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## Section1.4The 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 upon increasing a car's velocity by 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)}{h}\text{.} \end{equation*}

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

###### Average Rate of Change

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

\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.

###### Example1.49

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 Figure1.50 to answer the following questions. Figure1.50Plot of $y = f(x)$ for Example1.49.
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.

Hint
1. What are the coordinates of the two marked points on the graph of $f\text{?}$
2. Draw the triangle so that one of its legs is horizontal and the other is vertical.
3. Remember that the slope of a line segment can be thought of as rise over run.
4. How does the slope you found in (c) compare to the formula for the average rate of change of $f$ on the interval $[a,a+h]\text{?}$
1. The left marked point on the graph in Figure1.50 is the point $(a,f(a))\text{,}$ the right marked point on the graph is $(a+h,f(a+h))\text{.}$
2. The lengths of the legs of this triangle are $f(a+h)-f(a)$ and $h\text{.}$
3. $\frac{f(a+h)-f(a)}{h}$
4. The average rate of change of the function $f$ on the interval $[a,a+h]$ is the same as the slope of the line segment from $(a,f(a))$ to $(a+h,f(a+h))\text{.}$
Solution
1. The left marked point on the graph in Figure1.50 is the point $(a,f(a))\text{,}$ the right marked point on the graph is $(a+h,f(a+h))\text{.}$
2. The length of the vertical leg of this triangle is $f(a+h)-f(a)\text{,}$ and the length of the horizontal leg of this triangle is $a+h-a=h\text{.}$
3. The slope of the line connecting the two points is $\frac{f(a+h)-f(a)}{h}\text{,}$ as shown by the triangle drawn in the previous part: the rise is the length of the vertical leg, and the run is the length of the horizontal leg.
4. The average rate of change of the function $f$ on the interval $[a,a+h]$ is $AV_{[a,a+h]}=\frac{f(a+h)-f(a)}{h}\text{,}$ which is exactly what we found in the previous part as the slope of the line segment connecting the points $(a,f(a))$ and $(a+h,f(a+h))\text{.}$

### SubsectionThe 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 $f'(a)\text{.}$

###### Derivative at a Point

Let $f$ be a function and $x = a$ a value in the function's domain. The derivative of $f$ with respect to $x$ evaluated at $x = a$, denoted $f'(a)\text{,}$ is defined 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.

This is sometimes referred to as the limit definition of the derivative at a point.

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

###### Note1.51
• 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 depends on both the function $f$ and the point $x=a\text{.}$ Since this limit may not exist, 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 the position function of a moving object, $s'(a)$ tells us the instantaneous velocity of the object at time $t=a\text{.}$
• Because the units of $\frac{f(a+h)-f(a)}{h}$ are units of $f(x)$ per unit of $x\text{,}$ the derivative has these 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.
• 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 actually 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 at that value.

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.2One 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,3You can even consider building your own examples; the fantastic program Geogebra is available for free download and is easy to learn and use. see Marc Renault's library via Shippensburg University, with this example being especially fitting for our work in this section.

Figure1.52 below 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 $m = f'(a)\text{.}$ Figure1.52A 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 Figure1.53 below, we combine the four graphs from Figure1.52 into a single graph 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 $(a,f(a))\text{,}$ and how closely it resembles the curve near $x = a\text{.}$ Figure1.53A 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 $f'(a)\text{.}$
###### Note1.54

The instantaneous rate of change of $f$ with respect to $x$ at $x = a\text{,}$ which is the derivative value $f'(a)\text{,}$ 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.

###### Example1.55Using the limit definition of the derivative

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.

Hint

The definition of the derivative of $f$ with respect to $x$ evaluated at $x=a$ says

\begin{equation*} f'(a)=\lim_{h\to0}\frac{f(a+h)-f(a)}{h}\text{.} \end{equation*}

How does the value you find for $f'(2)$ manifest in the graph of $f(x)=x-x^2\text{,}$ particularly near the point $(2,f(2))\text{?}$

\begin{equation*} f'(2)=-3. \end{equation*}

The slope of the tangent line to $f$ through the point $(2,-2)$ is $-3\text{.}$

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 evaluate the limit, we must simplify the quotient. Expanding and distributing in the numerator gives us

\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, because $h\neq0$ within the limit, we may 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 $h$ approaches $0\text{,}$ 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 $(2,-2)\text{.}$ Figure1.56 below shows both the function and the line through $(2,-2)$ with slope $m = f'(2) = -3\text{.}$ Figure1.56The tangent line to $y = x - x^2$ at the point $(2,-2)\text{.}$

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

###### Example1.57

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?
1. $f$ is linear.
2. The average rate of change on $[1,4]\text{,}$ $[3,7]\text{,}$ and $[5,5+h]$ is $-2\text{.}$
3. $f'(1)=-2\text{.}$
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*}
###### Example1.58

A water balloon is tossed vertically in the air from a window. The balloon's height in feet $t$ 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 $y=s(t)$ for $t=0$ to $t=2\text{.}$ Label the scale on the axes carefully. You should be able to do this without using computing technology.

2. Compute the average rate of change of $s$ on the time interval $[1,2]\text{.}$ Include units in your answer and write one sentence to explain the meaning of the value you found.
3. Use the limit definition of the derivative 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 in 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.
1. The vertex is $(\frac{1}{2},36)\text{;}$ the full graph is below in (d).
2. $\frac{s(2)-s(1)}{2-1} = -32$ feet per second.
3. $s'(1) = -16\text{.}$
4. 5. $s'(a)$ is positive whenever $0 \le a \lt \frac{1}{2}\text{;}$ $s'(a)$ is negative whenever $\frac{1}{2} \lt a \lt 2\text{;}$ $s'(\frac{1}{2}) = 0\text{.}$
Solution
1. Since
\begin{align*} s(t) \amp= -16t^2 + 16t + 32 \\ \amp= -16(t^2 - t - 2) \\ \amp= -16(t-2)(t+1)\text{,} \end{align*}
$s$ 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 the full graph of $y=s(t)$ 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 balloon 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 balloon 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 balloon is falling). At the instant $a = \frac{1}{2}\text{,}$ the balloon is at its vertex and is neither rising nor falling, and at that point, $s'(\frac{1}{2}) = 0\text{.}$
###### Example1.59

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 $P(t) = 25000 e^{\frac{t}{5}}\text{.}$ Use this function to respond to the following questions.

1. Sketch an accurate graph of $y=P(t)$ for $t = 0$ to $t = 5\text{.}$ Label the scale on the axes carefully.

2. Compute the average rate of change of $P$ between 2030 and 2050. Include units in your answer and write one sentence to explain (in everyday language) the meaning of the value you found.
3. Use the limit definition of the derivative 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 value of the limit in (c) by using several small $h$ values. Once you have determined an accurate estimate of $P'(2)\text{,}$ include units in your answer, and write one sentence (using everyday language) to explain the meaning of the value you found.
5. On your graph from (a) 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 in the positive direction.
1. 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\to0}25000e^{2/5}\left(\frac{e^{h/5}-1}{h}\right)\text{.} \end{equation*}
4. $P'(2) \approx 7458.5$ 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. It appears that the tangent line's slope at the point $(a,P(a))$ will increase as $a$ increases.
Solution
1. 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)\text{.} \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\text{.} \end{equation*}
This means that at the rate the population is growing in 2030, the city's population would be expected to increase by about $7458.5$ people by 2040.
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.

### SubsectionSummary

• The average rate of change of a function $f$ on the interval $[a,b]$ is $\frac{f(b)-f(a)}{b-a}\text{.}$ The units on the average rate of change are units of $f(x)$ per unit of $x\text{,}$ and the numerical value of the average rate of change represents the slope of the secant line between the points $(a,f(a))$ and $(b,f(b))$ on the graph of $y = f(x)\text{.}$ If we view the interval as being $[a,a+h]$ instead of $[a,b]\text{,}$ the meaning is still the same, but the average rate of change is now computed by $\frac{f(a+h)-f(a)}{h}\text{.}$
• 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 of $a$) 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 this 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 $h$ approaches $0\text{.}$
• 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 $(a,f(a))\text{.}$
• Limits allow us to move from the rate of change over an interval to the rate of change at a single point.

### SubsectionExercises

Consider the graph of $y = f(x)$ provided in Figure1.60.

1. On the graph of $y = f(x)\text{,}$ sketch and label the following quantities:

• the secant line to $y = f(x)$ on the interval $[-3,-1]$ and the secant line to $y = f(x)$ on the interval $[0,2]\text{.}$
• the tangent line to $y = f(x)$ at $x = -3$ and the tangent line to $y = f(x)$ at $x = 0\text{.}$
2. What is the approximate value of the average rate of change of $f$ on $[-3,-1]\text{?}$ On $[0,2]\text{?}$ How are these values related to your work in (a)?
3. What is the approximate value of the instantaneous rate of change of $f$ at $x = -3\text{?}$ At $x = 0\text{?}$ How are these values related to your work in (a)? For each of the following prompts, sketch a graph on the provided axes in Figure1.61 of a function that has the stated properties. Figure1.61Axes for plotting $y = f(x)$ in (a) and $y = g(x)$ in (b).
1. $y = f(x)$ such that

• the average rate of change of $f$ on $[-3,0]$ is $-2$ and the average rate of change of $f$ on $[1,3]$ is 0.5, and
• the instantaneous rate of change of $f$ at $x = -1$ is $-1$ and the instantaneous rate of change of $f$ at $x = 2$ is 1.
2. $y = g(x)$ such that

• $\frac{g(3)-g(-2)}{5} = 0$ and $\frac{g(1)-g(-1)}{2} = -1\text{,}$ and
• $g'(2) = 1$ and $g'(-1) = 0$

Suppose that the population, $P\text{,}$ of China (in billions) can be approximated by the function $P(t) = 1.15(1.014)^t$ where $t$ is the number of years since the start of 1993.

1. According to the model, what was the total change in the population of China between January 1, 1993 and January 1, 2000? What will be the average rate of change of the population over this time period? Is this average rate of change greater or less than the instantaneous rate of change of the population on January 1, 2000? Explain and justify, being sure to include proper units on all your answers.
2. According to the model, what is the average rate of change of the population of China in the ten-year period starting on January 1, 2012?
3. Write an expression involving limits that, if evaluated, would give the exact instantaneous rate of change of the population on today's date. Then estimate the value of this limit (discuss how you chose to do so) and explain the meaning (including units) of the value you have found.
4. Find an equation for the tangent line to the function $y = P(t)$ at the point where the $t$-value is given by today's date.

The goal of this problem is to compute the value of the derivative at a point for several different functions, where for each one we do so in three different ways, and then to compare the results to see that each produces the same value.

For each of the following functions, use the limit definition of the derivative to compute the value of $f'(a)$ using three different approaches: strive to use the algebraic approach first (to compute the limit exactly), then test your result using numerical evidence (with small values of $h$), and finally plot the graph of $y = f(x)$ near $(a,f(a))$ along with the appropriate tangent line to estimate the value of $f'(a)$ visually. Compare your findings among all three approaches; if you are unable to complete the algebraic approach, still work numerically and graphically.

1. $f(x) = x^2 - 3x\text{,}$ $a = 2$
2. $f(x) = \frac{1}{x}\text{,}$ $a = 1$
3. $f(x) = \sqrt{x}\text{,}$ $a = 1$
4. $f(x) = 2 - |x-1|\text{,}$ $a = 1$
5. $f(x) = \sin(x)\text{,}$ $a = \frac{\pi}{2}$