Active Calculus · 1: Understanding the Derivative

1.3: The Derivative of a Function at a Point

Motivating Questions
  • How is the average rate of change of a function on a given interval defined, and what does this quantity measure?
  • How is the instantaneous rate of change of a function at a particular point defined? How is the instantaneous rate of change linked to average rate of change?
  • What is the derivative of a function at a given point? What does this derivative value measure? How do we interpret the derivative value graphically?
  • How are limits used formally in the computation of derivatives?

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,s\text{,} its average velocity on the time interval t=at = a to t=a+ht = a+h is given by the quotient

AV[a,a+h]=s(a+h)−s(a)h. AV_{[a,a+h]} = \frac{s(a+h)-s(a)}{h}\text{.}

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

Definition 1.3.11.3.1: Average Rate of Change

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

AV[a,a+h]=f(a+h)−f(a)h. AV_{[a,a+h]} = \frac{f(a+h)-f(a)}{h}\text{.}

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

AV[a,b]=f(b)−f(a)b−a. AV_{[a,b]} = \frac{f(b)-f(a)}{b-a}\text{.}

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

Preview Activity 1.3.11.3.1

Suppose that ff is the function given by the graph below and that aa and a+ha+h are the input values as labeled on the xx-axis. Use the graph in Figure 1.3.11.3.1 to answer the following questions.

1_3_PA1.svg

Figure 1.3.11.3.1. Plot of y=f(x)y = f(x) for Preview Activity 1.3.11.3.1.
  1. Locate and label the points (a,f(a))(a,f(a)) and (a+h,f(a+h))(a+h, f(a+h)) on the graph.
  2. Construct a right triangle whose hypotenuse is the line segment from (a,f(a))(a,f(a)) to (a+h,f(a+h)).(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))(a,f(a)) and (a+h,f(a+h))?(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.

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 ff over related intervals. This instantaneous rate of change of ff at aa is called “the derivative of ff at a,a\text{,}” and is denoted by f′(a).f'(a)\text{.}

Definition 1.3.21.3.2: Derivative

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

f′(a)=lim⁡h→0f(a+h)−f(a)h, f'(a) = \lim_{h \to 0} \frac{f(a+h)-f(a)}{h}\text{,}

provided this limit exists.

Aloud, we read the symbol f′(a)f'(a) as either “ff-prime at aa” or “the derivative of ff evaluated at x=a.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 1.3.11.3.1
  • The derivative of ff at the value x=ax = a is defined as the limit of the average rate of change of ff on the interval [a,a+h][a,a+h] as h→0.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=ax = a if it has a derivative at x=a.x = a\text{.}
  • The derivative is a generalization of the instantaneous velocity of a position function: if y=s(t)y = s(t) is a position function of a moving body, s′(a)s'(a) tells us the instantaneous velocity of the body at time t=a.t=a\text{.}
  • Because the units on f(a+h)−f(a)h\frac{f(a+h)-f(a)}{h} are “units of f(x)f(x) per unit of x,x\text{,}” the derivative has these very same units. For instance, if ss measures position in feet and tt measures time in seconds, the units on s′(a)s'(a) are feet per second.
  • Because the quantity f(a+h)−f(a)h\frac{f(a+h)-f(a)}{h} represents the slope of the line through (a,f(a))(a,f(a)) and (a+h,f(a+h)),(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][a,a+h] to shrink as h→0.h \to 0\text{.} We can think of one endpoint of the interval as “sliding towards” the other. In particular, provided that ff has a derivative at (a,f(a)),(a,f(a))\text{,} the point (a+h,f(a+h))(a+h,f(a+h)) will approach (a,f(a))(a,f(a)) as h→0.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 1.3.21.3.2 shows a sequence of figures with several different lines through the points (a,f(a))(a, f(a)) and (a+h,f(a+h)),(a+h,f(a+h))\text{,} generated by different values of h.h\text{.} These lines (shown in the first three figures in magenta), are often called secant lines to the curve y=f(x).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=f(a+h)−f(a)h,m = \frac{f(a+h) - f(a)}{h}\text{,} where the value of hh depends on the location of the point we choose. We can see in the diagram how, as h→0,h \to 0\text{,} the secant lines start to approach a single line that passes through the point (a,f(a)).(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)y = f(x) at the point (a,f(a))(a,f(a)) has slope m=f′(a).m = f'(a)\text{.}

1_3_SecToTanSeq.svg

Figure 1.3.21.3.2. A sequence of secant lines approaching the tangent line to ff at (a,f(a)).(a,f(a))\text{.}

If the tangent line at x=ax = a exists, the graph of ff looks like a straight line when viewed up close at (a,f(a)).(a,f(a))\text{.} In Figure 1.3.31.3.3 we combine the four graphs in Figure 1.3.21.3.2 into the single one on the left, and zoom in on the box centered at (a,f(a))(a,f(a)) on the right. Note how the tangent line sits relative to the curve y=f(x)y = f(x) at (a,f(a))(a,f(a)) and how closely it resembles the curve near x=a.x = a\text{.}

1_3_SecToTan.svg

Figure 1.3.31.3.3. A sequence of secant lines approaching the tangent line to ff at (a,f(a)).(a,f(a))\text{.} At right, we zoom in on the point (a,f(a)).(a,f(a))\text{.} The slope of the tangent line (in green) to ff at (a,f(a))(a,f(a)) is given by f′(a).f'(a)\text{.}
Note 1.3.21.3.2

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

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

Example 1.3.11.3.1

For the function f(x)=x−x2,f(x) = x - x^2\text{,} use the limit definition of the derivative to compute f′(2).f'(2)\text{.} In addition, discuss the meaning of this value and draw a labeled graph that supports your explanation.

Answer

From the limit definition, we know that

f′(2)=lim⁡h→0f(2+h)−f(2)h. f'(2) = \lim_{h \to 0} \frac{f(2+h)-f(2)}{h}\text{.}

Now we use the rule for f,f\text{,} and observe that f(2)=2−22=−2f(2) = 2 - 2^2 = -2 and f(2+h)=(2+h)−(2+h)2.f(2+h) = (2+h) - (2+h)^2\text{.} Substituting these values into the limit definition, we have that

f′(2)=lim⁡h→0(2+h)−(2+h)2−(−2)h. f'(2) = \lim_{h \to 0} \frac{(2+h) - (2+h)^2 - (-2)}{h}\text{.}

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

f′(2)=lim⁡h→02+h−4−4h−h2+2h. f'(2) = \lim_{h \to 0} \frac{2+h - 4 - 4h - h^2 + 2}{h}\text{.}

Combining like terms, we have

f′(2)=lim⁡h→0−3h−h2h. f'(2) = \lim_{h \to 0} \frac{ -3h - h^2}{h}\text{.}

Next, we remove a common factor of hh in both the numerator and denominator and find that

f′(2)=lim⁡h→0(−3−h). f'(2) = \lim_{h \to 0} (-3-h)\text{.}

Finally, we are able to take the limit as h→0,h \to 0\text{,} and thus conclude that f′(2)=−3.f'(2) = -3\text{.} We note that f′(2)f'(2) is the instantaneous rate of change of ff at the point (2,−2).(2,-2)\text{.} It is also the slope of the tangent line to the graph of y=x−x2y = x - x^2 at the point (2,−2).(2,-2)\text{.} Figure 1.3.41.3.4 shows both the function and the line through (2,−2)(2,-2) with slope m=f′(2)=−3.m = f'(2) = -3\text{.}

1_3_Ex1.svg
Figure 1.3.41.3.4. The tangent line to y=x−x2y=x - x^2 at the point (2,−2)(2,-2).

The following activities will help you explore a variety of key ideas related to derivatives.

Activity 1.3.21.3.2

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

  1. What familiar type of function is f?f\text{?} What can you say about the slope of ff at every value of x?x\text{?}
  2. Compute the average rate of change of ff on the intervals [1,4],[1,4]\text{,} [3,7],[3,7]\text{,} and [5,5+h];[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 ff with respect to xx at the value a=1.a = 1\text{.} That is, compute f′(1)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),f'(2)\text{,} f′(π),f'(\pi)\text{,} and f′(−2)?f'(-\sqrt{2})\text{?} Why?
Activity 1.3.31.3.3

A water balloon is tossed vertically in the air from a window. The balloon's height in feet at time tt in seconds after being launched is given by s(t)=−16t2+16t+32.s(t) = -16t^2 + 16t + 32\text{.} Use this function to respond to each of the following questions.

a. Sketch an accurate, labeled graph of ss on the axes provided in Figure 1.3.51.3.5. You should be able to do this without using computing technology.

1_3_Act2.svg

Figure 1.3.51.3.5. Axes for plotting y=s(t)y = s(t) in Activity 1.3.31.3.3.

b. Compute the average rate of change of ss on the time interval [1,2].[1,2]\text{.} Include units on your answer and write one sentence to explain the meaning of the value you found.

c. Use the limit definition to compute the instantaneous rate of change of ss with respect to time, t,t\text{,} at the instant a=1.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.

d. On your graph in (a), sketch two lines: one whose slope represents the average rate of change of ss on [1,2],[1,2]\text{,} the other whose slope represents the instantaneous rate of change of ss at the instant a=1.a=1\text{.} Label each line clearly.

e. For what values of aa do you expect s′(a)s'(a) to be positive? Why? Answer the same questions when “positive” is replaced by “negative” and “zero.”

Activity 1.3.41.3.4

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

a. Sketch an accurate graph of PP for t=0t = 0 to t=5t = 5 on the axes provided in Figure 1.3.61.3.6. Label the scale on the axes carefully.

1_3_Act3.svg

Figure 1.3.61.3.6. Axes for plotting y=P(t)y = P(t) in Activity 1.3.41.3.4.

b. Compute the average rate of change of PP 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.

c. Use the limit definition to write an expression for the instantaneous rate of change of PP with respect to time, t,t\text{,} at the instant a=2.a = 2\text{.} Explain why this limit is difficult to evaluate exactly.

d. Estimate the limit in (c) for the instantaneous rate of change of PP at the instant a=2a = 2 by using several small hh values. Once you have determined an accurate estimate of P′(2),P'(2)\text{,} include units on your answer, and write one sentence (using everyday language) to explain the meaning of the value you found.

e. On your graph above, sketch two lines: one whose slope represents the average rate of change of PP on [2,4],[2,4]\text{,} the other whose slope represents the instantaneous rate of change of PP at the instant a=2.a=2\text{.}

f. In a carefully-worded sentence, describe the behavior of P′(a)P'(a) as aa increases in value. What does this reflect about the behavior of the given function P?P\text{?}

Summary

  • The average rate of change of a function ff on the interval [a,b][a,b] is f(b)−f(a)b−a.\frac{f(b)-f(a)}{b-a}\text{.} The units on the average rate of change are units of f(x)f(x) per unit of x,x\text{,} and the numerical value of the average rate of change represents the slope of the secant line between the points (a,f(a))(a,f(a)) and (b,f(b))(b,f(b)) on the graph of y=f(x).y = f(x)\text{.} If we view the interval as being [a,a+h][a,a+h] instead of [a,b],[a,b]\text{,} the meaning is still the same, but the average rate of change is now computed by f(a+h)−f(a)h.\frac{f(a+h)-f(a)}{h}\text{.}
  • The instantaneous rate of change with respect to xx of a function ff at a value x=ax = a is denoted f′(a)f'(a) (read “the derivative of ff evaluated at aa” or “ff-prime at aa”) and is defined by the formula
    f′(a)=lim⁡h→0f(a+h)−f(a)h, f'(a) = \lim_{h \to 0} \frac{f(a+h)-f(a)}{h}\text{,}
    provided the limit exists. Note particularly that the instantaneous rate of change at x=ax = a is the limit of the average rate of change on [a,a+h][a,a+h] as h→0.h \to 0\text{.}
  • Provided the derivative f′(a)f'(a) exists, its value tells us the instantaneous rate of change of ff with respect to xx at x=a,x = a\text{,} which geometrically is the slope of the tangent line to the curve y=f(x)y = f(x) at the point (a,f(a)).(a,f(a))\text{.} We even say that f′(a)f'(a) is the “slope of the curve” at the point (a,f(a)).(a,f(a))\text{.}
  • Limits allow us to move from the rate of change over an interval to the rate of change at a single point.

Credit and licence

This page titled 1.3: The Derivative of a Function at a Point was last modified on Thu, 28 Sep 2023 14:38:16 GMT and is shared under a CC BY-SA 4.0 license and was authored, remixed, and/or curated by Matthew Boelkins, David Austin, and Steven Schlicker ( ScholarWorks @Grand Valley State University ) via source content that was edited to the style and standards of the LibreTexts platform.

Source: https://math.libretexts.org/Bookshelves/Calculus/Active_Calculus_(Boelkins_et_al.)/01%3A_Understanding_the_Derivative/1.03%3A_The_Derivative_of_a_Function_at_a_Point

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