object has traveled.

Let's let \(s(t)\) measure the distance traveled, in feet, of an object after \(t\) seconds of travel. Then \(s'(t)\) has units “feet per second,” and \(s'(t)\) measures the instantaneous rate of distance change with repsect to time — it measures velocity.

Now consider \(v(t)\text{,}\) a velocity function. That is, at time \(t\text{,}\) \(v(t)\) gives the velocity of an object. The derivative of \(v\text{,}\) \(v'(t)\text{,}\) gives the instantaneous rate of velocity change with respect to time — acceleration. (We often think of acceleration in terms of cars: a car may “go from \(0\) to \(60\) in \(4.8\) seconds.” This is an average acceleration, a measurement of how quickly the velocity changed.) If velocity is measured in feet per second, and time is measured in seconds, then the units of acceleration (i.e., the units of \(v'(t)\)) are “feet per second per second,” or \((\)ft/s\()\)/s. We often shorten this to “feet per second squared,” or ^ft⁄_s², but this tends to obscure the meaning of the units.

Perhaps the most well known acceleration is that of gravity. In this text, we use \(g=32\,\text{ft}/\text{s}^2\) or \(g=9.8\,\text{m}/\text{s}^2\text{.}\) What do these numbers mean?

A constant acceleration of \(32\,\frac{\text{ft}/\text{s}}{\text{s}}\) means that the velocity changes by \(32\,\text{ft}/\text{s}\) each second. For instance, let \(v(t)\) measure the velocity of a ball thrown straight up into the air, where \(v\) has units ft/s and \(t\) is measured in seconds. The ball will have a positive velocity while traveling upwards and a negative velocity while falling down. The acceleration is thus \(-32\,\text{ft}/\text{s}^2\text{.}\) If \(v(1) = 20\,\text{ft}/\text{s}\text{,}\) then \(1\) second later, the velocity will have decreased by \(32\,\text{ft}/\text{s}\text{;}\) that is, \(v(2) = -12\,\text{ft/s}\text{.}\) We can continue: \(v(3) = -44\,\text{ft/s}\text{.}\) Working backward, we can also figure that \(v(0) = 52\,\text{ft}/\text{s}\text{.}\)

These ideas are so important we write them out as a Key Idea.

We now consider the second interpretation of the derivative given in this section. This interpretation is not independent from the first by any means; many of the same concepts will be stressed, just from a slightly different perspective.

Given a function \(y=f(x)\text{,}\) the difference quotient \(\frac{f(c+h)-f(c)}{h}\) gives a change in \(y\) values divided by a change in \(x\) values; i.e., it is a measure of the “rise over run,” or “slope,” of the secant line that goes through two points on the graph of \(f\text{:}\) \((c, f(c))\) and \((c+h,f(c+h))\text{.}\) As \(h\) shrinks to \(0\text{,}\) these two points come close together; in the limit we find \(\fp(c)\text{,}\) the slope of a special line called the tangent line that intersects \(f\) only once near \(x=c\text{.}\)

Lines have a constant rate of change, their slope. Nonlinear functions do not have a constant rate of change, but we can measure their instantaneous rate of change at a given \(x\) value \(c\) by computing \(\fp(c)\text{.}\) We can get an idea of how \(f\) is behaving by looking at the slopes of its tangent lines. We explore this idea in the following example.

To demonstrate the accuracy of the tangent line approximation, we now state that in Example 2.2.10, \(f(x) = -x^3+7x^2-12x+4\text{.}\) We can evaluate \(f(3.1) = 4.279\text{.}\) Had we known \(f\) all along, certainly we could have just made this computation. In reality, we often only know two things:

1. What \(f(c)\) is, for some value of \(c\text{,}\) and

2. what \(\fp(c)\) is.

For instance, we can easily observe the location of an object and its instantaneous velocity at a particular point in time. We do not have a “function \(f\)” for the location, just an observation. This is enough to create an approximating function for \(f\text{.}\)

This last example has a direct connection to our approximation method explained above after Example 2.2.2. We stated there that \begin{equation*} f(c+h) \approx f(c)+\fp(c)\cdot h. \end{equation*}

If we know \(f(c)\) and \(\fp(c)\) for some value \(x=c\text{,}\) then computing the tangent line at \((c,f(c))\) is easy: \(y(x) = \fp(c)(x-c)+f(c)\text{.}\) In Example 2.2.10, we used the tangent line to approximate a value of \(f\text{.}\) Let's use the tangent line at \(x=c\) to approximate a value of \(f\) near \(x=c\text{;}\) i.e., compute \(y(c+h)\) to approximate \(f(c+h)\text{,}\) assuming again that \(h\) is “small.” Note: \begin{align*} y(c+h) \amp = \fp(c)\left((c+h)-c\right)+f(c)\\ \amp = \fp(c)\cdot h + f(c). \end{align*}

This is the exact same approximation method used above! Not only does it make intuitive sense, as explained above, it makes analytical sense, as this approximation method is simply using a tangent line to approximate a function's value.

The importance of understanding the derivative cannot be understated. When \(f\) is a function of \(x\text{,}\) \(\fp(x)\) measures the instantaneous rate of change of \(f\) with respect to \(x\) and gives the slope of the tangent line to \(f\) at \(x\text{.}\)