Rates Instantaneous rates of change The derivative function Simple rules of differentiation The derivatives of e The product rule The chain rule

5 Rates and derivatives cyan magenta yellow 95 Rates Instantaneous rates of change The derivative function Simple rules of differentiation The de...
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5 Rates and derivatives

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Rates Instantaneous rates of change The derivative function Simple rules of differentiation The derivatives of ex and ln_x The product rule The chain rule

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RATES AND DERIVATIVES

(Chapter 5)

The concept of a rate is central to the branch of mathematics called differential calculus. Differential calculus is one of the most useful branches of mathematical reasoning.

A

RATES A rate is a comparison of two quantities of different kinds.

We often judge performances using rates. For example: ² Sir Donald Bradman’s batting rate was 99:94 runs per innings. ² Matthew Lloyd’s AFL goal scoring rate was 4:17 goals per game. ² Jodie’s typing speed is 63 words per minute with an error rate of 2:3 errors per page. Note that a rate is not a ratio. A ratio is a comparison of two quantities of the same kind.

FINDING RATES First we will review the process of finding rates from given data. Example 1 Jason typed 213 words in 3 minutes and made 6 errors, whereas Sally typed 260 words in 4 minutes and made 7 errors. a Who is faster at typing? b Who types with greater accuracy? a

b

Jason’s typing rate

Jason’s error rate 6 errors 213 words ¼ 0:0282 errors per word

213 words 3 minutes = 71 words per minute

=

=

Sally’s error rate

Sally’s typing rate

7 errors 260 words ¼ 0:0269 errors per word

260 words 4 minutes = 65 words per minute =

=

Jason is faster at typing.

Sally types with greater accuracy.

SPEED One rate we use very frequently is speed. Speed is the rate of distance travelled per unit of time.

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Speed is usually measured in kilometres per hour (km h¡1 ) or in metres per second (m s¡1 ).

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(Chapter 5)

Example 2 To convert km h¡1 to

On metropolitan main roads, the speed limit is 60 km h¡1 . a What does this rate mean? b What rate is 60 km h¡1 in metres per second?

m s¡1 , we divide by 3:6.

a 60 km h¡1 means that in every hour we travel 60 km. 60 km b 60 km h¡1 = 1 hour (60 £ 1000) metres = (60 £ 60) seconds ¼ 16:7 m s¡1

EXERCISE 5A.1 1 Colin’s average heart rate is 67 beats per minute. a What does this rate mean? b How many heartbeats can we expect Colin to have in: i 1 hour ii 1 day? 2 Claire typed a 14 page document and made 8 errors. If an average page of typing consists of 380 words, find Claire’s error rate in: a errors per word b errors per 100 words. 3 Peter worked 12 hours for $148:20 whereas Marita worked 13 hours for $157:95. Who had the better hourly pay rate for their work? 4 A new model tyre has a tread depth of 8 mm. After driving for 32 178 km the tread depth was reduced to 2:3 mm. State the wearing rate of the tyres in: a mm per km b mm per 10 000 km. a 100 km h¡1 to m s¡1

5 Convert:

depth of tread tyre cross-section

b 25 cents per kg to $ per tonne.

6 The world records for athletes running various distances are given below for both men and women. Men’s Records Race Time Runner 100 metres 9:58 Usain Bolt

Women’s Records Race Time Runner 100 metres 10:49 F. G. Joyner

400 metres

43:18

Michael Johnson

400 metres

47:60

Marita Koch

800 metres

1:41:11

Wilson Kipketer

800 metres

1:53:28

J. Kratochvílová

1500 metres

3:26:00

H. el Guerrouj

1500 metres

3:50:46

Qu Yunxia

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i m s¡1 i m s¡1

a Convert each man’s record to: b Convert each woman’s record to:

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AVERAGE RATES OF CHANGE Consider a journey along the highway from Adelaide to Melbourne, passing various locations on the way. The following table gives the time taken to reach each location, and the distance travelled to that point. Location Adelaide tollgate Tailem Bend Bordertown Nhill Horsham Ararat Midland Highway Junction Melbourne

Time taken (min) 0 63 157 204 261 317 386 534

800

We plot distance travelled against time taken to obtain a graph of the journey.

Distance travelled (km) 0 98 273 356 431 527 616 729

distance travelled (km)

M

MHJ

600 A H

400

N 83 km

B 200

We can use the table to find the average speed between any two locations along the way.

47 min TB time taken (min) 500 600

A 100

200

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400

For example, the average speed from Bordertown to Nhill distance travelled = time taken (356 ¡ 273) km = (204 ¡ 157) min 83 km 47 min 83 km = (47 ¥ 60) h

Notice that the average speed

=

=

y-step x-step

= slope of the graph between B and N.

¼ 106 km h¡1 Example 3 For the Adelaide to Melbourne trip: a Find the average speed from Tailem Bend to Bordertown in:

i km h¡1

ii m s¡1 .

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b How can you tell from the graph that the drive from Tailem Bend to Bordertown was faster than the drive from Midland Highway Junction to Melbourne?

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distance travelled time taken (273 ¡ 98) km = (157 ¡ 63) min

i Average speed =

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(Chapter 5)

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ii speed ¼ 111:7 ¥ 3:6 ¼ 31:0 m s¡1

175 km 94 min 175 km = (94 ¥ 60) h =

¼ 111:7 km h¡1 b The faster the drive between two points, the greater the slope of the graph. The graph is steeper from Tailem Bend to Bordertown, so this trip was faster. Much information can be gained from displacement graphs. A typical displacement graph shows how far a person is away from home. Example 4 100s of metres

Paul walks to the newsagent to get the paper each morning. The graph alongside shows his distance from home during his walk. Use the graph to answer the following questions: a How far is the newsagent from Paul’s house?

8 7 6 5 talked to a friend

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b What is the slope of the line segment for the first 4 minutes of the walk? c What was Paul’s average walking speed during the first 4 minutes, in m min¡1 ?

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d What is the physical representation of the 0 slope in this problem? e How many minutes did Paul spend in the newsagent?

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8 12 16 20 24 28 32 minutes

f What was his average speed on the return journey, in km h¡1 ? g What was the total distance that Paul walked? a 800 metres y-step b slope = x-step

e 20 ¡ 12 = 8 minutes f Average return speed 800 m = 12 min (800 ¥ 1000) km = (12 ¥ 60) h

500 4

= 125 distance travelled time taken 500 m = 4 min = 125 m min¡1

c Average speed =

= 4 km h¡1

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g 800 m away + 800 m home = 1600 m

d The slope represents Paul’s speed.

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EXERCISE 5A.2 1 For the Adelaide to Melbourne data, find the average speed from: a Tailem Bend to Nhill

b Horsham to the Midland Highway Junction.

2 During December, Mount Bold reservoir was losing water at a constant rate due to usage and evaporation. On December 12, the estimate of water in the reservoir was 33:8 million kL, and on December 23 the estimate was 28:2 million kL. What is the average rate of water loss during this period? 3 When an oil tanker was damaged after hitting a reef, an oil slick appeared. Some days after the accident, the slick was roughly circular in shape and covered an area of radius 15 km. Four days later it was estimated that the radius of the oil slick had increased to 21 km. Find the average rate of change in the area of the slick during this period. 4 Water consumption in a laboratory depends on the number of scientists present. During a one month period, the following data was recorded for several laboratories: Number of scientists in lab

1

2

3

4

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6

Water consumption (kL)

10:1

16:3

20:9

24:8

27:7

30:1

a Graph water consumption against the number of scientists. b Comment on the shape of the graph. c Find the rate of water consumption per scientist for: i one scientist ii two scientists iv four scientists v five scientists 5 Water consumption invoices for 2009 show the following consumption figures:

iii three scientists vi six scientists.

Period

Consumption

First quarter (Jan 01 to Mar 31)

106:8 kL

Second quarter (Apr 01 to Jun 30)

79:4 kL

Third quarter (Jul 01 to Sept 30)

81:8 kL

Fourth quarter (Oct 01 to Dec 31)

115:8 kL

What is the average rate of water consumption for: a each quarter

b the first six months

c the whole year?

6 The graph shows the depth of water in a spa bath against the time in minutes.

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a What is the slope of OA?

depth of water (cm)

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b What is the significance of your answer in a? c How long did the person stay in the spa?

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d At what time did the person get out of the spa? e Does the spa empty at a constant rate? Explain your answer.

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7 Alongside is a typical travel graph. Write a story for this travel graph. Include numerical details such as the speed for various stages of the journey.

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(Chapter 5)

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distance (km)

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AVERAGE RATES FROM CURVED GRAPHS If the graph which compares two quantities is a straight line, there is a constant rate of change in one quantity with respect to the other. This constant rate is the slope of the straight line. If the graph is a curve, we can find the average rate of change between two points by finding the slope of the chord between them. Example 5 The number of mice in a colony over time is shown in the graph alongside.

350 300 250 200 150 100 50

a Estimate the average rate of increase in the population for: i the period from week 3 to week 6 ii the seven week period. b What is the overall trend with regard to population increase over this period? a

i

1

ii

rate of population increase increase in population = time (240 ¡ 110) mice ¼ (6 ¡ 3) weeks ¼ 43 mice per week

population size

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weeks 7

rate of population increase ¼

(315 ¡ 50) mice (7 ¡ 0) weeks

¼ 38 mice per week

b The graph is increasing by larger and larger amounts, so the population growth rate is increasing.

EXERCISE 5A.3 1 For the travel graph given alongside, estimate the average speed:

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a in the first 4 seconds b in the last 4 seconds c in the 8 second interval.

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(Chapter 5)

2 The number of lawn beetles per square metre surviving in a lawn after various doses of poison is given in the graph alongside.

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a Estimate the average rate of beetle decrease: i when the dose increases from 0 to 10 grams ii when the dose increases from 4 to 8 grams.

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b Describe the effect on the rate of beetle decline as the dose increases from 0 to 14 grams. 3 In a dry climate, the rate of evaporation is a very important statistic. The graph shows the depth of water in a shallow dish during a hot day over a 24 hour period. What was the average rate of evaporation between 6 am and 3 pm? Show how you arrived at your answer.

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mid. 3 a.m. 6 a.m. 9 a.m. noon 3 p.m. 6 p.m. 9 p.m. mid

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INSTANTANEOUS RATES OF CHANGE

A moving object such as a motor car, an aeroplane, or a runner, has variable speed. At a particular instant in time, the speed of the object is called its instantaneous speed. To examine this concept in greater detail, consider the following investigation.

INVESTIGATION 1

INSTANTANEOUS SPEED

Earlier we noticed that the average rate of change between two points on a graph is the slope of the chord connecting them. But what happens if these two points are extremely close together or in fact coincide? To discover what will happen, consider the following problem: A ball bearing is dropped from the top of a tall building and the distance fallen after t seconds is recorded. The following graph of distance against time is obtained.

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In this investigation we will try to measure the speed of the ball bearing at the instant when t = 2 seconds.

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(Chapter 5)

distance (m) 80

M(4, 80)

60

The average speed in the time interval 2 6 t 6 4

chord

distance travelled time taken (80 ¡ 20) m = (4 ¡ 2) s

=

curve

40 20

F(2, 20)

=

time (s) 2

4

6

60 2

m s¡1

= 30 m s¡1

What to do: DEMO

1 Click on the icon to start the demonstration. F is the point where t = 2 seconds, and M is another point on the curve. To start with, M is at t = 4 seconds. The value given by slope is the slope of the chord FM. This is the average speed of the ball bearing in the interval from F to M. For M at t = 4 seconds, you should see the average speed is 30 m s¡1 . 2 Click on M and drag it slowly towards F. Write down the slope of the chord when M is at the point where t is: a 3

b 2:5

c 2:1

d 2:01

3 When M reaches F observe and record what happens. Why is this so? 4 What do you suspect is the instantaneous speed of the ball bearing at the instant when t = 2? 5 Move M to the origin and then move it towards F from the other direction. Do you get the same result? From the investigation you should have discovered that: The instantaneous rate of change at a particular instant is given by the slope of the tangent to the graph at that point.

tangent P

For example, the graph alongside shows how a cyclist accelerates away from an intersection.

distance (m) 100

The average speed over the first 8 seconds is 100 m = 12:5 m s¡1 . 8s

80 60

Notice that the cyclist’s speed is quite small initially, but is increasing as time goes by.

40 20

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To find the instantaneous speed at any time instant, for example t = 4, we draw the tangent at that point and find its slope.

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(Chapter 5)

The tangent passes through (2, 0) and (7, 40)

distance (m) 100

) the instantaneous speed at t = 4

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= slope of tangent

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=

(40 ¡ 0) m (7 ¡ 2) s

=

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point where t=4

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tangent at t = 4 5

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m s¡1

= 8 m s¡1 We can use this method to find the speed of the cyclist at any instant in time.

7 8 time (s)

EXERCISE 5B.1 1 For each of the following graphs, find the rate of change at the point shown by the arrow. Make sure your answer contains the correct units. PRINTABLE a b distance (km)

distance (m) 4

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4

2 t=1

1 1

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time (s) 4

3

1

d 120

profit ($thousands)

40 number of items sold 30 40 50

2 1600

2 The graph alongside shows the height of a hot air balloon in metres for the first 50 minutes of its flight. The graphs over the page show more details of the curve. a At what speed was the balloon rising initially?

1000 800 600

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b When was the balloon neither rising nor falling? c What was the greatest rate of descent, and when did this occur?

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3 A safe is dropped from the top floor of a tall building, with the intent of cracking it open. The travel graph for the safe is given alongside, with the height measured in metres and the time in seconds. a What is the speed of the safe when it is initially dropped?

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b How fast is the safe travelling after 2 seconds? c When does the safe hit the ground?

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4 Water is leaking from a tank. The volume of water left in the tank over 4 hours is given in the graph alongside.

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FINDING THE SLOPE OF A TANGENT ALGEBRAICALLY

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In the previous exercise we found the instantaneous rate of change at a given point by drawing the tangent at that point and finding the slope of the tangent. The problem with this method is that it is difficult to draw an accurate tangent by hand, and the result will vary from one person to the next. So, we need a better method for finding the slope of a tangent.

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INVESTIGATION 2

THE SLOPE OF A TANGENT y

Given a curve y = x2 , we wish to find the slope of the tangent at the point F(1, 1).

y = x2

DEMO

F(1, 1) x

What to do: 1

y

Suppose M lies on y = x2 and M has coordinates (x, x2 ). Copy and complete the table below:

y = x2 M(x, x 2) F(1, 1)

x

Point M

5

(5, 25)

slope of FM 25¡1 5¡1

=6

3

x

2 1:5 1:1 1:01

2 Comment on the slope of FM as x gets closer to 1. 3 Repeat the process as x gets closer to 1, but from the left of F.

1:001

4 Click on the icon to view a demonstration of the process. 5 What do you suspect is the slope of the tangent at F?

We can find the slope of the tangent at F by observing the slope of the chord FM as M gets closer to F. Fortunately we do not have to use a graph and table of values each time we wish to find the slope of a tangent. Instead we can use an algebraic approach.

THE ALGEBRAIC METHOD y

To illustrate the algebraic method we will once again consider the curve y = x2 and the tangent at F(1, 1). Let the moving point M have x-coordinate 1+h, where h 6= 0.

y = x2

chord

So, M is at (1 + h, (1 + h)2 )

M(1 + h, (1 + h)2)

F(1, 1)

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187

(Chapter 5)

y-step (1 + h)2 ¡ 1 = x-step 1+h¡1

The slope of chord FM is

1 + 2h + h2 ¡ 1 h 2 2h + h = h h(2 + h) = h =2+h fas h 6= 0g =

Now as M approaches F, h approaches 0, and so 2 + h approaches 2. Therefore, the tangent at (1, 1) has slope 2. Example 6 Use the algebraic method to find the slope of the tangent to y = x2 at the point where x = 2. Let M(2 + h, (2 + h)2 ) be a point on y = x2 which is close to F(2, 4).

y

y = x2 M F(2, 4)

y-step Slope of FM = x-step (2 + h)2 ¡ 4 2+h¡2 4 + 4h + h2 ¡ 4 = fusing (a + b)2 = a2 + 2ab + b2 g h h(4 + h) = h =4+h fas h 6= 0g =

x

Now as M approaches F, h approaches 0, and 4 + h approaches 4. So, the tangent at (2, 4) has slope 4.

EXERCISE 5B.2 1 Use the algebraic method to find the slope of the tangent to: a y = x2 at the point where x = 1:5

b y = 3x2 at the point where x = 2

c y = x2 + 3x at the point where x = 1

d y = 2x ¡ x2 at the point where x = 3

e y = 2x + 1 at the point where x = 5. a Using (x + h)3 = (x + h)2 (x + h), show that (x + h)3 = x3 + 3x2 h + 3xh2 + h3 .

2

b Find (1 + h)3 in expanded form using a.

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c Consider finding the slope of the tangent to y = x3 at the point F(1, 1). If the x-coordinate of a moving point M, which is close to F, has value 1 + h, state the coordinates of M.

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d Find the slope of the chord FM in simplest form. e What is the slope of the tangent to y = x3 at the point (1, 1)? 3 Find the slope of the tangent to y = x3 at the point (2, 8). 1 alongside. x 1 ¡h 1 ¡ = . a Show that x+h x x(x + h)

4 Consider the graph of y =

y

2

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C

1 y=^ x F(1, 1) M

1 -2 -1

1

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-1 -2

THE DERIVATIVE FUNCTION y

For a non-linear function, the slope of the tangent changes as we move along the graph.

y = x2

For a given function, we can determine a corresponding slope function or derivative function which gives the slope of the tangent at any point. x

FUNCTION NOTATION REVIEW Recall that y = x2 can be written in function notation as f(x) = x2 . f (x) is read as “f of x”. When x = 3, y = 32 = 9. We write f (3) = 9. This means that the point (3, 9) lies on the graph of f(x) = x2 .

THE DERIVATIVE FUNCTION Following are graphs of f(x) = x2 with tangents drawn at A(1, 1), B(2, 4) and C(3, 9). y

y 8

f(x) = x2

6 4

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189

(Chapter 5)

At A(1, 1) the slope is 2, which is twice the x-coordinate. At B(2, 4) the slope is 4, which is twice the x-coordinate. At C(3, 9) the slope is 6, which is twice the x-coordinate. This suggests that the slope of the tangent to f (x) = x2 at any point is double the x-coordinate of the point. DEMO

So, if (x, y) lies on f (x) = x2 , the slope at this point is 2x. Click on the icon to check the validity of this statement. More generally, for a function f (x) we can define a slope function by f 0 (x). f 0 (x) is read “f dashed x”. For f (x) = x2 , we have f 0 (x) = 2x. We say that the derivative of x2 is 2x. Notice that when x = ¡1, f 0 (x) = 2(¡1) = ¡2. This means that the tangent to f(x) = x2 at the point where x = ¡1 has slope ¡2. We write f 0 (¡1) = ¡2. Alternatively, if we are given y in terms of x, the slope dy . function is represented by dx dy = 2x. So, if y = x2 , then dx dy is read ‘dee y by dee x’. dx

y 3

f(x) = x2

2 slope -2

1

-2 -1 -1

1

2

x

FINDING THE DERIVATIVE FUNCTION ALGEBRAICALLY dy algebraically, dx using a method similar to that given on page 186.

Given y = x2 , we can also find

Instead of finding the slope of the tangent at the specific point F(1, 1), we can find the slope of the tangent at the general point A(x, x2 ).

y

y = x2 M

A(x, x2)

Consider M(x + h, (x + h)2 ) which is close to A.

x 2

2

(x + h) ¡ x x+h¡x

The slope of chord MA is

x2 + 2hx + h2 ¡ x2 h h(2x + h) = h1

DEMO

=

= 2x + h

fas h 6= 0g

Now as M moves towards A, the slope of the chord MA approaches the slope of the tangent at A. Click on the demonstration to convince yourself of this.

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Also, as M approaches A, h approaches 0, and hence the slope of the chord MA approaches 2x.

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RATES AND DERIVATIVES

(Chapter 5)

Geometrically, the slope of chord MA approaches the slope of the tangent at A, which is

dy . dx

Algebraically, the slope of chord MA approaches 2x. dy = 2x. dx

Therefore, we conclude that

EXERCISE 5C.1 1 Use the algebraic method to find the derivative of: a y = 2x2

b y = x2 + 2x

c y = 3x2 ¡ 1

THE DERIVATIVE OF xn There are rules we can use to find the derivatives of functions without having to resort to the algebraic method each time. We will first determine a rule for finding the derivative of xn , where n is a real number.

INVESTIGATION 3

THE DERIVATIVE OF xn

What to do: 1 Click on the icon to open the derivative determiner. 2

2 Choose the function y = x , then press Play . By sliding the point along the graph, we can observe the changing slope of the tangent. We can hence graph the slope function. 3 Click on

DERIVATIVE DETERMINER

to find the equation of the slope function. You should

Find derivative function

dy find that for y = x2 , = 2x. dx 4 Find the slope function a y = x3 d y = 1 = x0

dy for: dx b y = x4 1 e y = = x¡1 x

c y=x 1 f y = 2 = x¡2 x

You should have obtained these results from the investigation:

4x3

3

x

3x2

x2

2x

x

x

0

x¡1

¡x¡2

x¡2

¡2x¡3

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Y:\HAESE\SA_12MET-2nd\SA_12MET2_05\190SA12MET2_05.CDR Wednesday, 15 September 2010 10:57:03 AM PETER

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RATES AND DERIVATIVES

191

(Chapter 5)

From these results, it appears that: The derivative of xn is nxn ¡ 1 . This result is true for any real n. Example 7 Find the slope function for: a y = x8

b f (x) =

y = x8

a

b

dy = 8x7 ) dx

1 x3

1 = x¡3 x3 ) f 0 (x) = ¡3x¡4 f (x) =

Remember that 1 = x¡n . xn

Example 8 p Consider f (x) = x x. a Find f 0 (x).

b Find and interpret f 0 (4).

p 3 f(x) = x x = x 2

a

y

1

) f 0 (x) = 32 x 2 p = 32 x p b f 0 (4) = 32 4

f(x) = x~`x

=3

p So, the tangent to f(x) = x x at the point where x = 4, has slope 3.

4 x slope = 3

EXERCISE 5C.2 1 Find the slope function a y = x6

dy for: dx b y=

1 x5

2 Find the slope function f 0 (x) for: 7

a f (x) = x 2 3 For f(x) =

b f (x) =

1 , find: x2

a f (2)

c y= p 3 x

b f 0 (2)

p x

p d y = x2 x

1 c f (x) = p x x

1 d f(x) = p x5

c f (¡3)

d f 0 (¡3)

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1 4 Consider f (x) = p . x b Find and interpret f 0 (1). a Find f 0 (x).

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Y:\HAESE\SA_12MET-2nd\SA_12MET2_05\191SA12MET2_05.CDR Wednesday, 15 September 2010 10:57:07 AM PETER

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192

RATES AND DERIVATIVES

(Chapter 5)

a Using technology to help, accurately graph the function f (x) =

5

b On your graph, draw the tangent to f(x) at the point (2, 12 ).

1 . x

c Estimate the slope of this tangent. d Check the accuracy of your estimate by finding f 0 (2).

D

SIMPLE RULES OF DIFFERENTIATION

Functions can be classed as: ² multiples of simpler functions, such as f (x) = 3x2 or g(x) = ¡5x ² sums of simpler functions, such as f (x) = x3 + x or g(x) = x4 ¡ 2x2 + 1 ² products of simpler functions, such as f (x) = x2 ex or g(x) = x ln x x2 2x ¡ 1 ² quotients of simpler functions, such as f (x) = 3 or g(x) = 2 . x +1 x +2 In the following investigation we will discover rules for differentiating multiples and sums of simpler functions.

INVESTIGATION 4

RULES OF DIFFERENTIATION DERIVATIVE DETERMINER

What to do: 1 Open the derivative determiner. 2 Find the derivative of: a x b 9x e x5 f 3x5

c x3 g x4

d 2x3 h 7x4

The derivative of c £ f (x) is .... .

3 Use 2 to copy and complete: 4 Find the derivative of: a x2 + x d 4x3 ¡ 3x

b x5 ¡ x2 e x2 + 6x ¡ 2

c 3x2 + 2x f 2x3 ¡ 5x2 + 3x + 1

The derivative of f (x) + g(x) is .... .

5 Use 4 to copy and complete:

a constant

Function c

Derivative 0

xn

xn

nxn¡1

multiples of functions

c £ f (x)

c £ f 0 (x)

sums of functions

f(x) + g(x)

f 0 (x) + g0 (x)

From the previous two investigations, we now have the following rules for differentiation:

These rules can be used to differentiate polynomials.

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For example, if f(x) = 2x3 ¡ 4x2 + 3x + 5, f 0 (x) = 2(3x2 ) ¡ 4(2x) + 3(1) + 0 = 6x2 ¡ 8x + 3

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Y:\HAESE\SA_12MET-2nd\SA_12MET2_05\192SA12MET2_05.CDR Wednesday, 15 September 2010 10:57:11 AM PETER

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RATES AND DERIVATIVES

193

(Chapter 5)

Example 9 dy for: dx

Find

a y = x4 ¡ 5x3 + 7x2 ¡ 6x + 3 a

b y = 2x ¡

3 5 + x x2

c y=

3x3 + x2 ¡ 4 x2

y = x4 ¡ 5x3 + 7x2 ¡ 6x + 3 dy = 4x3 ¡ 5(3x2 ) + 7(2x) ¡ 6(1) + 0 ) dx = 4x3 ¡ 15x2 + 14x ¡ 6 5 3 + 2 x x = 2x ¡ 3x¡1 + 5x¡2

y = 2x ¡

b

dy = 2 ¡ 3(¡x¡2 ) + 5(¡2x¡3 ) dx = 2 + 3x¡2 ¡ 10x¡3 3 10 =2+ 2 ¡ 3 x x

)

3x3 + x2 ¡ 4 x2 3 3x x2 4 = 2 + 2¡ 2 x x x = 3x + 1 ¡ 4x¡2

y=

c

)

dy = 3 ¡ 4(¡2x¡3 ) dx = 3 + 8x¡3 8 =3+ 3 x

Example 10 Find the slope of the tangent to f (x) = x3 ¡ 3x2 + x + 9 at the point where x = 2. y

f(x) = x3 ¡ 3x2 + x + 9 ) f 0 (x) = 3x2 ¡ 6x + 1 ) f 0 (2) = 3(2)2 ¡ 6(2) + 1

slope 1

f(x) = x3 - 3x2 + x + 9

=1 So, the tangent to f(x) at the point where x = 2 has slope 1.

2

You can also use technology to find the slope of a tangent to a function. Click on the appropriate icon for instructions.

x

Casio TI-84+ TI-nspire

EXERCISE 5D 1 Describe the following functions as either a sum, a product, or a quotient: p x p ex a xe b x + ex c p d x2 + 2x x

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g x2 + 5x3 + 2

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h x2 ln x

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RATES AND DERIVATIVES

(Chapter 5)

dy for: dx a y = 6x

b y = x3 ¡ x2 ¡ 7x + 2

c y = 5x ¡ 4

d y = 2x2 ¡ 5

e y = 4x3 ¡ 9x

f y = x ¡ 5x2

g y=7

h y = 11 ¡ 8x3

i y = 12 x2 ¡ 4x + 3

j y = 43 x3 ¡ 12 x ¡ 6

k y=

2 Find

x3 3x2 x ¡ + ¡1 2 4 5

l y = 8x ¡

2x3 7

m y = 0:7x4 ¡ 1:2x2 + 2:7x ¡ 1:9 3 Find f 0 (x) for: 3 x

a f (x) = 32 x2 ¡ 25 x + 4

b f(x) = 5x +

1 7 ¡ 3 x x 1 g f (x) = 9x ¡ 2 5x

p e f(x) = x3 + 3 ¡ x x

d f (x) = 4x2 +

c f (x) =

p 4 f f (x) = 2 x ¡ 5 x 3 2 x i f (x) = p ¡ p + x x x 7

4 h f(x) = p + 3x ¡ 2 x

dy for: dx 2x2 + 3 a y= x 4x2 + 5 d y= x3

p x¡x

4 Find

x2 + 3x + 4 x 8x ¡ 3 e y= p x

3x3 ¡ 5x2 + 7 x2 2 6x + 4x ¡ 7 p f y= x

b y=

5 For f (x) = x3 ¡ 2x +

c y=

4 , find and interpret f 0 (2). x

6 Consider f (x) = x2 + 3x ¡ 5. a Find f 0 (x). b Find the slope of the tangent to f(x) at the point where x = ¡2. c Using technology to help, draw the graph of y = f(x). Include the tangent found in b. 7 The graph of f (x) = 13 x3 + 34 x2 ¡ 52 x ¡ 14 is shown alongside. The tangents at A(¡3, 5) and B(1, ¡ 53 ) are also given.

f(x) = Qe x3 + Er x2 - Tw x - Qr y 8

A(-3, 5)

a Use the graph to find the slopes of the tangents at A and B. b Check your answers to a by calculating f 0 (¡3) and f 0 (1).

6 4 2

-4

-2

x 2

-2

B(1, -Te_)

8 Consider the function h(t) = t3 ¡ 3t2 + 2. a Find the average rate of change in h(t) from t = ¡1 to t = 3. b Find the instantaneous rate of change in h(t) at t = 2.

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c Find the slope of the tangent to h(t) at t = ¡1.

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Y:\HAESE\SA_12MET-2nd\SA_12MET2_05\194SA12MET2_05.CDR Wednesday, 15 September 2010 10:57:19 AM PETER

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RATES AND DERIVATIVES

(Chapter 5)

195

d Sketch the graph of h(t) = t3 ¡ 3t2 + 2 for ¡2 6 t 6 4. Check your answers to a, b and c. 9 Find the slope of the tangent to: a f (x) = 2x2 ¡ x + 5 at the point where x = 3 8 at the point where x = 2 b y = 6x + 1 ¡ x p c f (x) = 10 x ¡ 3x at the point (4, 8) 7x ¡ 4 at the point (1, 3). d y= p x

E

THE DERIVATIVES OF ex AND ln x

In Chapter 4 we encountered the natural exponential e and the natural logarithm ln. The graphs of f(x) = ex and f (x) = ln x are given below: y

y

f(x) = ex

f(x) = ln_x

(1, e)

(e, 1)

1

1

x

x

1

In the following investigation we will determine the derivatives of these functions.

INVESTIGATION 5

THE DERIVATIVES OF ex AND ln x

What to do:

DERIVATIVE DETERMINER

1 Click on the derivative determiner. 2 Make sure the function y = ex is selected, and press Observe the values of f(x) and

f 0 (x)

Play

.

for x = 0, 1, 2, 3, 4 and 5.

3 Compare the values of f(x) = ex with the values of f 0 (x). What do you notice? 4 Select the function y = ln x from the data entry box. Press values of

f 0 (x)

Play

, and observe the

for x = 1, 2, 3, 4 and 5.

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5 Write these values of f 0 (x) as fractions. What do you suspect is the derivative function of ln x?

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Y:\HAESE\SA_12MET-2nd\SA_12MET2_05\195SA12MET2_05.CDR Wednesday, 15 September 2010 10:57:23 AM PETER

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196

RATES AND DERIVATIVES

(Chapter 5)

For f(x) = ex , you should have observed the values of f (x) and f 0 (x) shown opposite. For each value of x, the values of f (x) = ex and f 0 (x) are the same. This suggests that f 0 (x) = f (x) = ex .

x

f (x)

f 0 (x)

0 1 2 3 4 5

1 2:7183 7:3891 20:086 54:598 148:41

1 2:7183 7:3891 20:086 54:598 148:41

For f (x) = ln x, you should have observed the values of f(x) and f 0 (x) shown opposite.

x

f (x)

For each value of x, the value of f 0 (x) is the reciprocal of x.

1

0

2

0:6931

3

1:0986

0:333 33 =

4

1:3863

0:25 =

5

1:6094

0:2 =

This suggests that f 0 (x) =

1 . x

f 0 (x) 1 1 = 12

1= 0:5

1 4 1 5

1 3

The derivative of ex is ex . The derivative of ln x is

1 . x

Example 11 dy for: dx a y = x2 + ex

Find

b y = 4 ln x ¡ 3x + 5

y = x2 + ex

a )

b

dy = 2x + ex dx

y = 4 ln x ¡ 3x + 5 µ ¶ dy 1 ¡3 ) =4 dx x 4 = ¡3 x

EXERCISE 5E 1 Find

dy for: dx

a y = ex + 5x

b y = x3 ¡ ln x

c y = 5ex ¡

d y = 6 ln x ¡ 2x4 + 3

p e y = 10 x ¡ ex ¡ 5 ln x

f y=

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b Find and interpret g0 (2) for g(x) =

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3 6 ¡ 7ex + 4 x x

a Find and interpret f 0 (0) for f(x) = 5ex ¡ 3x2 ¡ 2x + 1:

2

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1 x

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Y:\HAESE\SA_12MET-2nd\SA_12MET2_05\196SA12MET2_05.CDR Wednesday, 15 September 2010 10:57:26 AM PETER

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RATES AND DERIVATIVES

(Chapter 5)

197

3 Consider the function f (x) = x ¡ 2 ln x: a Use technology to help graph y = f (x). b Find f 0 (x): c Find the slope of the tangent to y = f (x) at the point where: i x=1 ii x = 2 iii x = 4 d Add the tangents in c to your graph in a. 4 Consider the function g(x) = 3ex ¡ 2x2 +

1 . x

a Find g0 (x): b Show that the slope of the tangent to y = g(x) at x = 1 is 3e ¡ 5: c Find the instantaneous rate of change in y = g(x) at x = 1:5, giving your answer to 3 significant figures.

F

THE PRODUCT RULE

The product rule is used to differentiate functions that are the product of two simpler functions, p for example f (x) = x3 (2x ¡ 1), g(x) = 3x2 ln x, or h(x) = ex x.

THE PRODUCT RULE If y = u(x) v(x) then

dy = u0 (x) v(x) + u(x) v 0 (x). dx

The proof of this rule is beyond the scope of this course. However, we can demonstrate this rule by considering the function y = 2x(x + 1). dy = 4x + 2. dx

Now, y = 2x2 + 2x, so

dy using the product rule, since we can write the function as dx y = u(x) v(x) where u(x) = 2x and v(x) = x + 1 ) u0 (x) = 2 and v 0 (x) = 1

However, we can also find

Using the product rule,

dy = u0 (x)v(x) + u(x)v 0 (x) dx = 2 £ (x + 1) + 2x(1) = 2x + 2 + 2x = 4x + 2 X

Example 12

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Find f 0 (x) for:

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Y:\HAESE\SA_12MET-2nd\SA_12MET2_05\197SA12MET2_05.CDR Wednesday, 15 September 2010 10:57:30 AM PETER

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198

RATES AND DERIVATIVES

(Chapter 5)

u(x) = 3x2

a f (x) = 3x2 ln x is the product of

) u0 (x) = 6x ) f 0 (x) = u0 (x)v(x) + u(x)v 0 (x) fproduct ruleg µ ¶ 1 2 = 6x ln x + 3x x = 6x ln x + 3x p b f (x) = ex x is the product of u(x) = ex and

v(x) = ln x 1 and v0 (x) = x

and

v(x) =

p 1 x = x2 1

) u0 (x) = ex

and v 0 (x) = 12 x¡ 2

) f 0 (x) = u0 (x)v(x) + u(x)v 0 (x) fproduct ruleg 1

1

= ex x 2 + ex ( 12 x¡ 2 ) p ex = ex x + p 2 x

EXERCISE 5F 1 Find the derivative of f(x) = x2 (3x ¡ 1): a by expanding f (x), then differentiating term by term dy for: dx a y = xex

b y=

d y = x2 ln x

e y = ex ln x

b using the product rule.

2 Find

p x(2x + 1)

c y = ex (x2 + 1) p f y = x ln x

3 Find f 0 (x) for: a f (x) = x3 ex

c f(x) = (x2 + 2)ex

d f (x) = x¡1 ex

b f(x) = x3 ln x p e f(x) = 3 x ln x

g f (x) = 6x + x4 ex

h f(x) = (x ¡ 2) ln x + ex

i f(x) = x3 ¡ ex (3x ¡ 1)

f f(x) = (x3 ¡ x) ln x

4 Consider z(t) = 3t2 et . a Find z 0 (t). c Find

z 0 (2)

b Find the instantaneous rate of change of z(t) when t = 1. and interpret its meaning.

5 Find the slope of the tangent to y = ¡x3 ln x at the point (1, 0). 6 Find the exact slope of the tangent to f(x) = (x + 5)ex at the point where x = 2. µ ¶ 1 y ln x is shown 10 7 The graph of f (x) = 5 + f(x) = (5 + Qv_)_ln_x x alongside. The tangent is drawn at the point P 5 where f(x) cuts the x-axis. P

a Find the coordinates of P. b Find the slope of the tangent.

2

4

x

-5

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Y:\HAESE\SA_12MET-2nd\SA_12MET2_05\198SA12MET2_05.CDR Friday, 17 September 2010 9:20:10 AM PETER

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RATES AND DERIVATIVES

G

(Chapter 5)

199

THE CHAIN RULE

The chain rule is used to differentiate more complicated functions known as composite functions.

COMPOSITE FUNCTIONS p 1 2x + 1 and f (x) = p are called composite 4 ¡ x2 functions because they are composed of simpler functions. Functions like f(x) = (2x + 3)4 , f(x) =

Given two functions f (x) and g(x), the composite function of f and g, f(g(x)), is obtained from f(x) by replacing x by g(x). Consider the functions f (x) = x4 and g(x) = 2x + 3. Now f (g(x)) = [g(x)]4

freplacing x by g(x) in f (x) = x4 g

= (2x + 3)4 So, f (g(x)) = (2x+3)4 is the composite of the simpler functions f (x) = x4 and g(x) = 2x+3. Also, g(f (x)) = 2[f (x)] + 3

freplacing x by f(x) in g(x) = 2x + 3g

= 2x4 + 3 We can see that in general, f(g(x)) 6= g(f (x)). Example 13 Find f(g(x)) for: a f(x) = 4x ¡ 3, g(x) =

p x

b f (x) = ex , g(x) = 7 ¡ x b f (g(x)) = eg(x) = e7¡x

a f(g(x)) = 4[g(x)] ¡ 3 p =4 x¡3

Example 14 Find f (x) and g(x) if f(g(x)) is equal to: a (7x ¡ 3)5

b ln(x ¡ 4)

a f (x) = x5 , g(x) = 7x ¡ 3 Check: f(g(x)) = [g(x)]5 = (7x ¡ 3)5

X

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b f (x) = ln x, g(x) = x ¡ 4 Check: f(g(x)) = ln[g(x)] = ln(x ¡ 4) X

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(Chapter 5)

EXERCISE 5G.1 1 Find f(g(x)) if: a f (x) = x2 and g(x) = 4x + 3

b f(x) = 4x + 3 and g(x) = x2

c f (x) = x2 and g(x) = ex 1 and g(x) = 3x + 2 e f (x) = x g f (x) = ex and g(x) = ¡x

d f(x) = ex and g(x) = x2 f f(x) = 3x + 2 and g(x) = h f(x) = ¡x and g(x) = ex

i f (x) = ln x and g(x) = x2 + 1

j f(x) = x2 + 1 and g(x) = ln x

2 Find f (x) and g(x) if f(g(x)) is equal to: a (4x + 7)3 p f 11 ¡ 3x

b (3x ¡ 2)4 1 g 2x + 5

2

k e¡x

1 x

c (5 ¡ 3x)5

d (x2 + 1)6 1 i p 3 ¡ 4x

h e4x+1

l ln(3x2 + 1)

1 (4x + 1)3

m

e

p 2x + 9

j e3¡x p o ln( x + 2)

n ln(6x)

THE CHAIN RULE If y = f (u), where u = g(x), then

dy du dy = £ . dx du dx

For example, consider y = (3x ¡ 1)2 . We can write this as y = u2 ) Using the chain rule,

dy du dy = £ dx du dx = 2u £ 3 = 2(3x ¡ 1) £ 3 = 6(3x ¡ 1) = 18x ¡ 6

dy = 2u du

where u = 3x ¡ 1 and

du =3 dx

fu = 3x ¡ 1g

We can check this rule by expanding y = (3x ¡ 1)2 , then differentiating term by term: y = (3x ¡ 1)2 = 9x2 ¡ 6x + 1 )

dy = 18x ¡ 6 X dx

The chain rule is extremely useful for differentiating functions like y = (3x ¡ 1)7

or

5 2

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y = (3x ¡ 1) , where expanding then differentiating term by term would be time consuming or even impossible.

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Y:\HAESE\SA_12MET-2nd\SA_12MET2_05\200SA12MET2_05.CDR Wednesday, 15 September 2010 10:57:42 AM PETER

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201

(Chapter 5)

Example 15 dy for: dx

Find

3

b y = e¡x

a y = (x2 ¡ 3x)5

c y = ln(x2 + 5x ¡ 3)

a y = (x2 ¡ 3x)5 can be written as y = u5 dy = 5u4 du

) So,

u = x2 ¡ 3x

where

du = 2x ¡ 3 dx

and

dy dy du = £ fchain ruleg dx du dx = 5u4 (2x ¡ 3) = 5(x2 ¡ 3x)4 (2x ¡ 3) 3

b y = e¡x

can be written as y = eu dy = eu du

) So,

dy dy du = £ dx du dx = eu (¡3x2 )

where u = ¡x3 and

du = ¡3x2 dx

fchain ruleg

3

= e¡x (¡3x2 ) c y = ln(x2 + 5x ¡ 3) can be written as y = ln u where u = x2 + 5x ¡ 3 dy 1 = du u

) So,

and

du = 2x + 5 dx

dy dy du = £ fchain ruleg dx du dx 1 = £ (2x + 5) u 1 = 2 £ (2x + 5) x + 5x ¡ 3 2x + 5 = 2 x + 5x ¡ 3

From these examples, we can establish some general rules for applying the chain rule to different types of composite functions:

then

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² if y = ln(f (x)) then

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² if y = ef (x)

dy = n[f (x)]n¡ 1 £ f 0 (x) dx dy = ef (x) £ f 0 (x) dx dy f 0 (x) = dx f (x)

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then

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² if y = [f (x)]n

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RATES AND DERIVATIVES

(Chapter 5)

EXERCISE 5G.2 a Expand y = (x2 ¡ 3)2 .

1

b Differentiate your expansion term by term to find

dy . dx

dy from the original function. dx d Show that your answers to b and c are the same. c Use the chain rule to find

dy for: dx a y = (3x + 4)2

b y = (3 ¡ 2x)2

c y = (4x + 1)3

d y = (1 ¡ 2x)4

e y = (x2 + 2)3

f y = (x2 ¡ 3x + 1)3

g y = (3x ¡ x2 )5

h y = (1 ¡ x3 )4

i y = (2x3 ¡ x2 )3

dy for: dx a y = e4x

b y = e¡x

c y = e¡3x

2 Find

3 Find

x

2

d y = e2

e y = e¡x

g y = e5x¡1

h y = ex

dy for: dx a y = ln(2x + 3)

b y = ln(x3 )

2

f y = e¡0:01x 2

+4

i y = e3x

4

¡x

4 Find

c y = ln(4x)

2

d y = ln(6 ¡ 5x)

f y = ln(4x3 ¡ x2 + 5)

e y = ln(x + 6x ¡ 3)

Example 16 Use the chain rule to find a y=

dy if: dx

1 2x + 1

b y=

p x2 ¡ 7x + 3

1 as y = (2x + 1)¡1 2x + 1 Now y = u¡1 where u = 2x + 1 du dy = ¡u¡2 and =2 ) du dx dy dy du So, = fchain ruleg dx du dx = ¡u¡2 £ 2

a First we write y =

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= ¡(2x + 1)¡2 £ 2 ¡2 = (2x + 1)2

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b First we write y =

(Chapter 5)

203

p 1 x2 ¡ 7x + 3 as y = (x2 ¡ 7x + 3) 2

1

Now y = u 2 where u = x2 ¡ 7x + 3. 1 dy du ) = 12 u¡ 2 and = 2x ¡ 7 du dx dy du dy = fchain ruleg So, dx du dx 1 = 12 u¡ 2 £ (2x ¡ 7) 1

= 12 (x2 ¡ 7x + 3)¡ 2 (2x ¡ 7) 2x ¡ 7 = p 2 x2 ¡ 7x + 3

5 For the following functions: i write the function in the form y = a £ un , stating u in terms of x dy . ii find dx 1 2 b y= 3x ¡ 4 7x + 1 3 ¡3 e y= f y= (2x + 1)4 (5x ¡ x2 )3 p 1 i y = x2 + x + 1 j y = p 3x + 4

10 1 ¡ 2x p g y = 3x + 4

a y=

c y=

1 k y=p 1 ¡ 2x

1 (x2 + 2)2 p h y = 3¡x

d y=

1 l y= p 2 x ¡2

MORE COMPLICATED FUNCTIONS Sometimes we must use a combination of rules to differentiate a function. For example, if y = x2 e¡2x , we need to use the product rule and the chain rule. The chain rule is required to differentiate e¡2x . Example 17 Use the rules of differentiation to find

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Now y = u(x)v(x) where u(x) = x2 and v(x) = e¡2x ) u0 (x) = 2x and v 0 (x) = ¡2e¡2x dy = u0 (x)v(x) + u(x)v 0 (x) Now dx = 2xe¡2x + x2 (¡2e¡2x ) = 2xe¡2x ¡ 2x2 e¡2x = 2xe¡2x (1 ¡ x)

dy if y = x2 e¡2x . dx

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Y:\HAESE\SA_12MET-2nd\SA_12MET2_05\203SA12MET2_05.CDR Wednesday, 15 September 2010 10:57:52 AM PETER

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RATES AND DERIVATIVES

(Chapter 5)

EXERCISE 5G.3 dy for: dx b y = 20e¡0:02x

1 Use the rules of differentiation to find a y = 50e0:04x d y = xe¡x

c y = x2 ln(3x ¡ 1)

e y = (x3 + 1) ln(x2 ) 20 h y= 1 + 2ex p k y = (2x ¡ 5)3 x

g y = 10(1 ¡ e¡2x ) j y = (x2 + 3)e4x¡1

f y = x2 e3x i y = e3x ln(x + 5) l y = (x2 ¡ 7)5 ln x

2 Find the rate of change of w(x) = (3x + 2) ln(5x) at the point where x = 1. 2

3 Find the slope of the tangent to y = ex ln x at the point (1, 0). 4 Consider P (t) = 800te¡4t . a Show that P 0 (t) = 800e¡4t (1 ¡ 4t): b Find P 0 (0) and interpret its meaning. c Find the value of t for which P 0 (t) = 0. What can be said about the tangent to P (t) at this value of t?

REVIEW SET 5A 1 Find

dy for: dx

a y = 2x3 ¡ 3ex

2 Find f 0 (x) for:

a f (x) =

b y = 5x2 + 3 ln x + 7

x3 + 2x ¡ 3 x

b f(x) = 5x3 ln x

3 Find f 0 (4) if: a f (x) = 3x2 ¡ 7

b f(x) =

4 “100% Top Hit Mix” is a new compilation album. Its total sales in Australia were recorded over the ten weeks after its release and are shown in the graph given. a Estimate: i the average number of sales per week over the first two weeks ii the average number of sales per day between weeks 5 and 8 iii the rate of sales per week at t = 3.

p x + ln x + 3

sales (thousands) 50 40 30 20 10 1

2

3

4

5

6

7

8 9 t (weeks)

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b Because of poor sales, music stores reduced the price from $40 to $19:95. When do you think this price reduction was made? Explain your answer.

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Y:\HAESE\SA_12MET-2nd\SA_12MET2_05\204SA12MET2_05.CDR Wednesday, 15 September 2010 10:57:56 AM PETER

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5 The graph of f (x) = ln(x6 + x + 4) is shown alongside.

y 4

a Copy the graph, and draw a tangent at P where x = ¡1. b Estimate the slope of the tangent drawn in a. c Check the accuracy of your estimate by finding f 0 (¡1).

3

205

(Chapter 5)

f(x) = ln_(x6 + x + 4)

2 P

1

-2 -1 -1

1

2

-2

x

PRINTABLE GRAPH

p 6 Consider w(z) = 2z 2 ¡ 3 z + 4. a Find the average rate of change in w from z = 4 to z = 9. b Find w0 (z). c Find the instantaneous rate of change in w when z = 4. d Find the slope of the tangent to w(z) when z = 9. 7 Use the rules of differentiation to find

dy for: dx

2

a y = e3x + 2x

b y = ln(x2 + 2) + 8

c y = e2x (3x2 + 2)4

d y = 7xe¡2x

e y = (2x2 ¡ 5) ln(1 ¡ 3x)

f y = 60(1 ¡ e¡2x )

REVIEW SET 5B 1 Find f 0 (x) for: a f (x) = 2x3 + x2 + 2 +

4 x2

b f (x) = 2 ln x + ex + 9

dy at x = 2 for: dx p a y = 2 x + ex

2 Find

3 b y = p + ln x + 5 x x

3 Consider f (x) = 2x2 ¡ 5x ¡ 4. a Find f 0 (x). b Find the slope of the tangent to f(x) at the point where x = 2. c Using technology to help, draw the graph of y = f(x). Include the tangent found in b.

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d f (x) = 3xe1¡2x

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c f (x) = (3x + 2)2 ln(3x + 2)

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b f (x) = 2ex + 8

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a f (x) = 8

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4 Find f 0 (x) for:

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RATES AND DERIVATIVES

5

(Chapter 5)

a Nikki is studying Civil Engineering height (cm) at university. While researching the 25 advantages of various kinds of speed humps for her honours project, she 20 comes across what she believes is 15 the perfect speed hump. Estimate: 10 i the average slope of the speed hump over the first 25 cm 5 ii the average slope of the speed hump over the last 25 cm 10 20 30 40 50 width (cm) iii the slopes when the speed hump is 10 cm above the road iv the maximum height of the speed hump and the slope at this point. b When solid potassium (K) is placed weight of K remaining (g) into a container of water (H2 O), 5 the following explosive reaction 4 occurs: 3 2K + 2H2 O ! 2KOH + H2 The amount of potassium 2 remaining t seconds after the reaction starts is shown in the 1 graph. Find: 0.2 0.4 0.6 0.8 1.2 1.4 1 i the starting weight of the potassium time (s) ii the time taken to complete the reaction iii the average rate at which the potassium disappears (in grams per second) iv the average rate at which the potassium disappears between t = 0:2 and t = 0:4 v the instantaneous rate at which the potassium disappears when t = 0:6. Hint: The rate at which the potassium is disappearing is the negative of the slope. a Sketch the curve y = x2 ¡ 2x: b Find the coordinates of point A on the curve at which x = 2. c Find the coordinates of point B on the curve at which x = 2 + h:

6

d Find the slope of the chord AB. e Use your answer to d to find the slope of the tangent at A. dy and evaluating it when x = 2. f Check your answer to e by finding dx 10 7 Consider H(t) = . 1 + 4et ¡40et : a Show that H 0 (t) = (1 + 4et )2

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b Find H 0 (1) and interpret its meaning.

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Y:\HAESE\SA_12MET-2nd\SA_12MET2_05\206SA12MET2_05.CDR Wednesday, 15 September 2010 10:58:04 AM PETER

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