Unit D: Differentiation

Foundational Objectives

• To demonstrate an understanding of secant and tangent lines. Supported by objectives 1, 2, 3, and 4.

• To recognize the slope of the tangent line as a measure of a rate of change. Supported by learning objectives 1, and 4 through 9.

• To demonstrate an understanding of differentiation. Supported by learning objectives 5 through 9.

• To demonstrate an understanding of implicit functions and the differentiation of implicit functions. Supported by learning objective 10.

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Objective

D.1 Back to Top

To interpret the slope of a line as a rate of change, and to find the equation of a line.

To read graphs for information in order to understand and analyze (NUM).

Instructional Notes

Students have met slope on many occasions but may not have interpreted slope as a rate of change.

You might begin by showing the students a graph of the population of two towns over a twelve-month period. See the example below.

Ask the students to draw some conclusions comparing the populations of the two towns based on the graph (NUM). The discussion should include questions such as which town experienced the greatest one-month growth, the least one-month growth, and when did these occur? What was the rate of growth of town Q between the 3 rd and 4 th months? Lead students, if they do not arrive there themselves, to see that the slope of each line segment gives the monthly rate of growth. You want students to recognize that no matter how the axes are labelled, the slope of a line gives the rate of change of the vertical units compared to the horizontal units.

Show students graphs of line segments similar to ones that follow and ask them to interpret the slope of each. What do the slope units represent?

Reacquaint students with the slope formula . Have students find slopes of line segments from data in a table giving the units of the slope with the correct interpretation. For example, in the table below if the x values represent the number of days spent holidaying and the y numbers represent the total amount of money spent on the holiday by the end of that day, then the slope of the line segment connecting the two points is: which would be the average daily rate of spending over that particular five day period.

(days travelled)

(amount spent)

$300

$1050

Give students the equation of a line such as and ask them to complete a table similar to the following. Give them only the x values and ask for help with the rest. Explain that and represent the change in x and the change in y as you move from any value to the one below.

Ask the students what patterns and conclusions can be made (NUM).

Have students determine how y would change if x increased by 10 units. How would x change if y decreased by 21 units?

Have the students consider how y changes if the slope of a line is positive and the value of x is increased. (Answer: y increases.) How does y change if the slope of a line is negative and the value of x is increased? ( Answer: y decreases.)

As students will frequently have to find equations of tangent lines, complete this objective by reviewing the point-slope formula and the slope-intercept formula. Be sure students can find equations of lines both parallel to and perpendicular to a given line.

Examples/Activities

What is the slope of each of the following lines?

(a) (b)

What is the rate of change of y compared to x along the line ?

For the linear function if:

(a) x increases by 3, how does y change?
(b) x increases by 21, how does y change?
(c) x decreases by 15, how does y change?

For the linear function if:

(a) x increases by 3, how does y change?
(b) x increases by 21, how does y change?
(c) x decreases by 15, how does y change?

Find the slope of the line passing through the points and .

Find the equation of the line passing through the point whose slope is .

The point lies on the line .

(a) Suppose a point Q whose x -coordinate is also lies on the line. What is the y -coordinate of Q ?

(b) As you move along the line from P to Q , what is and ?

(d) Does your answer in part (c) depend on the value of h ? In other words, is there an h in your answer to part (c)?

Find and interpret the slope of the line that passes through the pair of points in the table.

X                       y
(distance hiked) temperature
3 km
7 km

Find the equation of the line described. Write each answer in the form where a , b , and c are integers.

(a) The line passes through the point and has a slope of .
(b) The line contains the points and .
(c) The line passes through the point and is perpendicular to the line .
(d) The line has an x -intercept of 6 and is parallel to the line .

Water boils at or . Water freezes at or . The relationship between degrees Celsius and degrees Fahrenheit is linear. Write the function for the temperature in Fahrenheit with respect to the temperature in Celsius.
The landlord of an apartment block containing 40 units knows that all of them will be rented out if he charges $600 per month. For each $50 increase in rent, 2 of the units become vacant.

(a) Give an expression for the amount of monthly revenue, R , that the landlord will receive after the rent is increased by $50 x times.
(b) If his monthly expenses per occupied unit average $130, then find an expression for his total monthly expenses after x increases in rent.
(c) If his profit is found by subtracting his monthly expenses from his monthly revenue, find an expression for his monthly profit after x increases in rent.

A student, with an entrepreneurial bent, bought a lawn mower for $425. He decided to hire another student at $7.50 per hour to operate the lawn mower while he knocked on doors to drum up business. It costs $5.00 per hour to run the lawn mower (oil and gasoline). The student charges his customers $20 per hour for mowing the lawn.

(a) Give an expression for the total cost to the student of operating the lawn mower for x hours. Be sure to include the purchase price of the mower.
(b) Give an expression for the total revenue received by the student if the lawnmower is operated for x hours.
(c) Give an expression for the profit made by the student after the lawn mower is operated for x hours.
(d) How many hours must the lawn mower be operated before the student breaks even (the profit is $0)?

Objective

D.2 Back to Top

To introduce the concepts of secant line and tangent line to a curve at a point and to estimate the slope of a tangent line to a curve at a point by finding slopes of secant lines numerically.

To represent understandings through a variety of communication modes (COM).

To engage in activities that require exploration and manipulation in order to develop understandings of a concept (CCT).

To learn to interact, co-operate, and collaborate through classroom activities (PSVS).

Instructional Notes

This is the first in a series of four objectives engaging students in how to find the slope of a tangent line. For this objective, students will use their calculators extensively. They will only be able to estimate the slope of the tangent line to a curve at a point. They will not be able to give the actual slope of the tangent line with certainty. This introductory phase is critically important to later developing a full understanding of the limit definition of the slope of the tangent line.

Students have dealt with tangent lines to circles in Math 20 and polynomial functions that are tangent to the x- axis in Math B30.

You could begin by dividing the students into groups and give each group a different overhead transparency containing the graph of a function with a point P marked on the graph. Without as yet having discussed the term “tangent to”, ask the students to draw a line that they think is tangent to the curve at P . Place the students' responses on the overhead and try to arrive at a class consensus as to whether or not each group was successful. See if the students can come up with a definition of “tangent to a curve at a point P ” (COM, CCT).

The discussion should point out that a tangent line could not be defined as a line that touches a curve in only one point. Although this definition is adequate for circles it is inadequate for other functions (some polynomial functions and trig functions for example – see below).

Define the term secant line as a line that intersects the graph of function in two (or more) points. The word secant comes from the Latin secare which means to cut. It is not related to the secant function . Lead students to the understanding that a tangent line to a curve at a point P is the limiting position of a sequence of secant lines drawn through points P and Q as Q approaches P provided this limiting position is the same no matter whether Q approaches P from the left or from the right .

Instructional Notes

Point out that in order for the tangent line to exist, the limiting positions must coincide whether the sequence of points approach P from the left or from the right. If this is not the case, as below, there is no tangent line.

When students have a clear understanding of the tangent line to a curve at a point on the curve, see if they can estimate the slope of the tangent line to the curve drawn at the point . Present them with that challenge, telling them that their calculators, and secant lines, will be of assistance. Have students work in groups for a while (PSVS). The following steps should emerge from the small group work.

• Sketch the situation.

Students should sketch the function in the vicinity of P showing the tangent line and a typical secant line. See right.

• Create a table.

The table should show the coordinates of P , the coordinates of a sequence of points each getting closer to P , and the calculation for the slope of the secant lines . The students' table should be similar to the one below.

• Estimate the slope of the tangent line.

Based on the table an estimate of the tangent line slope is 4.

Point	x	y	slope of
P	2	4
	3	9
	2.5	6.25
	2.1	4.41
	2.01	4.0401
	2.001	4.004001

Note: If the student can justify that the graph of the function has no sharp corners and is continuous, approaching P from both sides is not necessary. Certainly approaching from both sides is safer. At first, have half the class approach P from the left using values for Q such as 1 , 1.5, 1.9, 1.99, and 1.999. While the other half of the class uses values approaching P from the left as seen in the given table.

Examples/Activities

At points P and R, draw what you consider to be the tangent line to the curve and estimate its slope by determining the rise and run by counting squares. Assume each square represents 1 unit.

There are two points on the graph above at which the tangent line's slope would appear to be zero. Label those points J and K .

Sketch the graph of a function so the slope of any tangent line will be positive.

Sketch the graph of a function so the slope of any tangent line will be negative.

Sketch the graph of a function so the slope of any tangent line will be zero.

Estimate the slope of the tangent line to the given curve at the given point P by taking a sequence of points each of which is closer to P than the one before. Sketch the situation showing the tangent line and a typical secant line, complete a table showing the coordinates of P , , and the slopes of .

1. ;

2. ;

3. ;

4. ;

Repeat the exercise of estimating the slope of the tangent line to at by drawing a series of secant lines through points on either side of P . To clarify, find the slope of the secant lines through:

and .

A student was attempting to estimate the slope of the tangent line to the graph of at the point . The student found the slope of the secant through P and the point . Next the student found the slope of the secant through P and the point . Which secant had the greater slope, or ? You will need knowledge of the graph of this logarithmic function. (See B.2.)

Sketch the graph of a function that has a jump discontinuity at . In light of our understanding of tangent lines, explain why you cannot draw a tangent line to the function at .

Sketch the graph of a continuous split function that does not have a tangent line at the point where the function is split.

Suggestions/Extensions

In the May 1986 issue of the journal Mathematics Teacher , there is an article entitled “Mathematical Firsts—Who Done It?” by Richard H. Williams and Roy D. Mazzagatti. Read this article and find out who contributed to solving the tangent line problem.

Why do we care about the slope of a tangent line?

Students may already know from their study of physics that the slope of the tangent line to a distance-time graph gives the velocity of the object.

In the business setting, if we know the profit to be a function of the number of objects produced, then the profit is maximized at a place where the tangent line has a slope of zero.

Have the students look for other situations that involve tangent line determining slopes. Any situation concerned with instantaneous rate of change will provide examples (e.g., medicine, biology, chemistry, etc.). It is very important that students develop a thorough understanding of the tangent line and what its slope tells us before they start working with derivatives.

Students can use the List feature of a TI-83 to perform the calculations. Consider, again, the task of estimating the slope of the tangent line to at .

In enter the x -coordinates of the Q points to be used.

Use to define the function. Arrow up to highlight and type in equation. Remember is 2 nd 1.

Press ENTER to have the

y - coordinates of Q displayed.

Use to define the slope of the secant lines. Remember that P had coordinates .

Press ENTER to see the secant slopes.

The secant slopes appear to be moving to a value of 4.

Additional values, approaching 2 from the left could then be added to .

Objective

D.3 Back to Top

To determine the slope of the tangent line drawn to a curve at a particular point.

Instructional Notes

In the preceding section the slope of a tangent line was estimated. In this section the slope of a tangent line is calculated using a limit process. This requires knowledge of limits.

Using the same introductory problem as in D.2 is very helpful. The problem was to find the slope of the tangent line to the function drawn at the point , in other words . It may not be necessary to guide the students through the whole process. You might begin by giving them the coordinates of point Q .

First, sketch the situation.

Then, assign general coordinates to Q .

Place Q to the right of P as before. How far to the right? Since Q is to the right of P , let its coordinates be .

We assume h is positive. h is however far you want Q to be to the right of P .

Thus, h is the horizontal distance between P 's first coordinate and Q 's. We write Q 's second coordinate using function notation . Using function notation highlights its importance and it provides a generalizable approach to the process with respect to h .

Next, determine the slope of the secant line .

Some students may want to see the information in table form before determining the slope.

Then

, provided . This is the time to examine the result carefully. It tells us that the slope of the secant line will be h larger than 4, where h is how far Q is to the right of P . Refer the students back to D.2. When Q was at , one unit to the right of P ( ), we found the secant slope to be 5. That is because if . When Q was at , units to the right of P ( ), we found the secant slope to be 4.1. That is because if . Repeat this with the other Q locations from D.2.

Finally, determine the slope of the tangent line. As Q moves along the curve towards P , its horizontal separation from Q approaches 0. Thus, the tangent line slope is just as the calculator investigation indicated.

You may want to guide the students through a similar exercise using a different function.

In this section, the slope of the secant line is found and then the slope of the tangent line is found. In the next section, we will use the limit notation at the outset to express the slope of the tangent line (i.e., combine two steps into one).

Students struggle with expressing Q 's coordinates using function notation and determining the expression for Q 's second coordinate.

This is a major building block. Do not continue until this is clear. You will get into some interesting limits if you do not choose the functions carefully.

Again, in this section take Q to the right of P . You must point out to the students that this is dangerous unless we know that the function is continuous and has no corners. To emphasize that point, try and determine the slope of the tangent line to at using this strategy. Take Q to the right of P and find the slope of the tangent line. Then take Q to the left of P and find the slope of the tangent line. As the two answers are different , the tangent line does not exist at P .

The absolute value function has one sharp corner at which the tangent line cannot be drawn.

In closing, point out that a tangent line to the curve drawn at the point P is that straight line that most closely follows the curve at P . In other words, if you erase the graph of the function near P and replace it with a line, the tangent line is the best replacement. The tangent line is a linear approximation to the function near P . The slope of a curve at a point is said to be the same as the slope of the tangent line to the curve at that point. Thus, the curve has a slope of 4 at as the slope of the tangent line at is 4.

Examples/Activities

Exercises similar to the following should be used to see if the students really understand the process outlined in this section.

For a particular function, the slope of the secant line through points P and Q was found to be . What is the slope of the secant line if:

(a) ?

(b) ?

What limit must be evaluated to determine the slope of the tangent line? Evaluate the limit.

A secant line is drawn through points and . The slope of the secant is found to be .

Find the slope of the secant if Q is the point with coordinates:

(a)

(b)

(c)

What limit must be evaluated to determine the slope of the tangent line? Evaluate the limit.

Use the four-step process outlined in this section to find the slope of the tangent line to the given function at point P . After determining the slope of the tangent line, find the equation of the tangent line.

(a) ;

(b) ;

(d) ;

(e) ;

What limit would a student have to set up in order to find the slope of the tangent line to the graph of at the point ? You do not have to evaluate the limit.

Consider the function . Find the slope of the tangent line to the function at by:

(a) considering the slope of the secant line through P and Q where Q is to the right of P and then taking an appropriate limit.

(b) considering the slope of the secant line through P and Q where Q is to the left of P and then taking an appropriate limit.

On a particular day, the temperature x hours past 12:00 noon was given by the function .

(a) What was the temperature at noon ( )?

(b) What was the temperature at 2:00 p.m.?

(d) You know that the slope of a line is a measure of a rate of change. You also know that the slope of the tangent line is the same as the slope of the curve. Put units to your answer in part (c) and interpret the answer.

Suggestions/Extensions

The graphing calculator can be used to draw and find equations of tangent lines. Consider the problem of drawing and finding the equation of the tangent line to the function at the point .

Enter the function.

Graph the function using a window that will include point P .

Press 2nd PRGM to select the DRAW menu.

Press 5.

Press 2 (for , P 's first coordinate) followed by ENTER.

You can see the tangent line with its equation. Notice from the equation that you also have the slope of the tangent line.

There are limitations. The calculator does not correctly handle situations where the tangent line does not exist.

Consider the function with tangent line drawn at . Find an expression for the slope of the secant line through the points and . What do you notice? Why do you think this happens?

The saw-tooth function has infinitely many sharp corners. Have the students explore this function on their graphic calculators and discuss where a tangent line would not exist.

It is important for the students to develop the understanding of the slope of the tangent line being the same as the instantaneous rate of change of one variable with respect to the other. As this unit progresses, these two ideas should then be presented as equivalent to the derivative. Students should be encouraged to think about, explore, and share real-life situations where the instantaneous rate of change has importance and meaning. Often in calculus, physics connections of distance, velocity, and acceleration are made, but students should be experiencing examples from business, medicine, biology, chemistry and other areas.

Objective

D.4 Back to Top

To determine the slope of a tangent line at the general point .

Instructional Notes

Remind the students that they now know how to find the slope of the tangent line to curves such as at specific points like or . In each case, a secant slope was found and then a limit was determined. For every point, the process had to be repeated. In this section, a “once for all points” method of determining the slope of the tangent line at any point on the curve is introduced. You may want to challenge the students to see if they can develop a limit formula that will find the slope of the tangent line to any point on the graph of the function . Then guide them through the following steps.

If P is any point on the graph, let P have coordinates of . If Q lies a bit to the right of P , let its coordinates be . Now the slope of the secant line is

As Q approaches P (the horizontal distance betweeen them is decreasing so ), the secant line becomes more and more like the tangent line. Thus, the slope of the tangent line is . This is an amazing result, because you can now declare that wherever you are along the graph of , the slope of the tangent line is twice your x-coordinate, namely . Thus, the slope of the tangent line at is and the slope of the tangent line at is .

In addition to guiding the students to find the slope of one tangent line at any point on , repeat the experience with other functions such as , , and .

Define for the students the derivative of a function at a point as being the instantaneous rate of change at that point, or in other words the slope of the tangent line at that point. The derivative can be determined by applying the following limit definition: . In general, the derivative of for any point x is given by . In short-hand notation, the derivative can be represented by , read “ f prime of x ”.

Using one or more of the previously determined derivatives, have the students determine , , , , and .

Also ask the students to find the coordinates of the point at which the tangent line has a specific slope and then to find the equation of that tangent line.

Now that you have shown that the slope of the tangent line to the function at any point is , develop the formula to find the slope of the tangent line to any point on the graph of any function .

Consider the function whose graph might look like the one at right. If P is any point on the graph of , we can represent its coordinates as .

If Q is on the graph of and however far to the right of P as we wish, then Q 's coordinates can be represented by . Thus, the slope of the secant is given by .

As Q approaches P (the horizontal separation between them decreases towards 0), the secant line becomes more and more like the tangent line, and the slope of the tangent line is .

Introduce the two notations that are commonly used to represent the words “the slope of the tangent line to the curve at the point ”, namely “dee y by dee x ” and “ f prime x ”. Indicate that and likewise . Point out that for the function , we thus have or .

Have students practise saying and interpreting statements such as , , . To use the notation at a specific point, students must write , , and . The notation is not as convenient at this time as the notation. It does, however, become useful in the chain rule, in implicit differentiation, and in multi-variable calculus.

Return to a statement such as , which is true for the function , and ask students what the statement means. Then interact with a series of true or false statements about such as: ; ; ; ; if , then ; if , then .

The notation arises from the fact that the numerator in the limit quotient results from taking a difference of y-coordinates and the denominator results from taking a difference of x-coordinates.

Introduce the terms differentiation and derivative . The process of going from to or is known as differentiation . The answer we obtain after going through the process is known as the derivative of . Thus and are each known as the derivative of . In particular, is the derivative of the function .

The summary below is useful to refer to throughout the course.

Summary

As you find derivatives of functions in this objective, be these examples that you do with the students in class or assignment questions, keep track of the functions and their derivatives on a side chalkboard. You may want to tell the students that there is a shortcut to arriving at the derivative, but don't reveal what that is in this objective. In the next objective you will prove the power rule for differentiation.

Examples/Activities

1. Use the definition of the derivative, namely , to find for each of the following functions.

(a)
(b)
(c)
(d)
(e)
(f)
(g)

2. Use the definition of the derivative, namely , to find if . Use this result to help you find the equation of the tangent line that passes through , , and .

3. Where is the slope of the tangent line to the function equal to:

(a) 14?

(b) ?

4. Set up, but do not evaluate, the limit that would be required to find the derivative of one or more of the following functions. The purpose of this is for students to realize that not all derivatives can be easily found and to set the stage for upcoming developments.

(a)

(b)

(c)

The temperature, y , inside a car with its windows closed, x hours after sunrise can be modelled by the function for .

Find:

(a) the temperature in the car at sunrise ( ).
(b) the temperature in the car 6 hours after sunrise.
(c) using the definition of the derivative.
(d) and interpret the result with appropriate units.
(e) the number of hours after sunrise until .
(f) the number of hours after sunrise until the temperature in the car reached a maximum.
(g) and interpret the result with appropriate units. In particular, what does the negative number indicate?

The number of live grasshoppers per square metre x hours after spraying with a pesticide is modelled by the function .

(a) Find the number of live grasshoppers per square metre 8 hours after spraying.

(b) Find the number of live grasshoppers per square metre 58 hours after spraying.

(d) Find and interpret the result giving appropriate units.

(e) Based on your answer in (c), when are the grasshoppers dying at the greatest rate, immediately after spraying or a long time after spraying?

Suggestions/Extensions

Find and for the functions and using the definition of the derivative. Based on these results, find two values of x at which tangent lines drawn to these functions would be parallel. Find one value of x at which tangent lines drawn to these two functions would be perpendicular.

A tangent line is drawn to the graph of at the point . Find the size of the acute angle made by the tangent line and the x -axis. Give your answer in degrees to one decimal place.

Explain why the formula could also give .

Objective

D.5 Back to Top

To find derivatives of power functions and polynomial functions without the use of limits.

Instructional Notes

Begin by looking at the data you have gathered on the side chalkboard in section D.4 as derivatives were found (such as those given above).

Ask the students to see if they can detect a quick way to find the derivative from knowledge of the function. Hopefully, they will observe that if the function is of the form , then the derivative is . Similarly, if the function consists of more than one term, the derivative can be found by taking the derivative of each term in this quick way. There is potential to save a great deal of time. Guide students through a proof of the power rule. Students should not have to memorize how to prove this theorem, but they certainly should be capable of understanding the proof. Different textbooks have different proofs. The one shown below is consistent with our factoring efforts in A.2.

Theorem If , where n is a positive integer, then .

Proof

We know that . Here . Thus, . From A.2, we know that . Replacing T by , we have

Thus (1) above becomes:

The sum of n terms, each of which is , is .

Work out proof, extend the power rule to negative integers, or rational numbers. Note: The proof for negative integers is easily accomplished when the quotient rule is proved, and likewise the proof for rational numbers can be accomplished following implicit differentiation.

Next, orally practise finding derivatives of functions of the form where n is a positive or negative integer or a rational number. Students should not say, or write, “ equals ” because that is not the case (except for or ). Rather they should say “the derivative of is ”. Suggestions for the oral practise include:

, , , , , , , , , , .

By studying the table of functions and derivatives at the beginning of this section, students likely already have a strong and natural feeling that the derivative of a sum (difference) of terms of the form is the sum (difference) of the derivatives of the individual terms. Consolidate student understanding by proving the following theorem.

Theorem If , then .

Proof

Let . Now we know that . Thus

In a similar fashion, it can be proven that if , then .

You may also want to prove the constant multiple rule, namely if then where c is a constant. The proof follows.

If , then

We have proven the following three important rules.

1. The derivative of a sum of functions is the sum of the derivatives of each individual function.

or

2. The derivative of a difference of functions is the difference of the derivatives of each individual function.

or

3. The derivative of a constant multiple of a function is the constant multiplied by the derivative of the function.

With these rules in hand, we are now more informed compared to where we were in sections D.2, D.3, and D.4.

It is important to stress that before you can differentiate using the power rule, the function must be written in the form .

Initially, students are confused by questions such as . Remind students of how they can convert radicals to exponents— . Thus can be written as whereupon .

Some students think that is already in the form – not so! It is vital that the variable and coefficient not be to a power, but rather separate as .

It is important to emphasize that the derivative of a constant is zero. Appeal to the visual aspect of this. The graph of , for example, is a horizontal line. Any tangent line to this function would coincide with the function and have a slope of 0. Similarly, the derivative of a linear function such as will be 4 as the tangent to this function will coincide with the function and have its same slope. Obviously, these functions can be differentiated using the power rule by writing 4 as and by writing as .

Examples/Activities

In the proof of the derivative of , what do we obtain if we try to evaluate the limit by directly substituting at the beginning of the proof?

How do we get out of the situation?

What limit laws were used in the proof of the theorem?

Put the power rule into words. Complete the sentence: “To get the derivative of a power function, take the exponent ….”

Give the derivative to each of the following functions orally.

Find the derivatives of the following functions at the indicated points.

at and at .

at .

Find the derivatives of each of the following functions.

Find the slope of the tangent line to the graph of at the point where:

(a)

(b) (be careful, the answer is not )

(c)

(d)

(e) (be particularly careful as does not exist—there is a sharp corner there)

Find the equation of the tangent line to each function at the given point.

(a) at (b) at
(c) at

Find the coordinates of all points on the graph of at which the tangent line is horizontal.

Introduce students to other ways of asking for the derivative.

(a) If , find .
(b) If , find .
(c) If , find .
(d) If , find .
(e) Find .
(f) Find .

Find the equations of two lines, each of slope 21, that are tangent to the function .

Find the point of intersection of the lines drawn tangent to the function at the points where and .

Explain why a line with slope 4 could never be tangent to the function .

Find the x -intercept of the line drawn tangent to at the point .

Find the equation of two lines that pass through the point and are each tangent to .

A motorist followed the path of the function heading in a direction of increasing x values. He fell asleep and left the road at the point following the tangent to the curve from thereon. A tree is located at the point . Did the motorist crash into the tree? If not, on which side of the tree did the motorist pass by?

Find if .

Find the derivative of and the derivative of . Examine your answers. How are the two derivatives related?

Suggestions/Extensions

If , find where a is any real number. If , find where a is any real number. Examine these two answers. If tangent lines are drawn at and , what special relationship will they have?

Ask the students to prove that if , then .

Suppose that the profit obtained by producing x skateboards is given by the function . To the nearest cent, find the profit in producing:

(a) 50 skateboards.
(b) 75 skateboards.
(c) 100 skateboards.

Have the students reflect on any patterns they see. The rate at which the profit is changing, namely , is called the marginal profit . In this example, , meaning that if 10 skateboards are produced, the profit per skateboard is $11.09.

In the context of this example, find:

(d)
(e)
(f) .

Explain why the marginal profit is increasing as more skateboards are produced.

A motorist drives 24 000 km per year and gets x km per litre of gasoline used.

(a) Write an expression for the number of litres of gasoline used by the motorist each year.

(b) If gasoline costs 80 cents per litre, write a function that gives the annual driving cost in terms of x .

(d) Find and .

(e) Interpret your answers in part (d) explaining the negative sign.

Objective

D.6 Back to Top

To use the product rule to find the derivative of a product of two or more functions.

Instructional Notes

Remind the students of previous learning – from having to take limits to find the slope of the tangent line – to being able to apply the power rule (which saves several minutes). At the same time, point out the limitations in finding some derivatives. If students want to find the derivative of a function such as , students must first expand the right side writing each term of the expansion in the form and then find the derivative of each term using the power rule. It would be convenient to be able to find the derivative without first having to expand.

As students have seen that the derivative of a sum of functions is the sum of the derivatives of the individual functions, it is natural to assume that the derivative of a product of two functions would be the product of the derivatives of each individual function. Use an example to show that such a procedure does not work. You could use the function above:

. Thus .

On the other hand, if you regard as where and , then . Thus, if , .

Theorem If , then .

The derivative of the product of two functions is found by adding the product of the derivative of the first function with the second to the product of the first function with the derivative of the second.

Proof

If , then

In short we say that .

It is worthwhile to extend the product rule to find the derivative of the product of three functions.
If , then . This is easily established as shown.

. Now let . Then . Thus

We have established that . This result could be extended to the derivative of the product of any number of functions.

Having established the product rule, go back and have another look at the introductory example and show that does give the correct result.

. Thus,

This agrees with the result we obtained when we first expanded the product of with and then found the derivative.

A student may ask, what is the point of the product rule, as I can find the product of the two functions and then differentiate later. Not so. What if ? At this point students cannot give the derivative of , but that will come.

Another may ask, what is the point of the product rule, as all you are doing is delaying multiplying. It is not always necessary to multiply. Consider the following question.

Find the equation of the line tangent to the function at the point . Note that and we require in order to know the slope of the tangent line. We can substitute directly into the derivative without first expanding the derivative. We obtain . Thus the equation of the tangent line through is or .

Be sure to include in your examples, product rules that require students to deal with negative integer exponents and rational exponents.

Examples/Activities

Find each of the following derivatives using the product rule. Leave your answer in simplified form.

8. Hint: Think

Find each of the following derivatives using the product rule. Leave your answer in unsimplified form (stop after the first line of differentiation).

Find the slope of the tangent line to the given curve at the given point. Use the product rule to find the derivative.

1. ,

2. ,

3. ,

Find the equation of the tangent line to the curve at the given point.

1. ,

2. ,

Write each of the following quotients as a product and then use the product rule to find .

1.
2.

Write each of the following powers as a product. Then use the product rule to find .

Marie's annual salary over the past 25 years of her working life can be modelled by the function , where y is her salary in dollars in her x th year of work.

(a) How much was she earning when she started work, that is when ?

(b) How much was she earning in her 20 th year?

(d) Find as a function of x .

(e) Find and . Interpret the results.

(f) Is increasing or decreasing?

(g) What do you think Marie's occupation is? Explain.

Suggestions/Extensions

Ask the students where they have seen the algebraic strategy of adding and subtracting off the same term before. Examples include completing the square and solving linear equations.

The price, p , that is charged in order to sell x units of a product is usually a decreasing function as shown in the graph below.

(To sell more units, drop the price.)

This relationship is known as a price function or a demand function and we write . The revenue from the sale of x units of the product is given by or . Suppose that the demand function for a certain product is .

(a) How many units will be sold if the price per unit is $50?
(b) How many units will be sold if the price per unit is $100?
(c) Find the revenue function .
(d) How much revenue is generated if the price per unit is $75?
(e) Find . This is known as the marginal revenue. Of course it is the rate of change of with respect to x . That is, it is the rate of change of total revenue with respect to the number of units x that are sold..
(f) Find , , and and interpret your results.
(g) Draw a graph of the revenue function using your graphing calculator.
(i) Use the Maximum feature in the CALCULATE menu to find the number of units that should be produced to maximize the revenue.
(j) What is the maximum revenue?

Use mathematical induction to prove the product rule for n functions.

Provide the students with a variety of problems based on different real-life contexts.

Objective

D.7 Back to Top

To use the quotient rule to find the derivative of a quotient of functions.

Instructional Notes

Begin by asking students if the derivative of a quotient of functions is the quotient of the derivatives. Give students time to experiment, although they may have trouble choosing functions that are suitable. Suggest appropriate functions so that students can carry out the division before trying to differentiate.

Because , students are likely to be cautious about claiming that . Use an example, such as the following, to take away any tendency for students to think that the derivative of a quotient of functions is the quotient of the derivatives. The following is a suggestion.

Let . Then, by division, , and . But . Note that this is not the same as the correct result of . Thus . Challenge the students to develop the quotient rule stated in the theorem below. If needed, provide the hint to first isolate .

Theorem: If , then .

Proof: If , then . Using the product rule, we have . Rearranging, . Thus, . As , . Obtaining a common denominator, . Multiplying we obtain the desired result .

In short, we write .

Ask the students to try out this result on the introductory example that was used to show that .

If , then . Thus, the quotient rule is credible.

Examples/Activities

Find two ways getting the same answer both times:

simplify to rewrite the function, then differentiate.
use the quotient rule.

1.
2.
3.
4.
5.

Use the quotient rule to find the derivative of the function. Do not rewrite the question and use the product rule. Simplify to remove all parentheses from the numerator of your answer.

1.
2.
3.
4.
5.
6.
7.
8.

Find the equation of the tangent line to the function drawn at the point .

1. Find the coordinates of two points on the graph of at which the slope of the tangent line is .

2. If , , and , find:

(a)

(b)

(c)

(d)

3. Below are the graphs of and . Find each of the following.

(a) (b) (c)

4. Consider the function . Find:

(a) . Graphically, what have you found?

(b) . What does this answer confirm?

Suggestions/Extensions

A curve of the form is known as a witch of Agnesi. Research to find out about Maria Agnesi. Find the slope of the line tangent to the curve at the point . Confirm with a graphing calculator.

A curve like

is known as Newton’s serpentine. Draw the graph of this function on your graphing calculator. Find the slope of the tangent line to the curve at

and at

. Prove that the function is odd.

x minutes after being poured, the temperature, y , of the soup in a bowl of wide diameter is approximated by the function . Find:

(a) the initial temperature of the soup.

(b) .

(d) and interpret your result.

(e) Draw the graph of this function on your graphing calculator. By looking at the graph and by using your answers to (c) and (d) what happens to as x increases?

(f) Find . Interpret this result.

A musical group started playing at the Fringe Festival. The size of the audience x minutes after the group began is given by the function .

(a) How many spectators were initially there?

(b) Find . What would this measure?

(d) Based on the answers in part (c), approximately when did the group have the most spectators?

Have the students use the quotient rule to prove that if where n is a negative integer and a is a real number then .

Objective

D.8 Back to Top

To use the chain rule to differentiate composite functions.

Instructional Notes

It takes great care to teach the chain rule with understanding. This rule is most critical to so many parts of calculus. Nearly all students will be able to use it correctly with sufficient practice, but understanding takes time. To attempt to prove it may only confuse students. Use illustrative examples to support the statement of the chain rule and then use many practice examples to be sure students know how to use it.

Point out to the students that although we have dramatically extended our ability to differentiate, there are certain types of functions that would still confuse us. We can differentiate and , but how would we differentiate ? We can differentiate and , but how would we differentiate ? We can differentiate and , but how would we differentiate ?
The following example will help to lead students to the chain rule. Ask them to find if by

(1) not expanding . Take a guess!
(2) first expanding .

Method (1)

If students are not allowed to expand, most will guess that the derivative of would be coming from their experience with the derivative of being . Let them hold on to that opinion for now.

Method (2) Based on

. Thus, .

Note that the correct result differs from the conjecture by only a factor of . Note also that is the derivative of .

Theorem: If is a differentiable function of u and is a differentiable function of x , then or .

This is known as the chain rule . No proof will be provided.
What is the chain rule saying? Consider the introductory example where we were given . By letting , we have . Thus, y is a function of u and u is a function of x, exactly what the chain rule is suited for. Thus .

Use two or three more examples to show students that the chain rule works. For each of the following examples, have students find (a) y as a function of x , (b) , (c) , (d) , (e) . Lead students to see that the result in (b) is the same as the result in (e).

Example 1: ,

Example 2: ,

Example 3: ,

Mention that the name chain rule is appropriate since there is a linkage of symbols in the statement of the theorem.

We have . The chain rule extended would say that if and , and , and are all differentiable functions, then . The name is also appropriate because there is also a “chain of events” that must occur in order for the derivative to be found.

After completing the three illustrative examples above, introduce three or four examples such as the following. Do not let students jump right to the derivative until they have demonstrated that they understand the role of u . They need to become adept at decomposition, a skill they were to have developed in B.7.

Example 1: If , find .

Solution: Let . Then . By the chain rule or .

Example 2: If , find .

Solution: Let . Then . By the chain rule or .

Note the final form of the answer in these two examples. Answers are usually arranged with the monomials out front, and the factors with the greatest or rational exponents, towards the end.

When the role of u is understood, have students proceed more directly to the derivative, as the following examples illustrate.

Example 1: If , then .

Example 2: If , then . Thus,

In what form should the answer be left? Different resources have different requirements. In the subsequent section we will expect the answers to be in complete factored form. That would mean that example 1 above would have to be taken further. Study the resource you are using and ask the students to meet those requirements. Certainly factored form is very useful when we need the first derivative to analyze the function.

We cannot fully appreciate the importance of the chain rule until we learn to differentiate transcendental functions. At this point, the chain rule just allows us to generalize the power rule.

Generalized Power Rule: If , then

Whether you emphasize the generalized power rule is up to you. Probably it is best to just keep talking about the chain rule, every time you use the generalized power rule. That way the term “chain rule” will be better remembered.

Examples/Activities

Write y as a function of u in the form and write u as a function of x . See the sample.

Find for each of the functions above going through the chain rule step by step. The sample is done below.

Regard each of the following functions as being in the form . Then use the chain rule to find .
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
12.

The function is the top half of the circle with centre and radius 5.
(a) Find .
(b) Find .
(c) Find the equation of the tangent line passing through .

Find the equation of the line tangent to the curve at the point . Prove that the function is even. What then is the equation of the tangent line through the point ?

If , find and .

, find

and

1. If $10 000 is invested at an interest rate of x % compounded annually, then it will grow to an amount A in 10 years given by the relationship .

(a) Find .
(b) Find and to two decimal places. Interpret the two results.
(c) Find and to two decimal places. Interpret the two results.
(d) Would an increase in the interest rate of 1% benefit investors more if interest rates were low or if interest rates were high?

2. If the probability that you will get a speeding ticket on your way to work on a particular day is x , then the probability you will not get a speeding ticket on your way to work that day is . The probability, P , that you will not get a speeding ticket on any of 10 consecutive workdays is given by the function .

(a) Find .

(b) Find and to four decimal places. Interpret your results.

3. Each edge of a cube is shrinking at a rate of 2 cm per hour. If each edge of the cube is initially 20 cm in length, the volume of the cube after x hours will be given by the function .

(a) Find .
(b) Find and and interpret your answers.
(c) Find and and interpret your answers.

4. The radius, r , of a sphere whose volume is V is given by the function .

(a) Find .
(b) Find r if .
(c) Find and interpret the result.

5. If you live on the side of a grid road, 3 km from a highway (see the figure), then your distance, y , from a car that is x km from the intersection of the highway and the grid road is given by the function .

(a) Find .
(b) Find and interpret your result.

Suggestions/Extensions

By using the product rule for n functions, show that the derivative of the function is .

Have students work in pairs creating and differentiating functions that they would use the chain rule to differentiate (PSVS).

Have the students discuss why

and

and how they would differentiate each (COM).

Provide application problems from a variety of contexts including commerce/economics, biology, physics and medicine to provide the students a better understanding of the relevance and usefulness of the derivative.

Objective

D.9 Back to Top

To differentiate functions requiring a selection from, or combinations of, the power, product, quotient, and chain rule and to write the answers in required forms.

Instructional Notes

It is relatively easy to apply individual differentiation rules if all the questions in a particular section use that same rule. It is more difficult, however, to differentiate functions if you first must identify the appropriate rule that should be applied. The level of difficulty increases further if asked to express the answer in factored form. Of course, both of these skills are important in order to use differentiation to analyze the graphs of functions and to solve problems.

Students should be given a few questions that ask for the first line of the derivative and then stop. Then you can see if they are applying the differentiation laws appropriately. Most of the questions, however, should require students to write their answers in factored form. Students' experience with factoring in A.2 will be an asset.

Often one can save considerable work simply by rewriting the function in a form most appropriate for differentiation.

Carefully guide students through several representative examples like the ones that follow. With practice, factoring out common factors containing non-monomial bases and negative rational exponents becomes easier. Fractional coefficients will also be factoring out during the process.

Example 1: Find if .

Solution: Recognize that the function contains a product of functions so the product rule will be needed, as well as the chain rule. We must rewrite the radical term in order to differentiate it during the product rule. We write . Using the product rule and the chain rule, we have – the first line.

– reduce the 2 and .

– the factoring line. Note that is a common monomial factor of both parts. We see the base in both parts and the smaller exponent is , so we remove as the common factor.

– simplify the expression in the square brackets.

or . The monomials are usually placed in front, followed by the other terms in order of increasing complexity of their exponents (whole numbers exponents before rational exponents). Note that the radical term could also have been factored, but because of the overall exponent of –1/2 it is not of real benefit.

Example 2: Find if .

Solution:

Option 1 – The radical term will have to be written with a rational exponent. We have a quotient of functions, so we will use the quotient rule although we could transform the question to a product by bringing the term in the denominator to the numerator.

– rewriting the question.

The quotient rule has been applied.

This is the difficult step for students. Note the factoring of , , and as common factors.

The in the numerator was divided by the in the denominator.

This answer is adequate. Further factoring and simplification can take place as shown.

Option 2 – write f(x) as a product of functions first.

Example 3: Find if .

Solution: There are many options as to how to proceed.

Option 1 – write the function in the form ; use the chain rule, followed by the quotient rule. Thus, . Now simplify.

Option 2 – write the function in the form and apply the quotient rule.

Option 3 – write the function in the form and use the product rule.

We will proceed with option 3.

Example 4: Find if .

Solution: Rewrite the function in the form and then apply the product rule to the first and the second parts.

Students frequently make a sign error in applying the product rule following a subtraction sign. Encourage them to use parentheses or to develop their own strategy for the situation.

Example 5: Find if .

Solution: If one were to apply the quotient rule, much more work would be created than is necessary. Instead simplify the question by dividing each term in the numerator by . Thus . Then .

This section provides a tremendous review in basic algebraic skills. So often students will have done the differentiation step correctly but will make an error in factoring or simplification. You could easily spend 3 or 4 class periods on this objective. Most resources have appropriate assignments. Be sure to select questions containing radicals so the students have an opportunity to strengthen their factoring skills.

Examples/Activities

Start the assignment with an assortment of power, product, quotient, and chain rule questions. These introductory questions should not require the students to use more than one rule. You want to see that the students can recognize which rule to apply. A few suggestions follow.

1.
2.
3.
4.

Next create some questions requiring students to give the first line of the derivative only – no simplification is required. Discuss how to rewrite the question to make differentiation easiest.

In the final part of the assignment, create many questions that contain radical expressions and require students to put their answers in factored form similar to examples 1 to 5 shown in the Instructional Notes of this section.

A few more suggestions follow.

Point P on the island is 3 km from the nearest point A on the shoreline (see figure). An injured camper is to be transported by canoe from point P to point B on the shore, and then from point B to the doctor's cabin at point C by an all-terrain vehicle. Point B is x km down shore from point A . The doctor's cabin is 10 km from point A.
(a) If the canoe can travel at 3 km/h and the all-terrain vehicle can travel at 5 km/h, show that the time it will take to get the injured camper from point P to the doctor, as a function of x , is given by .
(b) Use your calculator to find , , , , and to two decimal places.
(c) Based on only the five calculations in part (b), where should the canoe be beached in order to minimize the travel time to the doctor?
(d) Find .
(e) How would knowledge of enable us to find the point where the canoe should be beached to minimize the travel time to the doctor's cabin?

Suggestions/Extensions

Encourage the students to discuss the rule or rules that students apply to find a derivative. In example 1 to the left, students may choose to use the quotient rule, or they may choose to rewrite f(x) as a product with the second factor having a negative exponent.

Also encourage the students to share their strategies in determining the factored form of the derivative. If necessary, discuss why one approach works better than another, but in general encourage the sharing of different approaches. It is important for the students to explore what strategies work best for them (CCT).

Have the students, working in pairs, discuss their strategies in carrying out the differentiation of different functions. As part of their individual roles, the partners should be checking for missing parts and other errors. In this way, both students are engaged in solidifying their understanding of finding a derivative.

The demand (the number of units sold) for a product is a function of the price. This relationship is known as a demand function. Suppose that for a certain product the demand function is given by

(a) Use your calculator to find the demand for the product if the price is $1. Let .

(b) What is the demand if the price is $4?

Objective

D.10 Back to Top

To use implicit differentiation to differentiate implicitly defined functions.

Instructional Notes

Begin by asking students to give their understanding of the words explicit and implicit as used in a non-mathematical setting. If no response is forthcoming, mention that the words implicate, implication, and imply are related to the word implicit. Indicate that sometimes movies give explicit language warnings.

In the mathematical setting, a function or relation is defined explicitly if y has been completely isolated. That is, we know exactly what y is like. Relations in which y is not isolated, and hence its “workings” are hidden, may contain one or more explicitly defined functions. Showing students that is an explicitly defined function because y has been isolated, whereas the relation is not an explicit function because y has not been isolated. It is not always possible to isolate y and convert from implicit to explicit form.

Because we cannot always solve for y , it is important to develop a strategy to find the derivative of functions that are given implicitly. This technique is called implicit differentiation.

Have students practise changing from implicit to explicit form showing them exactly how much effort can be involved. The following two examples are ones the students can relate to if they have taken Math C30.

Example 1: Consider the equation . What conic does it represent? Sketch its graph. Show that the equation implicitly defines two functions. Find these explicit functions, and label each function on your graph.

Solution: The equation is that of a circle with radius 13 and centre at the origin. If , then , and . Thus the two implicitly defined functions are (the top semi-circle) and (the bottom semi-circle).

Example 2: Consider the equation . What conic does it represent? Sketch its graph. Show that the equation implicitly defines two functions. Find these explicit functions, and label each function on your graph.

Solution: The form of the equation ( y is squared, x is not) indicates that this is a parabola that opens either left or right. If x is isolated, we have . Since the coefficient of is negative, the parabola opens to the left. Complete the square to aid in sketching the parabola. . The vertex has coordinates .

To isolate y we have: