Activity 3.3 The First Derivative

We have seen two applications of expressions of the form \(\lim\limits_{h \to 0}\frac{\fe{f}{a+h} - \fe{f}{a}}{h}\text{.}\) It turns out that this expression is so important in mathematics, the sciences, economics, and many other fields that it deserves a name in its own right. We call the expression the first derivative of \(f\) at \(a\).

So far we’ve always fixed the value of \(a\) before making the calculation. There’s no reason why we couldn’t use a variable for \(a\text{,}\) make the calculation, and then replace the variable with specific values; in fact, it seems like this might be a better plan all around. This leads us to a definition of the first derivative function.

Definition 3.3.1 The First Derivative Function.

If \(f\) is a function of \(x\text{,}\) then we define the first derivative function, \(\fd{f}\text{,}\) as

\begin{equation*} \fe{\fd{f}}{x} = \lim_{h \to 0}\frac{\fe{f}{x+h} - \fe{f}{x}}{h}\text{.} \end{equation*}

The symbols \(\fe{\fd{f}}{x}\) are read aloud as “\(f\) prime of \(x\)” or “\(f\) prime at \(x\text{.}\)”

As we’ve already seen, \(\fe{\fd{f}}{a}\) gives us the slope of the tangent line to \(f\) at \(a\text{.}\)

We’ve also seen that if \(s\) is a position function, then \(\fe{\fd{s}}{a}\) gives us the instantaneous velocity at \(a\text{.}\) It’s not too much of a stretch to infer that the velocity function for \(s\) would be \(\fe{v}{t} = \fe{\fd{s}}{t}\text{.}\)

Let’s find a first derivative.

Example 3.3.2.

Let \(\fe{f}{x} = \frac{3}{2-x}\text{.}\) We can find \(\fe{\fd{f}}{x}\) as follows.

\begin{align*} \fe{\fd{f}}{x} \amp= \lim_{h \to 0}\frac{\fe{f}{x+h} - \fe{f}{x}}{h}\\ \amp= \lim_{h \to 0}\frac{\frac{3}{2 - (x+h)} - \frac{3}{2-x}}{h}\\ \amp= \lim_{h \to 0}\left(\frac{\frac{3}{2-x-h} - \frac{3}{2-x}}{h}\cdot\frac{(2-x-h)(2-x)}{(2-x-h)(2-x)}\right)\\ \amp= \lim_{h \to 0}\frac{3(2-x) - 3(2-x-h)}{h(2-x-h)(2-x)}\\ \amp= \lim_{h \to 0}\frac{6-3x-6+3x+3h}{h(2-x-h)(2-x)}\\ \amp= \lim_{h \to 0}\frac{3h}{h(2-x-h)(2-x)}\\ \amp= \lim_{h \to 0}\frac{3}{(2-x-h)(2-x)}\\ \amp= \frac{3}{(2-x-0)(2-x)}\\ \amp= \frac{3}{(2-x)^2} \end{align*}

the plot of a function curve with two segments; the first segment enters the graph from the left at just above the x-axis; it moves mostly to the right and slightly upward, but becomes steeper as it goes; it crosses the y-axis at 1.5, passes through (1,3) (which is emphasized with a solid dot) and bends upward so steeply that it exits the graph on its top just to the left of the vertical line x=2; a second segment enters the graph from its bottom, just to the right of the line x=2; at first it moves mostly upward and slightly to the right; as it continues, it moves more and more to the right and less and less upward; it passes through (5,-1), which is emphasized with a solid dot; it exits the graph on its right side, moving almost entirely rightward, exiting a little beneath the x-axis — Figure 3.3.3. \(y = \fe{f}{x} = \frac{3}{2-x}\)

Exercises Exercises

Exercise Group.

A graph of the function \(f\) defined by \(\fe{f}{x} = \frac{3}{2-x}\) is shown in Figure 3.3.3 and the formula for \(\fe{\fd{f}}{x}\) is derived in Example 3.3.2.

1.

Use the formula \(\fe{\fd{f}}{x} = \frac{3}{(2-x)^2}\) to calculate \(\fe{\fd{f}}{1}\) and \(\fe{\fd{f}}{5}\text{.}\)

2.

Draw onto Figure 3.3.3 a line through the point \(\point{1}{3}\) with a slope of \(\fe{\fd{f}}{1}\text{.}\) Also draw a line through the point \(\point{5}{-1}\) with a slope of \(\fe{\fd{f}}{5}\text{.}\) What are the names for the two lines you just drew? What are their equations?

3.

Showing work consistent with that shown in Example 3.3.2, find the formula for \(\fe{\fd{g}}{x}\) where \(\fe{g}{x} = \frac{5}{2x+1}\text{.}\)

Exercise Group.

Suppose that the elevation of an object (measured in ft) is given by \(\fe{s}{t} = -16t^2 + 112t + 5\) where \(t\) is the amount of time that has passed since the object was launched into the air (measured in s).

4.

Use the equations below to find the formula for the velocity function associated with this motion. The first two lines of your presentation should be an exact copy of the two lines below.

\begin{align*} \fe{v}{t} \amp= \fe{\fd{s}}{t}\\ \amp= \lim_{h \to 0}\frac{\fe{s}{t+h} - \fe{s}{t}}{h} \end{align*}

5.

Find the values of \(\fe{v}{2}\) and \(\fe{v}{5}\text{.}\) What is the unit on each of these values? What do the values tell you about the motion of the object? Don’t just say “the velocity”—describe what is actually happening to the object \(2\) seconds and \(5\) seconds into its travel.

6.

Use the velocity function to determine when the object reaches its maximum elevation. (Think about what must be true about the velocity at that instant.) Also, what is the common mathematical term for the point on the parabola \(y = \fe{s}{t}\) that occurs at that value of \(t\text{?}\)

7.

Use (3.3.1) to find the formula for \(\fe{\fd{v}}{t}\text{.}\) The first line of your presentation should be an exact copy of (3.3.1).

\begin{align} \fe{\fd{v}}{t} \amp= \lim_{h \to 0}\frac{\fe{v}{t+h} - \fe{v}{t}}{h}\tag{3.3.1} \end{align}

8.

What is the common name for the function \(\fe{\fd{v}}{t}\text{?}\) Is its formula consistent with what you know about objects in free fall on Earth?

9.

What is the constant slope of the function \(w\) defined by \(\fe{w}{x} = 12\text{?}\) Verify this by using Definition 3.3.1 to find the formula for \(\fe{\fd{w}}{x}\text{.}\)

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