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Section 1.2 The derivative

The instantaneous rate of change of a function is an idea that sits at the foundation of calculus. It is a generalization of the notion of instantaneous velocity and measures how fast a particular function is changing at a given point. If the original function represents the position of a moving object, this instantaneous rate of change is precisely the velocity of the object. In other contexts, instantaneous rate of change could measure the number of cells added to a bacteria culture per day, the number of additional gallons of gasoline consumed by increasing a car's velocity one mile per hour, or the number of dollars added to a mortgage payment for each percentage point increase in interest rate. The instantaneous rate of change can also be interpreted geometrically on the function's graph, and this connection is fundamental to many of the main ideas in calculus.

Recall that for a moving object with position function \(s\text{,}\) its average velocity on the time interval \(t = a\) to \(t = a+h\) is given by the quotient

\begin{equation*} AV_{[a,a+h]} = \frac{s(a+h)-s(a)}{(a+h)-a} =\frac{s(a+h)-s(a)}{h}\text{.} \end{equation*}

In a similar way, we make the following definition for an arbitrary function \(y = f(x)\text{.}\)

Definition 1.2.1.

For a function \(f\text{,}\) the average rate of change of \(f\) on the interval \([a,a+h]\) is given by the value

\begin{equation*} AV_{[a,a+h]} = \frac{f(a+h)-f(a)}{h}\text{.} \end{equation*}

Equivalently, if we want to consider the average rate of change of \(f\) on \([a,b]\text{,}\) we compute

\begin{equation*} AV_{[a,b]} = \frac{f(b)-f(a)}{b-a}\text{.} \end{equation*}

It is essential that you understand how the average rate of change of \(f\) on an interval is connected to its graph.

Preview Activity 1.2.1.

Suppose that \(f\) is the function given by the graph below and that \(a\) and \(a+h\) are the input values as labeled on the \(x\)-axis. Use the graph in Figure 1.2.2 to answer the following questions.

Figure 1.2.2. Plot of \(y = f(x)\) for Preview Activity 1.2.1.

  1. Locate and label the points \((a,f(a))\) and \((a+h, f(a+h))\) on the graph.

  2. Construct a right triangle whose hypotenuse is the line segment from \((a,f(a))\) to \((a+h,f(a+h))\text{.}\) What are the lengths of the respective legs of this triangle?

  3. What is the slope of the line that connects the points \((a,f(a))\) and \((a+h, f(a+h))\text{?}\)

  4. Write a meaningful sentence that explains how the average rate of change of the function on a given interval and the slope of a related line are connected.

Subsection 1.2.1 The Derivative of a Function at a Point

Just as we defined instantaneous velocity in terms of average velocity, we now define the instantaneous rate of change of a function at a point in terms of the average rate of change of the function \(f\) over related intervals. This instantaneous rate of change of \(f\) at \(a\) is called “the derivative of \(f\) at \(a\text{,}\)” and is denoted by \(f'(a)\text{.}\)

Definition 1.2.3.

Let \(f\) be a function and \(x = a\) a value in the function's domain. We define the derivative of \(f\) with respect to \(x\) evaluated at \(x = a\), denoted \(f'(a)\text{,}\) by the formula

\begin{equation*} f'(a) = \lim_{h \to 0} \frac{f(a+h)-f(a)}{h}\text{,} \end{equation*}

provided this limit exists.

Aloud, we read the symbol \(f'(a)\) as either “\(f\)-prime at \(a\)” or “the derivative of \(f\) evaluated at \(x = a\text{.}\)” Much of the next several chapters will be devoted to understanding, computing, applying, and interpreting derivatives. For now, we observe the following important things.

Note 1.2.4.
  • The derivative of \(f\) at the value \(x = a\) is defined as the limit of the average rate of change of \(f\) on the interval \([a,a+h]\) as \(h \to 0\text{.}\) This limit may not exist, so not every function has a derivative at every point.
  • We say that a function is differentiable at \(x = a\) if it has a derivative at \(x = a\text{.}\)
  • The derivative is a generalization of the instantaneous velocity of a position function: if \(y = s(t)\) is a position function of a moving body, \(s'(a)\) tells us the instantaneous velocity of the body at time \(t=a\text{.}\)
  • Because the units on \(\frac{f(a+h)-f(a)}{h}\) are “units of \(f(x)\) per unit of \(x\text{,}\)” the derivative has these very same units. For instance, if \(s\) measures position in feet and \(t\) measures time in seconds, the units on \(s'(a)\) are feet per second.
  • Because the quantity \(\frac{f(a+h)-f(a)}{h}\) represents the slope of the line through \((a,f(a))\) and \((a+h, f(a+h))\text{,}\) when we compute the derivative we are taking the limit of a collection of slopes of lines. Thus, the derivative itself represents the slope of a particularly important line.

We first consider the derivative at a given value as the slope of a certain line.

When we compute an instantaneous rate of change, we allow the interval \([a,a+h]\) to shrink as \(h \to 0\text{.}\) We can think of one endpoint of the interval as “sliding towards” the other. In particular, provided that \(f\) has a derivative at \((a,f(a))\text{,}\) the point \((a+h,f(a+h))\) will approach \((a,f(a))\) as \(h \to 0\text{.}\) Because the process of taking a limit is a dynamic one, it can be helpful to use computing technology to visualize it. One option is a java applet in which the user is able to control the point that is moving. For a helpful collection of examples, consider the work of David Austin of Grand Valley State University, and this particularly relevant example. For applets that have been built in Geogebra 1 , see Marc Renault's library via Shippensburg University, with this example being especially fitting for our work in this section.

You can even consider building your own examples; the fantastic program Geogebra is available for free download and is easy to learn and use.

Figure 1.2.5 shows a sequence of figures with several different lines through the points \((a, f(a))\) and \((a+h,f(a+h))\text{,}\) generated by different values of \(h\text{.}\) These lines (shown in the first three figures in magenta), are often called secant lines to the curve \(y = f(x)\text{.}\) A secant line to a curve is simply a line that passes through two points on the curve. For each such line, the slope of the secant line is \(m = \frac{f(a+h) - f(a)}{h}\text{,}\) where the value of \(h\) depends on the location of the point we choose. We can see in the diagram how, as \(h \to 0\text{,}\) the secant lines start to approach a single line that passes through the point \((a,f(a))\text{.}\) If the limit of the slopes of the secant lines exists, we say that the resulting value is the slope of the tangent line to the curve. This tangent line (shown in the right-most figure in green) to the graph of \(y = f(x)\) at the point \((a,f(a))\) has slope \(m = f'(a)\text{.}\)

Figure 1.2.5. A sequence of secant lines approaching the tangent line to \(f\) at \((a,f(a))\text{.}\)

If the tangent line at \(x = a\) exists, the graph of \(f\) looks like a straight line when viewed up close at \((a,f(a))\text{.}\) In Figure 1.2.6 we combine the four graphs in Figure 1.2.5 into the single one on the left, and zoom in on the box centered at \((a,f(a))\) on the right. Note how the tangent line sits relative to the curve \(y = f(x)\) at \((a,f(a))\) and how closely it resembles the curve near \(x = a\text{.}\)

Figure 1.2.6. A sequence of secant lines approaching the tangent line to \(f\) at \((a,f(a))\text{.}\) At right, we zoom in on the point \((a,f(a))\text{.}\) The slope of the tangent line (in green) to \(f\) at \((a,f(a))\) is given by \(f'(a)\text{.}\)
Note 1.2.7.

The instantaneous rate of change of \(f\) with respect to \(x\) at \(x = a\text{,}\) \(f'(a)\text{,}\) also measures the slope of the tangent line to the curve \(y = f(x)\) at \((a,f(a))\text{.}\)

The following example demonstrates several key ideas involving the derivative of a function.

For the function \(f(x) = x - x^2\text{,}\) use the limit definition of the derivative to compute \(f'(2)\text{.}\) In addition, discuss the meaning of this value and draw a labeled graph that supports your explanation.

Solution

From the limit definition, we know that

\begin{equation*} f'(2) = \lim_{h \to 0} \frac{f(2+h)-f(2)}{h}\text{.} \end{equation*}

Now we use the rule for \(f\text{,}\) and observe that \(f(2) = 2 - 2^2 = -2\) and \(f(2+h) = (2+h) - (2+h)^2\text{.}\) Substituting these values into the limit definition, we have that

\begin{equation*} f'(2) = \lim_{h \to 0} \frac{(2+h) - (2+h)^2 - (-2)}{h}\text{.} \end{equation*}

In order to let \(h \to 0\text{,}\) we must simplify the quotient. Expanding and distributing in the numerator,

\begin{equation*} f'(2) = \lim_{h \to 0} \frac{2+h - 4 - 4h - h^2 + 2}{h}\text{.} \end{equation*}

Combining like terms, we have

\begin{equation*} f'(2) = \lim_{h \to 0} \frac{ -3h - h^2}{h}\text{.} \end{equation*}

Next, we remove a common factor of \(h\) in both the numerator and denominator and find that

\begin{equation*} f'(2) = \lim_{h \to 0} (-3-h)\text{.} \end{equation*}

Finally, we are able to take the limit as \(h \to 0\text{,}\) and thus conclude that \(f'(2) = -3\text{.}\) We note that \(f'(2)\) is the instantaneous rate of change of \(f\) at the point \((2,-2)\text{.}\) It is also the slope of the tangent line to the graph of \(y = x - x^2\) at the point \((2,-2)\text{.}\) Figure 1.2.9 shows both the function and the line through \((2,-2)\) with slope \(m = f'(2) = -3\text{.}\)

Figure 1.2.9. The tangent line to \(y = x - x^2\) at the point \((2,-2)\text{.}\)
Activity 1.2.2.

Consider the function \(f\) whose formula is \(\displaystyle f(x) = 3 - 2x\text{.}\)

  1. What familiar type of function is \(f\text{?}\) What can you say about the slope of \(f\) at every value of \(x\text{?}\)
  2. Compute the average rate of change of \(f\) on the intervals \([1,4]\text{,}\) \([3,7]\text{,}\) and \([5,5+h]\text{;}\) simplify each result as much as possible. What do you notice about these quantities?
  3. Use the limit definition of the derivative to compute the exact instantaneous rate of change of \(f\) with respect to \(x\) at the value \(a = 1\text{.}\) That is, compute \(f'(1)\) using the limit definition. Show your work. Is your result surprising?
  4. Without doing any additional computations, what are the values of \(f'(2)\text{,}\) \(f'(\pi)\text{,}\) and \(f'(-\sqrt{2})\text{?}\) Why?
Hint
  1. If \(f(x) = 3x^2 + 2x - 4\text{,}\) we say “\(f\) is quadratic.” If \(f(x) = 5 e^{2x-1}\text{,}\) we say “\(f\) is exponential.” What do we say about \(f(x) = 3-2x\text{?}\)
  2. Remember that to compute the average rate of change of \(f\) on \([a,b]\text{,}\) we calculate \(\frac{f(b)-f(a)}{b-a}\text{.}\)
  3. Observe that \(f(1+h) = 3 - 2(1+h) = 3 - 2 - 2h = 1 - 2h\text{.}\)
  4. Think about the how the graph of \(f\) appears. What is the same at every point?
Answer
  1. \(f\) is linear.
  2. \(f'(1)=-2\text{.}\)
  3. The average rate of change on \([1,4]\text{,}\) \([3,7]\text{,}\) and \([5,5+h]\) is \(-2\text{.}\)
  4. \(f'(2)=-2\text{,}\) \(f'(\pi)=-2\text{,}\) and \(f'(-\sqrt{2})=-2\text{,}\) since the slope of a linear function is the same at every point.
Solution
  1. Because \(f(x) = 3 - 2x\) is of the form \(f(x) = mx + b\text{,}\) we call \(f\) a linear function.
  2. The average rate of change on \([1,4]\) is \(\frac{f(4)-f(1)}{4-1} = \frac{-5 - 1}{3} = -2\text{.}\) Similar calculations show the average rate of change on \([3,7]\) is also \(-2\text{.}\) On \([5,5+h]\text{,}\) observe that
    \begin{align*} \frac{f(5+h)-f(5)}{h} \amp = \frac{3-2(5+h) - (3-10)}{h}\\ \amp = \frac{3 - 10 - 2h + 7}{h}\\ \amp = \frac{-2h}{h}\\ \amp = -2\text{.} \end{align*}
  3. Using the limit definition of the derivative, we find that
    \begin{align*} f'(1) = \amp \lim_{h \to 0} \frac{f(1+h) - f(1)}{h}\\ = \amp \lim_{h \to 0} \frac{(3 - 2(1+h)) - (3-2)}{h}\\ = \amp \lim_{h \to 0} \frac{3 - 2 - 2h - 1}{h}\\ = \amp \lim_{h \to 0} \frac{-2h}{h}\\ = \amp \lim_{h \to 0} -2\\ = \amp -2\text{.} \end{align*}
Activity 1.2.3.

A water balloon is tossed vertically in the air from a window. The balloon's height in feet at time \(t\) in seconds after being launched is given by \(s(t) = -16t^2 + 16t + 32\text{.}\) Use this function to respond to each of the following questions.

  1. Sketch an accurate, labeled graph of \(s\) on the axes provided in Figure 1.2.10. You should be able to do this without using computing technology.

    Figure 1.2.10. Axes for plotting \(y = s(t)\) in Activity 1.2.3.
  2. Compute the average rate of change of \(s\) on the time interval \([1,2]\text{.}\) Include units on your answer and write one sentence to explain the meaning of the value you found.
  3. Use the limit definition to compute the instantaneous rate of change of \(s\) with respect to time, \(t\text{,}\) at the instant \(a = 1\text{.}\) Show your work using proper notation, include units on your answer, and write one sentence to explain the meaning of the value you found.
  4. On your graph in (a), sketch two lines: one whose slope represents the average rate of change of \(s\) on \([1,2]\text{,}\) the other whose slope represents the instantaneous rate of change of \(s\) at the instant \(a=1\text{.}\) Label each line clearly.
  5. For what values of \(a\) do you expect \(s'(a)\) to be positive? Why? Answer the same questions when “positive” is replaced by “negative” and “zero.”
Hint
  1. Observe that \((t^2 - t - 2) = (t-2)(t+1)\) and that \(s(t)\) has its vertex at \(t = \frac{1}{2}\text{.}\)
  2. Recall the formula for average rate of change.
  3. Note that \(s(1+h) = -16(1+h)^2 + 16(1+h) + 32\text{.}\)
  4. Think about a secant line and a tangent line.
  5. A line with positive slope is one that is rising; a line with negative slope is one that is falling.
Answer
  1. The vertex is \((\frac{1}{2},36)\text{.}\)
  2. \(\frac{s(2)-s(1)}{2-1} = -32\) feet per second.
  3. \(s'(1) = -16\text{.}\)
  5. \(s'(a)\) is positive whenever \(0 \le a \lt \frac{1}{2}\text{;}\) \(s'(a)\) to be negative whenever \(\frac{1}{2} \lt a \lt 2\text{;}\) \(s'(\frac{1}{2}) = 0\text{.}\)
Solution
  1. Since \(s(t) = -16t^2 + 16t + 32 = -16(t^2 - t - 2) = -16(t-2)(t+1)\text{,}\) \(s\) has \(t\)-intercepts at \((2,0)\) and \((-1,0)\text{;}\) the \(s\)-intercept is clearly \((0,32)\text{;}\) and the vertex is \((\frac{1}{2},36)\text{.}\) See more in part (d).
  2. Observe that \(\frac{s(2)-s(1)}{2-1} = \frac{0 - 32}{1} = -32\) feet per second. This value represents the average rate at which the ball is falling over the time interval from \(t = 1\) to \(t = 2\text{.}\)
  3. We compute \(s'(1)\) as follows:
    \begin{align*} s'(1) = \amp \lim_{h \to 0} \frac{s(1+h)-s(1)}{h}\\ = \amp \lim_{h \to 0} \frac{(-16(1+h)^2 + 16(1+h) + 32) - (-16(1)^2 + 16(1) + 32)}{h}\\ = \amp \lim_{h \to 0} \frac{-16 - 32h - 16h^2 + 16 + 16h + 32 - 32}{h}\\ = \amp \lim_{h \to 0} \frac{-16h - 16h^2}{h}\\ = \amp \lim_{h \to 0} (-16-16h)\\ = \amp -16\text{.} \end{align*}

  4. We plot and label the secant line through \((1,s(1))\) and \((2,s(2))\text{,}\) as well as the tangent line through \((1,s(1))\) with slope \(s'(1)\text{.}\)

  5. Observe that whenever the ball is rising, its position function is rising, and thus the slope of its tangent line at any such point will be positive. This means that we should find \(s'(a)\) to be positive whenever \(0 \le a \lt \frac{1}{2}\text{,}\) and similarly \(s'(a)\) to be negative whenever \(\frac{1}{2} \lt a \lt 2\) (which is when the ball is falling). At the instant \(a = \frac{1}{2}\text{,}\) the ball is at its vertex and is neither rising nor falling, and at that point, \(s'(\frac{1}{2}) = 0\text{.}\)
Activity 1.2.4.

A rapidly growing city in Arizona has its population \(P\) at time \(t\text{,}\) where \(t\) is the number of decades after the year 2010, modeled by the formula \(P(t) = 25000 e^{t/5}\text{.}\) Use this function to respond to the following questions.

  1. Sketch an accurate graph of \(P\) for \(t = 0\) to \(t = 5\) on the axes provided in Figure 1.2.11. Label the scale on the axes carefully.

    Figure 1.2.11. Axes for plotting \(y = P(t)\) in Activity 1.2.4.
  2. Compute the average rate of change of \(P\) between 2030 and 2050. Include units on your answer and write one sentence to explain the meaning (in everyday language) of the value you found.
  3. Use the limit definition to write an expression for the instantaneous rate of change of \(P\) with respect to time, \(t\text{,}\) at the instant \(a = 2\text{.}\) Explain why this limit is difficult to evaluate exactly.
  4. Estimate the limit in (c) for the instantaneous rate of change of \(P\) at the instant \(a = 2\) by using several small \(h\) values. Once you have determined an accurate estimate of \(P'(2)\text{,}\) include units on your answer, and write one sentence (using everyday language) to explain the meaning of the value you found.
  5. On your graph above, sketch two lines: one whose slope represents the average rate of change of \(P\) on \([2,4]\text{,}\) the other whose slope represents the instantaneous rate of change of \(P\) at the instant \(a=2\text{.}\)
  6. In a carefully-worded sentence, describe the behavior of \(P'(a)\) as \(a\) increases in value. What does this reflect about the behavior of the given function \(P\text{?}\)
Hint
  1. \(P(t)\) is the standard exponential function, scaled by \(25000\text{.}\)
  2. Use the formula for the average rate of change of a function.
  3. Because of the exponential nature of \(P(t)\text{,}\) we're not able to simplify \(\frac{P(2+h)-P(2)}{h}\) in a way that removes \(h\) from the denominator.
  4. Try using \(h = 0.001, 0.0001, 0.00001\) and \(h = -0.001, -0.0001, -0.00001\text{.}\) Be careful not to round or use computing precision that is too limited.
  5. For the first line, think about the points \((2,P(2))\) and \((4,P(4))\text{.}\)
  6. Visualize the slope of the tangent line and how it changes as a point moves along the curve.
Answer
  2. \(AV_{[2,4]} \approx 9171\) people per decade is expected to be the average rate of change of the city's population over the two decades from 2030 to 2050.
  3. \begin{equation*} P'(2) = \lim_{h \to 0} \frac{P(2+h)-P(2)}{h} \approx 7458.5 \end{equation*}
    which is measured in people per decade.
  4. See the graph provided in (a) above. The magenta line has slope equal to the average rate of change of \(P\) on \([2,4]\text{,}\) while the green line is the tangent line at \((2,P(2))\) with slope \(P'(2)\text{.}\)
  5. It appears that the tangent line's slope at the point \((a,P(a))\) will increase as \(a\) increases.
Solution
  2. \(AV_{[2,4]} = \frac{P(4)-P(2)}{4-2} = \frac{25000e^{4/5} - 25000e^{2/5}}{2} \approx 9171\) people per decade is expected to be the average rate of change of the city's population over the two decades from 2030 to 2050.
  3. Note that
    \begin{align*} P'(2) = \amp \lim_{h \to 0} \frac{P(2+h)-P(2)}{h} = \lim_{h \to 0} \frac{25000 e^{(2+h)/5}-25000e^{2/5}}{h}\\ = \amp \lim_{h \to 0} \frac{25000 e^{2/5} e^{h/5} -25000e^{2/5}}{h} = \lim_{h \to 0} 25000e^{2/5}\left( \frac{e^{h/5} - 1}{h}\right) \end{align*}
    Because there is no way to remove a factor of \(h\) from the numerator, we cannot eliminate the \(h\) that is making the denominator go to zero, so it appears we need to be content estimating the limit with small values of \(h\text{.}\)
  4. Using \(h = 0.00001\text{,}\) we find \(\frac{P(2+0.00001)-P(2)}{0.00001} \approx 7457\text{;}\) using \(h = -0.00001\text{,}\) we find \(\frac{P(2-0.00001)-P(2)}{-0.00001} \approx 7460\text{.}\) Averaging these two results, we find that
    \begin{equation*} P'(2) = \lim_{h \to 0} \frac{P(2+h)-P(2)}{h} \approx 7458.5 \end{equation*}
    which is measured in people per decade.
  5. See the graph provided in (a) above. The magenta line has slope equal to the average rate of change of \(P\) on \([2,4]\text{,}\) while the green line is the tangent line at \((2,P(2))\) with slope \(P'(2)\text{.}\)
  6. If we consider the point where \(t = a\) and let \(a\) start at 0 and then increase, it appears that the tangent line's slope at the point \((a,P(a))\) will increase as \(a\) increases.

Subsection 1.2.2 How the derivative is itself a function

We now know that the instantaneous rate of change of a function \(f(x)\) at \(x = a\text{,}\) or equivalently the slope of the tangent line to the graph of \(y = f(x)\) at \(x = a\text{,}\) is given by the value \(f'(a)\text{.}\) In all of our examples so far, we have identified a particular value of \(a\) as our point of interest: \(a = 1\text{,}\) \(a = 3\text{,}\) etc. But it is not hard to imagine that we will often be interested in the derivative value for more than just one \(a\)-value, and possibly for many of them. In this section, we explore how we can move from computing the derivative at a single point to computing a formula for \(f'(a)\) at any point \(a\text{.}\) Indeed, the process of “taking the derivative” generates a new function, denoted by \(f'(x)\text{,}\) derived from the original function \(f(x)\text{.}\)

We observe that the particular value of \(a\) has very little effect on the process of computing the value of the derivative through the limit definition. To see this more clearly, we compute \(f'(a)\text{,}\) where \(a\) represents a number to be named later. Following the process of using the limit definition of the derivative,

\begin{align*} f'(a) =\mathstrut \amp \lim_{h \to 0} \frac{f(a + h) - f(a)}{h}\\ =\mathstrut \amp \lim_{h \to 0} \frac{4(a + h) - (a + h)^2 - (4a-a^2)}{h}\\ =\mathstrut \amp \lim_{h \to 0} \frac{4a + 4h - a^2 - 2ha - h^2 - 4a+a^2}{h}\\ =\mathstrut \amp \lim_{h \to 0} \frac{4h - 2ha - h^2}{h}\\ =\mathstrut \amp \lim_{h \to 0} \frac{h(4 - 2a - h)}{h}\\ =\mathstrut \amp \lim_{h \to 0} (4 - 2a - h)\text{.} \end{align*}

Here we observe that neither \(4\) nor \(2a\) depend on the value of \(h\text{,}\) so as \(h \to 0\text{,}\) \((4 - 2a - h) \to (4 - 2a)\text{.}\) Thus, \(f'(a) = 4 - 2a\text{.}\)

This result is consistent with the specific values we found above: e.g., \(f'(3) = 4 - 2(3) = -2\text{.}\) And indeed, our work confirms that the value of \(a\) has almost no bearing on the process of computing the derivative. We note further that the letter being used is immaterial: whether we call it \(a\text{,}\) \(x\text{,}\) or anything else, the derivative at a given value is simply given by “4 minus 2 times the value.” We choose to use \(x\) for consistency with the original function given by \(y = f(x)\text{,}\) as well as for the purpose of graphing the derivative function. For the function \(f(x) = 4x - x^2\text{,}\) it follows that \(f'(x) = 4 - 2x\text{.}\)

In Section 1.2 when we first defined the derivative, we wrote the definition in terms of a value \(a\) to find \(f'(a)\text{.}\) As we have seen above, the letter \(a\) is merely a placeholder, and it often makes more sense to use \(x\) instead. For the record, here we restate the definition of the derivative.

Definition 1.2.12.

Let \(f\) be a function and \(x\) a value in the function's domain. We define the derivative of \(f\), a new function called \(f'\text{,}\) by the formula

\begin{equation*} f'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}\text{,} \end{equation*}

provided this limit exists.

Subsection 1.2.3 Summary

  • The instantaneous rate of change with respect to \(x\) of a function \(f\) at a value \(x = a\) is denoted \(f'(a)\) (read “the derivative of \(f\) evaluated at \(a\)” or “\(f\)-prime at \(a\)”) and is defined by the formula
    \begin{equation*} f'(a) = \lim_{h \to 0} \frac{f(a+h)-f(a)}{h}\text{,} \end{equation*}
    provided the limit exists. Note particularly that the instantaneous rate of change at \(x = a\) is the limit of the average rate of change on \([a,a+h]\) as \(h \to 0\text{.}\)
  • Provided the derivative \(f'(a)\) exists, its value tells us the instantaneous rate of change of \(f\) with respect to \(x\) at \(x = a\text{,}\) which geometrically is the slope of the tangent line to the curve \(y = f(x)\) at the point \((a,f(a))\text{.}\) We even say that \(f'(a)\) is the “slope of the curve” at the point \((a,f(a))\text{.}\)

  • The limit definition of the derivative, \(f'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}\text{,}\) produces a value for each \(x\) at which the derivative is defined, and this leads to a new function \(y = f'(x)\text{.}\) It is especially important to note that taking the derivative is a process that starts with a given function (\(f\)) and produces a new, related function (\(f'\)).

Exercises 1.2.4 Exercises

1. The derivative function graphically.
2. Applying the limit definition of the derivative.
3. Sketching the derivative.
4. Comparing function and derivative values.
5. Limit definition of the derivative for a rational function.
6.

Let \(f\) be a function with the following properties: \(f\) is differentiable at every value of \(x\) (that is, \(f\) has a derivative at every point), \(f(-2) = 1\text{,}\) and \(f'(-2) = -2\text{,}\) \(f'(-1) = -1\text{,}\) \(f'(0) = 0\text{,}\) \(f'(1) = 1\text{,}\) and \(f'(2) = 2\text{.}\)

  1. On the axes provided at left in Figure 1.2.13, sketch a possible graph of \(y = f(x)\text{.}\) Explain why your graph meets the stated criteria.

  2. Conjecture a formula for the function \(y = f(x)\text{.}\) Use the limit definition of the derivative to determine the corresponding formula for \(y = f'(x)\text{.}\) Discuss both graphical and algebraic evidence for whether or not your conjecture is correct.

Figure 1.2.13. Axes for plotting \(y = f(x)\) in (a) and \(y = f'(x)\) in (b).
Answer

  1. See the figure below.

  2. See the figure below.

  3. One example of a formula for \(f\) is \(f(x) = \frac{1}{2}x^2 - 1\text{.}\)

Solution

  1. The fact that \(f\) is differentiable everywhere means that the graph of \(f\) is smooth everywhere. The slopes of the tangent lines to \(f\) are negative but increasing on the interval \((-\infty,0)\) and positive and increasing on the interval \((0,\infty)\text{,}\) with a slope of \(0\) when \(x=0\text{.}\) This is the kind of behavior that a quadratic function possesses, so we could guess that \(f\) has a graph something like that shown in the figure below.

  2. Because the change in the derivative values is constant, it looks like \(f'\) is linear with a slope of \(1\text{,}\) passing through the point \((0,0)\text{,}\) so it is reasonable to guess that \(f'(x) = x\text{.}\) A plot of \(f'\) is shown at right in the figure below.

  3. A natural guess is \(f(x) = x^2\text{;}\) since we need the function to pass through the point \((1,-2)\text{,}\) we might try \(f(x) = x^2 - 3\text{.}\) Using the limit definition, we have

    \begin{align*} f'(x) &= \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}\\ &= \lim_{h \to 0} \frac{\left[(x+h)^2-3\right] - \left[x^2-3\right]}{h}\\ &= \lim_{h \to 0} \frac{\left[x^2+2xh+h^2-3\right] - \left[x^2-3\right]}{h}\\ &= \lim_{h \to 0} \frac{2xh+h^2}{h}\\ &= \lim_{h \to 0} 2x+h\\ &= 2x\text{.} \end{align*}

    So this guess is close, but is off by a factor of \(2\text{,}\) since we want \(f'(x) = x\text{.}\) Instead, if we use \(f(x) = \frac{1}{2}x^2-1\) (note that we chose the ``\(-1\)'' so that \(f(-2)=1\text{,}\) then we have

    \begin{align*} f'(x) &= \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}\\ &= \lim_{h \to 0} \frac{\left[\frac{1}{2}(x+h)^2-1\right] - \left[\frac{1}{2}x^2-1\right]}{h}\\ &= \lim_{h \to 0} \frac{\frac{1}{2}\left[x^2+2xh+h^2\right] - \frac{1}{2}\left[x^2\right]}{h}\\ &= \lim_{h \to 0} \frac{1}{2}\frac{2xh+h^2}{h}\\ &= \lim_{h \to 0} x+\frac{h}{2}\\ &= x\text{.} \end{align*}

    This appears to be the correct function \(f\text{.}\)

7.

Consider the function \(g(x) = x^2 - x + 3\text{.}\)

  1. Use the limit definition of the derivative to determine a formula for \(g'(x)\text{.}\)

  2. Use a graphing utility to plot both \(y = g(x)\) and your result for \(y = g'(x)\text{;}\) does your formula for \(g'(x)\) generate the graph you expected?

  3. Use the limit definition of the derivative to find a formula for \(p'(x)\) where \(p(x) = 5x^2 - 4x + 12\text{.}\)

  4. Compare and contrast the formulas for \(g'(x)\) and \(p'(x)\) you have found. How do the constants 5, 4, 12, and 3 affect the results?

Answer

  1. \(g'(x) = 2x - 1\text{.}\)

  3. \(p'(x) = 10x - 4\text{.}\)

  4. The constants \(3\) and \(12\) don't seem to affect the results at all. The coefficient \(-4\) on the linear term in \(p(x)\) appears to make the ``\(-4\)'' appear in \(p'(x)= 10x - 4\text{.}\) The leading coefficient \(5\) in \((x) = 5x^2 - 4x + 12\) leads to the coefficient of ``\(10\)'' in \(p'(x) = 10x -4\text{.}\)

Solution

  1. By definition,

    \begin{align*} g'(x) &= \lim_{h \to 0} \frac{g(x+h)-g(x)}{h}\\ &= \lim_{h \to 0} \frac{(x+h)^2 - (x+h) + 3 - (x^2 - x + 3)}{h}\\ &= \lim_{h \to 0} \frac{x^2 + 2xh + h^2 - x-h + 3 - x^2 + x - 3}{h}\\ &= \lim_{h \to 0} \frac{2xh + h^2 - h}{h}\\ &= \lim_{h \to 0} \frac{h(2x + h - 1)}{h}\\ &= \lim_{h \to 0} (2x + h - 1)\\ &= 2x - 1\text{.} \end{align*}

  2. In the above figure, we see plots of both \(g\) and \(g'\text{.}\) We observe that the point \((0.5, 2.75)\) is the vertex of the quadratic function \(g\text{,}\) and at this point the slope of the tangent line to \(g(x)\) is zero. This aligns with the point \((0.5, 0)\) where \(y=g'(x)\) crosses the \(x\)-axis. In addition, we note that \(g'(x)\) is negative for \(x \lt 0.5\text{,}\) which corresponds to where \(g\) is decreasing and has tangent lines with slopes that are negative. Similarly, the values of \(g'\) are positive for \(x \gt 0.5\text{,}\) which align with the values of slopes we see on the original function \(g\text{.}\)

  3. By definition,

    \begin{align*} p'(x) &= \lim_{h \to 0} \frac{p(x+h)-p(x)}{h}\\ &= \lim_{h \to 0} \frac{5(x+h)^2 - 4(x+h) + 12 - (5x^2 - 4x + 12)}{h}\\ &= \lim_{h \to 0} \frac{5x^2 + 10xh + 5h^2 - 4x-4h + 12 - 5x^2 + 4x - 12}{h}\\ &= \lim_{h \to 0} \frac{10xh + 5h^2 - 4h}{h}\\ &= \lim_{h \to 0} \frac{h(10x + 5h - 4)}{h}\\ &= \lim_{h \to 0} (10x + 5h - 4)\\ &= 10x - 4\text{.} \end{align*}

  4. For \(g(x) = x^2 - x + 3\text{,}\) we found that \(g'(x) = 2x - 1\text{.}\) For \(p(x) = 5x^2 - 4x + 12\text{,}\) we determined that \(p'(x) = 10x - 4\text{.}\) The constants \(3\) and \(12\) don't seem to affect the results at all, and that makes sense because those numbers only serve to shift the graphs of \(g\) and \(p\) vertically, which does nothing to change the slope. The coefficient \(-4\) on the linear term in \(p(x)\) appears to make the ``\(-4\)'' appear in \(p'(x)= 10x - 4\text{.}\) That, too, makes sense in light of the fact that if we considered only the linear function \(L(x) = -4x\text{,}\) the slope would everywhere be \(-4\text{,}\) in contrast to the coefficent \(-1\) found on the linear term in \(g(x) = x^2 - x + 3\text{,}\) which leads to the constant \(-1\) in \(g'(x) = 2x - 1\text{.}\) Finally, the leading coefficient \(5\) in \((x) = 5x^2 - 4x + 12\) leads to the coefficient of ``\(10\)'' in \(p'(x) = 10x -4\text{.}\) This makes sense because if we considered only \(y = 5x^2\text{,}\) the \(5\) would make the graph \(5\) times as steep as the graph of \(y = x^2\text{,}\) and thus it affects the derivative proportionately.

8.

Let \(g\) be a continuous function (that is, one with no jumps or holes in the graph) and suppose that a graph of \(y= g'(x)\) is given by the graph on the right in Figure 1.2.14.

Figure 1.2.14. Axes for plotting \(y = g(x)\) and, at right, the graph of \(y = g'(x)\text{.}\)

  1. Observe that for every value of \(x\) that satisfies \(0 \lt x \lt 2\text{,}\) the value of \(g'(x)\) is constant. What does this tell you about the behavior of the graph of \(y = g(x)\) on this interval?

  2. On what intervals other than \(0 \lt x \lt 2\) do you expect \(y = g(x)\) to be a linear function? Why?

  3. At which values of \(x\) is \(g'(x)\) not defined? What behavior does this lead you to expect to see in the graph of \(y=g(x)\text{?}\)

  4. Suppose that \(g(0) = 1\text{.}\) On the axes provided at left in Figure 1.2.14, sketch an accurate graph of \(y = g(x)\text{.}\)

Answer

  1. \(g\) is linear.

  2. On \(-3.5 \lt x \lt -2\text{,}\) \(-2 \lt x \lt 0\) and \(2 \lt x \lt 3.5\text{.}\)

  3. At \(x = -2, 0, 2\text{;}\) \(g\) must have sharp corners at these points.

Solution

  1. Since \(g'(x)\) is constant (with value \(1\)) on the interval \(0 \lt x \lt 2\text{,}\) it follows that \(g\) is linear on that same interval, since \(g\) is increasing at a constant rate.

  2. On \(-3.5 \lt x \lt -2\text{,}\) we also expect \(g\) to be linear with slope \(1\text{,}\) while on \(-2 \lt x \lt 0\) and \(2 \lt x \lt 3.5\text{,}\) \(g\) will be linear with slope \(-1\text{;}\) in each case this is true because the value of \(g'(x)\) is constant with the noted value on the interval.

  3. From the given graph of \(g'(x)\text{,}\) we observe that \(g'\) is undefined at \(x = -2, 0, 2\text{.}\) Since we have been given that \(g\) is a continuous function, we can conclude that \(g\) must have sharp corners on its graph at these points. Moreover, that makes sense in light of our earlier observations that show \(g\) has constant slope on the intervals that connect at \(x = -2, 0, 2\) and the graph jumps from having slope \(1\) to \(-1\text{,}\) and then back, and so on.

9.

For each graph that provides an original function \(y = f(x)\) in Figure 1.2.15, your task is to sketch an approximate graph of its derivative function, \(y = f'(x)\text{,}\) on the axes immediately below. View the scale of the grid for the graph of \(f\) as being \(1 \times 1\text{,}\) and assume the horizontal scale of the grid for the graph of \(f'\) is identical to that for \(f\text{.}\) If you need to adjust the vertical scale on the axes for the graph of \(f'\text{,}\) you should label that accordingly.

Figure 1.2.15. Graphs of \(y = f(x)\) and grids for plotting the corresponding graph of \(y = f'(x)\text{.}\)
Answer
Solution

At any point where there is a jump in the graph of the derivative, the derivative is undefined. Normally we would draw an open circle at each end of the graph, but those are omitted here for convenience of plotting.

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