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Section2.1Instantaneous Rates of Change: The Derivative

A common amusement park ride lifts riders to a height then allows them to freefall a certain distance before safely stopping them. Suppose such a ride drops riders from a height of \(150\) feet. Students of physics may recall that the height (in feet) of the riders, \(t\) seconds after freefall (and ignoring air resistance, etc.) can be accurately modeled by \(f(t) = -16t^2+150\text{.}\)

Using this formula, it is easy to verify that, without intervention, the riders will hit the ground when \(f(t)=0\) so at \(t=2.5\sqrt{1.5} \approx 3.06\) seconds. Suppose the designers of the ride decide to begin slowing the riders' fall after \(2\) seconds (corresponding to a height of \(f(2)=86\) ft). How fast will the riders be traveling at that time?

We have been given a position function, but what we want to compute is a velocity at a specific point in time, i.e., we want an instantaneous velocity. We do not currently know how to calculate this.

However, we do know from common experience how to calculate an average velocity. (If we travel \(60\) miles in \(2\) hours, we know we had an average velocity of \(30\) mph.) We looked at this concept in Section 1.1 when we introduced the difference quotient. We have \begin{equation*} \frac{\text{change in distance}}{\text{change in time}} = \frac{\text{“rise”}}{\text{“run”}} = \text{average velocity.} \end{equation*}

We can approximate the instantaneous velocity at \(t=2\) by considering the average velocity over some time period containing \(t=2\text{.}\) If we make the time interval small, we will get a good approximation. (This fact is commonly used. For instance, high speed cameras are used to track fast moving objects. Distances are measured over a fixed number of frames to generate an accurate approximation of the velocity.)

Consider the interval from \(t=2\) to \(t=3\) (just before the riders hit the ground). On that interval, the average velocity is \begin{equation*} \frac{f(3)-f(2)}{3-2} = \frac{6-86}{1} =-80\,\text{ft/s} , \end{equation*} where the minus sign indicates that the riders are moving down. By narrowing the interval we consider, we will likely get a better approximation of the instantaneous velocity. On \([2,2.5]\) we have \begin{equation*} \frac{f(2.5)-f(2)}{2.5-2} = \frac{50-86}{0.5} =-72\,\text{ft/s}\text{.} \end{equation*}

Units in Calculations

In the above calculations, we left off the units until the end of the problem. You should always be sure that you label your answer with the correct units. For example, if \(g(x)\) gave you the cost (in $) of producing \(x\) widgets, the units on the difference quotient would be $/widget.

We can do this for smaller and smaller intervals of time. For instance, over a time span of one tenth of a second, i.e., on \([2,2.1]\text{,}\) we have \begin{equation*} \frac{f(2.1)-f(2)}{2.1-2} = \frac{79.44-86}{0.1} =-65.6\,\text{ft/s}\text{.} \end{equation*}

Over a time span of one hundredth of a second, on \([2,2.01]\text{,}\) the average velocity is \begin{equation*} \frac{f(2.01)-f(2)}{2.01-2} = \frac{85.3584-86}{0.01} =-64.16\,\text{ft/s}\text{.} \end{equation*}

What we are really computing is the average velocity on the interval \([2,2+h]\) for small values of \(h\text{.}\) That is, we are computing \begin{equation*} \frac{f(2+h) - f(2)}{h} \end{equation*} where \(h\) is small.

What we really want is for \(h=0\text{,}\) but this, of course, returns the familiar \(“0/0”\) indeterminate form. So we employ a limit, as we did in Section 1.1.

We can approximate the value of this limit numerically with small values of \(h\) as seen in Table 2.1.1. It looks as though the velocity is approaching \(-64\) ft/s.

\(h\) Average Velocity (ft/s)
\(1\) \(\phantom{Average}{-80}\)
\(0.5\) \(\phantom{Average}{-72}\)
\(0.1\) \(\phantom{Average}{-65.6}\)
\(0.01\) \(\phantom{Average}{-64.16}\)
\(0.001\) \(\phantom{Average}{-64.016}\)
Table2.1.1Approximating the instantaneous velocity with average velocities over a small time period \(h\text{.}\)

Computing the limit directly gives \begin{align*} \lim_{h\to 0} \frac{f(2+h)-f(2)}{h} \amp = \lim_{h\to 0}\frac{-16(2+h)^2+150 - (-16(2)^2+150)}{h}\\ \amp = \lim_{h\to 0}\frac{-16(4+4h+h^2)+150 - 86}{h}\\ \amp = \lim_{h\to 0}\frac{-64-64h-16h^2+64}{h}\\ \amp = \lim_{h\to 0}\frac{-64h-16h^2}{h}\\ \amp = \lim_{h\to 0}-64 -16h\\ \amp =-64. \end{align*}

Graphically, we can view the average velocities we computed numerically as the slopes of secant lines on the graph of \(f\) going through the points \((2,f(2))\) and \((2+h,f(2+h))\text{.}\) In Figures 2.1.22.1.4, the secant line corresponding to \(h=1\) is shown in three contexts. Figure 2.1.2 shows a “zoomed out” version of \(f\) with its secant line. In Figure 2.1.3, we zoom in around the points of intersection between \(f\) and the secant line. Notice how well this secant line approximates \(f\) between those two points — it is a common practice to approximate functions with straight lines.

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Figure2.1.2The function \(f(x)\) and its secant line corresponding to \(t=2\) and \(t=3\text{.}\)
Figure2.1.3The function \(f(x)\) and a secant line corresponding to \(t=2\) and \(t=3\text{,}\) zoomed in near \(t=2\text{.}\)

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Figure2.1.4The function \(f(x)\) with the same secant line, zoomed in further.
Figure2.1.5The function \(f(x)\) with its tangent line at \(t=2\text{.}\)

As \(h\to 0\text{,}\) these secant lines approach the tangent line, a line that goes through the point \((2,f(2))\) with the special slope of \(-64\text{.}\) In Figure 2.1.4 and Figure 2.1.5, we zoom in around the point \((2,86)\text{.}\) We see the secant line, which approximates \(f\) well, but not as well the tangent line shown in Figure 2.1.5.

We have just introduced a number of important concepts that we will flesh out more within this section. First, we formally define two of them.

Definition2.1.6Derivative at a Point

Let \(f\) be a continuous function on an open interval \(I\) and let \(c\) be in \(I\text{.}\) The derivative of \(f\) at \(c\), denoted \(\fp(c)\text{,}\) is \begin{equation*} \lim_{h\to 0}\frac{f(c+h)-f(c)}{h}, \end{equation*} provided the limit exists. If the limit exists, we say that \(f\) is differentiable at \(c\); if the limit does not exist, then \(f\) is not differentiable at \(c\). If \(f\) is differentiable at every point in \(I\text{,}\) then \(f\) is differentiable on \(I\).

Definition2.1.7Tangent Line

Let \(f\) be continuous on an open interval \(I\) and differentiable at \(c\text{,}\) for some \(c\) in \(I\text{.}\) The line with equation \(\ell(x) = \fp(c)(x-c)+f(c)\) is the tangent line to the graph of \(f\) at \(c\text{;}\) that is, it is the line through \((c,f(c))\) whose slope is the derivative of \(f\) at \(c\text{.}\)

Some examples will help us understand these definitions.

Example2.1.8Finding derivatives and tangent lines

Let \(f(x) = 3x^2+5x-7\text{.}\) Find:

  1. \(\fp(1)\)

  2. The equation of the tangent line to the graph of \(f\) at \(x=1\text{.}\)

  3. \(\fp(3)\)

  4. The equation of the tangent line to the graph \(f\) at \(x=3\text{.}\)

Solution

Another important line that can be created using information from the derivative is the normal line. It is perpendicular to the tangent line, hence its slope is the opposite–reciprocal of the tangent line's slope.

Definition2.1.10Normal Line

Let \(f\) be continuous on an open interval \(I\) and differentiable at \(c\text{,}\) for some \(c\) in \(I\text{.}\) The normal line to the graph of \(f\) at \(c\) is the line with equation \begin{equation*} n(x) =\frac{-1}{\fp(c)}(x-c)+f(c), \end{equation*} when \(\fp(c)\neq 0\text{.}\) (When \(\fp(c)=0\text{,}\) the normal line is the vertical line through \(\left(c,f(c)\right)\text{;}\) that is, \(x=c\text{.}\))

Example2.1.11Finding equations of normal lines

Let \(f(x) = 3x^2+5x-7\text{,}\) as in Example 2.1.8. Find the equations of the normal lines to the graph of \(f\) at \(x=1\) and \(x=3\text{.}\)

Solution

Linear functions are easy to work with; many functions that arise in the course of solving real problems are not easy to work with. A common practice in mathematical problem solving is to approximate difficult functions with not-so-difficult functions. Lines are a common choice. It turns out that at any given point on the graph of a differentiable function \(f\text{,}\) the best linear approximation to \(f\) is its tangent line. That is one reason we'll spend considerable time finding tangent lines to functions.

One type of function that does not benefit from a tangent-line approximation is a line; it is rather simple to recognize that the tangent line to a line is the line itself. We look at this in the following example.

Example2.1.13Finding the Derivative of a Line

Consider \(f(x) = 3x+5\text{.}\) Find the equation of the tangent line to \(f\) at \(x=1\) and \(x=7\text{.}\)

Solution

We often desire to find the tangent line to the graph of a function without knowing the actual derivative of the function. While we will eventually be able to find derivatives of many common functions, the algebra and limit calculations on some functions are complex. Until we develop futher techniques, the best we may be able to do is approximate the tangent line. We demonstrate this in the next example.

Example2.1.14Numerical Approximation of the Tangent Line

Approximate the equation of the tangent line to the graph of \(f(x)=\sin(x)\) at \(x=0\text{.}\)

Solution

Recall from Section 1.3 that \(\lim\limits_{x\to 0}\frac{\sin(x)}x =1\text{,}\) meaning for values of \(x\) near 0, \(\sin(x) \approx x\text{.}\) Since the slope of the line \(y=x\) is \(1\) at \(x=0\text{,}\) it should seem reasonable that “the slope of \(f(x)=\sin(x)\)” is near \(1\) at \(x=0\text{.}\) In fact, since we approximated the value of the slope to be \(0.9983\text{,}\) we might guess the actual value is 1. We'll come back to this later.

Consider again Example 2.1.8. To find the derivative of \(f\) at \(x=1\text{,}\) we needed to evaluate a limit. To find the derivative of \(f\) at \(x=3\text{,}\) we needed to again evaluate a limit. We have this process: \begin{equation*} \begin{array}{c}\text{input specific}\\\text{number }c\end{array}\longrightarrow\begin{array}{|c|}\hline\text{do something}\\\text{to }f\text{ and }c\\\hline\end{array}\longrightarrow\begin{array}{c}\text{return}\\\text{number }f'(c)\end{array} \end{equation*}

This process describes a function; given one input (the value of \(c\)), we return exactly one output (the value of \(\fp(c)\)). The “do something” box is where the tedious work (taking limits) of this function occurs.

Instead of applying this function repeatedly for different values of \(c\text{,}\) let us apply it just once to the variable \(x\text{.}\) We then take a limit just once. The process now looks like: \begin{equation*} \begin{array}{c}\text{input}\\\text{variable }x\end{array}\longrightarrow\begin{array}{|c|}\hline\text{do something}\\\text{to }f\text{ and }x\\\hline\end{array}\longrightarrow\begin{array}{c}\text{return}\\\text{function }f'(x)\end{array} \end{equation*}

The output is the derivative function, \(\fp(x)\text{.}\) The \(\fp(x)\) function will take a number \(c\) as input and return the derivative of \(f\) at \(c\text{.}\) This calls for a definition.

Definition2.1.16Derivative Function

Let \(f\) be a differentiable function on an open interval \(I\text{.}\) The function \begin{equation*} \fp(x) = \lim_{h\to 0} \frac{f(x+h)-f(x)}{h} \end{equation*} is the derivative of \(f\).

Let \(y = f(x)\text{.}\) The following notations all represent the derivative: \begin{equation*} \fp(x)\ =\ y'\ =\ \frac{dy}{dx}\ =\ \frac{df}{dx}\ =\ \frac{d}{dx}(f)\ =\ \frac{d}{dx}(y). \end{equation*}

Important: The notation \(\frac{dy}{dx}\) is one symbol; it is not the fraction “\(dy/dx\)”. The notation, while somewhat confusing at first, was chosen with care. A fraction-looking symbol was chosen because the derivative has many fraction-like properties. Among other places, we see these properties at work when we talk about the units of the derivative, when we discuss the Chain Rule, and when we learn about integration (topics that appear in later sections and chapters).

Examples will help us understand this definition.

Example2.1.17Finding the derivative of a function

Let \(f(x) = 3x^2+5x-7\) as in Example 2.1.8. Find \(\fp(x)\text{.}\)

Solution
Example2.1.18Finding the derivative of a function

Let \(f(x) = \frac{1}{x+1}\text{.}\) Find \(\fp(x)\text{.}\)

Solution
Example2.1.19Finding the derivative of a function

Find the derivative of \(f(x) = \sin(x)\text{.}\)

Solution
Example2.1.20Finding the derivative of a piecewise defined function

Find the derivative of the absolute value function, \begin{equation*} f(x) = \abs{x} = \begin{cases} -x \amp x\lt 0 \\ x \amp x\geq 0\end{cases}\text{.} \end{equation*}

See Figure 2.1.21.

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Figure2.1.21The absolute value function, \(f(x) = \abs{x}\text{.}\) Notice how the slope of the lines (and hence the tangent lines) abruptly changes at \(x=0\text{.}\)
Solution

The point of non-differentiability came where the piecewise defined function switched from one piece to the other. Our next example shows that this does not always cause trouble.

Example2.1.23Finding the derivative of a piecewise defined function

Find the derivative of \(f(x)\text{,}\) where \(f(x) = \begin{cases}\sin(x) \amp x\leq \pi/2 \\ 1 \amp x\gt\pi/2\end{cases}\text{.}\) See Figure 2.1.24.

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Figure2.1.24A graph of \(f(x)\) as defined in Example 2.1.23.
Solution

Recall we pseudo-defined a continuous function as one in which we could sketch its graph without lifting our pencil. We can give a pseudo-definition for differentiability as well: it is a continuous function that does not have any “sharp corners” or a vertical tangent line. One such sharp corner is shown in Figure 2.1.21. Even though the function \(f\) in Example 2.1.23 is piecewise-defined, the transition is “smooth” hence it is differentiable. Note how in the graph of \(f\) in Figure 2.1.24 it is difficult to tell when \(f\) switches from one piece to the other; there is no “corner.”

This section defined the derivative; in some sense, it answers the question of “What is the derivative?” Section 2.2 addresses the question “What does the derivative mean?”

Subsection2.1.1Exercises

In the following exercises, use the definition of the derivative to compute the derivative of the given function.

In the following exercises, a function and an \(x\)-value \(c\) are given. (Note: these functions are the same as those given in Exercises 2.1.1.6 through 2.1.1.12.) Find the tangent line to the graph of the function at \(c\text{,}\) and find the normal line to the graph of the function at \(c\text{.}\)

In the following exercises, a function \(f\) and an \(x\)-value \(a\) are given. Approximate the equation of the tangent line to the graph of \(f\) at \(x=a\) by numerically approximating \(\fp(a)\text{,}\) using \(h=0.1\text{.}\)

In the following exercises, a graph of a function \(f(x)\) is given. Using the graph, sketch \(\fp(x)\text{.}\)

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