Pg 225

3.6 Finding Some Shortcuts —
The Sum and Difference Rules
SETTING THE STAGE
A cliff diver in Acapulco, Mexico, dives from about 17 m above the water.
The function f (t) 4.9t 2 1.5t 17 models the diver’s height above the
water, in metres, at t seconds. Determine the diver’s instantaneous rate of
descent with respect to time at intervals of 0.1 s during the first second the diver
is in the air.
The instantaneous rate of change of height is given by the derivative of the
height function. Can the differentiation rules from the previous section be
applied to the polynomial function that models the diver’s height?
In this section we will develop the rules to differentiate any polynomial
function.
EXAMINING THE CONCEPT
Height (m)
y
30
20
10
0
–10
–20
–30
d(t) = 17
c(t) = 1.5t
Developing the Sum and Difference Rules
The function f (t) 4.9t 2 1.5t 17 is the sum of three different functions.
By defining b(t) 4.9t 2 , c(t) 1.5t, and d(t) 17, you can write the original
function as f (t) b(t) c(t) d(t). Examine the graph of all four functions.
The function f (t) 4.9t 2 1.5t 17 is equal to the sum
of its three component functions, b(t) 4.9t 2 , c(t) 1.5t,
and d(t) 17, for all real values of t.
But how are the derivatives of the functions related? Is f ′(t)
equal to the sum of b′(t), c′(t), and d′(t)?
Using the differentiation rules from the previous section,
f(t) = – 4.9t 2 + 1.5t + 17
0.4 0.8 1.2 1.6 2.0 2.4
Time (s)
f (t) =b(t) +c(t) +d(t)
b(t) = –4.9t 2
t
d
b′(t) (4.9t 2 dt
) c′(t) dt
(1.5t) d′(t) dt
(17)
2(4.9)t 2 1 1.5 0
9.8t
Summing the derivatives together gives
b′(t) c′(t) d′(t) 9.8t 1.5 0
9.8t 1.5
Compare this sum and the derivative of f (t).
Determine the derivative f ′(t) from first principles.
d
d
3.6 FINDING SOME SHORTCUTS — THE SUM AND DIFFERENCE RULES 225

h(t) 4.9t 2 1.5t 17
h′(t) lim
h → 0
lim
h → 0
lim
h → 0
lim
h → 0
lim
h → 0
lim
h → 0
lim
f (t h) f (t)
h
[4.9(t h) 2 1.5(t h) 17] [4.9t 2 1.5t 17]
h
4.9(t 2 2th h 2 ) 1.5(t h) 17 [4.9t 2 1.5t 17]
h
4.9t 2 9.8th 4.9h 2 1.5t 1.5h 17 4.9t 2 1.5t 17
h
9.8th 4.9h 2 1.5h
h
h 1 (9.8t 4.9h 1.5)
h
1
20.0
(9.8t 4.9h 1.5) 10.0
h → 0
c′(t) = 1.5t
0.0
d′(t) = 0
9.8t 4.9(0) 1.5
9.8t 1.5
The derivative of the original
function, f (t), is the sum of the
derivatives of the component
functions, b(t), c(t), and d(t).
f (t) b(t) c(t) d(t)
4.9t 2 1.5t 17
f ′(t) b′(t) c′(t) d′(t)
9.8t 1.5
Value of Derivative
–10.0
–20.0
–30.0
0.4 0.8 1.2 1.6 2.0 2.4
Time (s)
f ′(t) = b′(t) + c′(t) + d′(t)
f′(t) = b′(t) + c′(t) + d′(t)
= –9.8t + 1.5
b′(t) = – 9.8t
The derivative of a sum is
the sum of the derivatives.
The Sum Rule
If h(x) f (x) g(x) and f and g are both differentiable, then
h′(x) f ′(x) g′(x). In Leibniz notation,
d
d
[ f (x) g(x)] d
[ f (x)] d
[g(x)].
d
d
x
x
x
Proof
Let F(x) f(x) g(x).
h ′(x) lim
h → 0
lim
h → 0
lim
h → 0
F(x h) F(x)
h
f (x h) f (x)
h
lim
h → 0
f ′(x) g′(x)
[ f (x h) g(x h)] [ f (x) g(x)]
h
[f (x h) f (x)] [g(x h) g(x)]
h
lim
h → 0
g(x h) g(x)
h
Rearrange the numerator.
Rewrite using the
sum rule for limits.
226 CHAPTER 3 RATES OF CHANGE IN POLYNOMIAL FUNCTION MODELS

The sum rule can be extended to any number of functions. A corresponding rule
also exists for differences.
The derivative of a
difference is the difference
of the derivatives.
The Difference Rule
If h(x) f (x) g(x) and f and g are both differentiable, then
h′(x) f ′(x) g′(x). In Leibniz notation,
d
d
[ f (x) g(x)] d
[ f (x)] d
[g(x)].
d
d
x
x
x
The proof for the difference rule is similar to the proof for the sum rule.
Example 1
Using the Sum and Difference Rules
Determine the derivative of each function.
(a) g(x) 8x 4 6x 2 12
(b) h(x) 4x 2 5x 2 4x
Solution
Use the sum and difference rules. Differentiate term by term.
d
(a) g′(x) d
(8x
x
4 6x 2 12)
d
d
(8x
x
4 d
) d
(6x
x
2 d
) d
(12)
x
32x 3 12x 0
32x 3 12x
d
(b) h′(x) d
(4x
x
2 5x 2 4x)
d
d
(4x
x
2 d
) d
(5x
x
2 d
) d
(4x)
x
8x 3 10x 4
A polynomial function is a
sum of functions in the
form ax k , where k ∈ N. To
find the derivative of any
polynomial function, add
the derivatives of the
individual terms.
Example 2
The Derivative of Any Polynomial Function
For any polynomial function,
P(x) a n
x n a n 1
x n 1 … a 2
x 2 a 1
x 1 a 0
, where n ∈ N,
P′(x) na n
x n 1 (n 1)a n 1
x n 2 … 2a 2
x 1 a 1
.
Determining the Derivative of a Polynomial Function
Differentiate y 5x 4 2x 3 6x 2 9x 18.
Solution
d y d
dx
d x
d
d x
(5x 4 2x 3 6x 2 9x 18)
(5x 4 d
) d
(2x 3 d
) d
(6x 2 d d
) d
(9x) d
(18)
20x 3 6x 2 12x 9 0
20x 3 6x 2 12x 9
x
x
3.6 FINDING SOME SHORTCUTS — THE SUM AND DIFFERENCE RULES 227
x
x

Example 3
Using the Sum and Difference Rules to Determine an
Instantaneous Rate of Change
Recall the problem in Setting the Stage.
A cliff diver in Acapulco, Mexico, dives from about 17 m above the water.
The function f (t) 4.9t 2 1.5t 17 models the diver’s height above the
water, in metres, at t seconds. Determine the diver’s instantaneous rate of
descent with respect to time at intervals of 0.1 s during the first second the diver
is in the air.
Solution
Given f (t), which is the height at time t, the instantaneous rate of change of
height with respect to time — the rate of descent — is f ′(t). In this case,
d
f ′(t) d
(4.9t
t
2 1.5t 17)
9.8t 1.5
The table shows the rate of descent every 0.1 s during the first second.
Time, t (s) 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0
Instantaneous
Rate of Descent, 1.5 0.5 –0.5 –1.4 –2.4 –3.4 –4.4 –5.4 –6.3 –7.3 –8.3
f ′(t) = –9.8t + 1.5 (m/s)
A positive rate means the diver is moving up. A negative rate means the diver is
moving down. The rate is initially positive because the diver jumps up off the
cliff. Between 0.1 s and 0.2 s, the diver starts to fall, and the pull of gravity
steadily increases the rate of descent until the diver enters the water.
Height (m)
f(t)
18
16
f(t) = – 4.9t 2 + 1.5t + 17
14
12
10
8
6
4
2
t
0 1 2 3 4
Time (s)
Rate of Descent (m/s)
8.0
4.0
0.0
–4.0
–8.0
–12.0
–16.0
–20.0
f′(t)
t
0.5 1 1.5 2 2.5
f′(t) = – 9.8t + 1.5
Time (s)
Diver’s rate of descent versus time
Diver’s height versus time
228 CHAPTER 3 RATES OF CHANGE IN POLYNOMIAL FUNCTION MODELS

CHECK, CONSOLIDATE, COMMUNICATE
d
1. Show graphically that d
(x
x
2 d
2x) d
(x
x
2 d
) d
(2x).
x
2. If h(x) g(x) f (x), where g(x) 4x 3 and f (x) 5x 2 , does
h′(2) g′(2) f ′(2)? Explain.
3. Explain why any polynomial function can be differentiated term by term.
KEY IDEAS
The table summarizes the differentiation rules developed in this section.
Rule Function Notation Leibniz Notation
d
d
d
Sum Rule If h(x) = f (x) + g(x) and d
[f (x) + g(x)] =
x
d
[f (x)] +
x d
[g(x)]
x
f and g are both
differentiable, then
h′(x) = f ′(x) + g′(x).
d
d
d
Difference If h(x) = f (x) – g(x) and d
[f (x) – g(x)] =
x
d
[f (x)] –
x d
[g(x)]
x
Rule f and g are both
differentiable, then
h′(x) = f ′(x) – g′(x).
Derivative of If P(x) = a n
x n + a n – 1
x n – 1 + … + a 2
x 2 + a 1
x 1 + a 0
, where n ∈ N,
a Polynomial then P′(x) = na n
x n – 1 + (n – 1)a n – 1
x n – 2 + … + 2a 2
x 1 + a 1
.
3.6 Exercises
A
1. For y 3x 2 2x 1,
(a) find d y
from first principles
dx
(b) confirm your results for (a) using the rules for differentiation
2. Differentiate.
(a) y 4x 2 5x 2 (b) y x 3 5x 2 2x 8
(c) y 3x 2 (d) y 4x 2 5x 1
(e) f (x) 8x 3 (f) f (x) 3x 2 2x 5
3
(g) f (x) x 4 2 x 5 (h) f (x) 6x4 3x 3 9x 2 5x 8
3. Determine the slope of the tangent line at the given point for each function.
(a) y 3x 5 at (2, 11) (b) y 4x 2 3x 7 at (1, 14)
(c) y 2x 3 at (2, 16)
(d) y 5x 4 4x 3 3x 2 6x 2 at (0, 2)
(e) y 5x 2 6x 3 at (0, 3) (f) y 7x 4 x 2 6x at (1, 2)
3.6 FINDING SOME SHORTCUTS — THE SUM AND DIFFERENCE RULES 229

B
4. (a) Find the equation of the tangent to the curve of y 2x 2 5x 7
where x 1.
(b) Draw a sketch of the function and the tangent.
5. Communication: Express in your own words the sum and difference rules
for differentiation. Use an example to illustrate each rule.
6. Knowledge and Understanding: Determine d for the function
dx
y 5x 4 8x 3 3x 2 6x 9.
7. Determine the equation of the tangent to the curve of y x 2 3x 1 at
each point.
(a) (1, 5) (b) (2, 11) (c) (0, 1) (d) 1 2 , 1 4
8. For f (x) 3x 2 8x 5 and g(x) 5x 3 4x 2 5x 7, show that
(a) the derivative of the sum equals the sum of the derivatives
(b) the derivative of the difference equals the difference between the
derivatives
9. Find the equation of the tangent to each curve.
(a) y x 2 5x 4 where x 3
(b) y 4x 2 x 5 where x 1 2
(c) f (x) x 3 5x 2 6x 7 where x 1
(d) f (x) 5x 4 x 3 6x where x 3
10. Find the equations of all the tangents to the graph of f (x) x 2 4x 25
that pass through the origin.
11. Application: Liquid is flowing out of a tank. The volume, V, in litres
remaining after t minutes is given by V(t) 1000(20 t 2 ).
(a) What is the initial volume of liquid in the tank?
(b) Over the first two minutes, what is the average rate at which the tank is
being emptied?
(c) At exactly what time is this rate in effect?
(d) How fast is the liquid leaving the tank at 3 min?
(e) How long, to the nearest half minute, will the liquid take to drain
completely from the tank?
(f) What is the average rate, to the nearest litre per minute, at which the
liquid drains?
12. A business report determines that a company’s profit, P, in dollars per
month can be expressed as a function of the number of items
manufactured, x:
P(x) x 3 32x 2 560x 9600, 0 ≤ x ≤ 40
(a) Explain why the y-intercept of the graph is negative.
y
230 CHAPTER 3 RATES OF CHANGE IN POLYNOMIAL FUNCTION MODELS

(b) At what rate is the profit changing when 15 items are manufactured?
35? 26?
(c) What is the profit when 15 items are manufactured? 35? 26?
(d) For what levels of production is the company profitable?
13. Kathy has diabetes. Her blood sugar level, B, one hour after an insulin
injection, depends on the amount of insulin, x, in milligrams injected.
B(x) 0.2x 2 500, 0 ≤ x ≤ 40
(a) Find B(0) and B(30).
(b) Find B′(0) and B′(30).
(c) Interpret your results.
(d) Consider the values of B′(50) and B(50). Comment on the significance
of these values. Why are restrictions given for the original function?
14. (a) Find coordinates of the points, if any, where each function has a
horizontal tangent line.
i. f (x) 2x 5x 2
ii. f (x) 4x 2 2x 3
iii. f (x) x 3 8x 2 5x 3
(b) Suggest a graphical interpretation for each of these points.
15. Thinking, Inquiry, Problem Solving: Find numbers a, b, and c so that the
graph of f (x) ax 2 bx c has x-intercepts at (0, 0) and (8, 0), and a
tangent with slope 16 where x 2.
16. The population, P, of a bacteria colony at t hours can be modelled by
P(t) 100 120t 10t 2 2t 3
(a) What is the initial population of the bacteria colony?
(b) What is the population of the colony at 5 h?
(c) What is the growth rate of the colony at 5 h?
17. Coffee consumption in the United States can be modelled by
C(x) 2.767 75 0.084 794 3x 0.008 320 58x 2 0.000 144 017x 3 ,
where C represents the number of cups consumed per day by the average
adult and x represents the number of years since 1955.
(a) How many cups of coffee did the average American adult consume each
day in 2000?
(b) What was the rate of change in the number of cups of coffee consumed
per adult per day in 2000?
18. Check Your Understanding
(a) Determine f ′(3), where f (x) 6x 3 4x 5x 2 10.
(b) Give two interpretations of the meaning of f ′(3).
3.6 FINDING SOME SHORTCUTS — THE SUM AND DIFFERENCE RULES 231

C
19. Prove that the tangent to the curve y x 2x 2 x 4 at point (1, 0) is
also tangent to the curve at (1, 2).
20. For any point (x, y) on the graph of y 3x 2 15x 3, determine the
coordinates of the point on y 2x 2 5x 4 such that the tangents at the
two points are parallel.
21. Show that f (x) 8x 3 7x 5 has no tangent line with slope 5.
22. Show that the x- and y-intercepts for any tangent to the curve
y 16 8x x have a sum of 16.
ADDITIONAL ACHIEVEMENT CHART QUESTIONS
Knowledge and Understanding: Consider the function f (x) x 4 7x 12.
(a) Write the equation of the tangent line at x 1.
(b) Find the coordinates of the point on the function so that the tangent line is
perpendicular to x 25y 175 0.
Application: An ant colony was treated with an insecticide and the number of
survivors, A, in hundreds at t hours is A(t) t 3 5t 750.
(a) Find A′(t).
(b) Find the rate of change of the number of living ants in the colony at 10 h.
(c) How many ants were in the colony before it was treated with the insecticide?
(d) How many hours after the insecticide was applied were no ants remaining in
the colony?
Thinking, Inquiry, Problem Solving: Suppose that f (x) is a quadratic polynomial
function. Where would you find a horizontal tangent line? a vertical tangent
line?
Communication: You are writing a self-help manual in calculus. Write the steps
for helping readers find the equation of the line that is tangent to the graph of
g(x) x 4 2x 3 5x 2 where x 2. Explain each step.
The Chapter Problem
Average Salaries in Professional Sports
Apply what you learned to answer this question about The Chapter
Problem on page 168.
CP9.
Use the cubic regression equations you created for question CP7.
Then apply the differentiation rules to verify the derivatives you
found for question CP7.
232 CHAPTER 3 RATES OF CHANGE IN POLYNOMIAL FUNCTION MODELS

3.7 Polynomial Function Models
and the First Derivative
SETTING THE STAGE
EXAMINING THE CONCEPT
Displacement
f(t)
secant
s = f(t)
tangent
Time (s)
Displacement versus time
This chapter has discussed using the derivative to determine the instantaneous
rate of change of a quantity. The derivative has wide application in fields such as
economics, science, engineering, and the social sciences. Here is an example:
Suppose a rock is thrown into the air from a bridge 15 m above the water.
As it rises and then falls, its height above the water is a function of the time
since it was thrown. The height of the rock, in metres, above the water at
t seconds can be modelled by the function h(t) 4.9t 2 12t 15.
What is the velocity of the rock when it enters the water?
In this section we will examine applications of the first derivative of a
polynomial model.
Polynomial Models Involving Velocity and Speed
In many situations, an object’s position, s, can be described by a
function of time, s f (t). Average velocity is defined as the rate of
change of displacement over an interval of time. Instantaneous velocity
is the rate of change of displacement at a specific point in time.
On a displacement-time graph, the slope of a secant represents
average velocity, while the slope of a tangent represents instantaneous
velocity.
t
average velocity ∆
∆t
instantaneous velocity lim
s
∆
s
∆t → 0 ∆t
lim
∆t → 0
lim
h → 0
f ′(t)
As a result, the derivative of the position function, s′ f ′(t), represents the
instantaneous velocity of the object at time t. So
s
v(t) d
dt
s′
f ′(t)
f (t ∆t) f (t)
∆t
f (t h) f (t)
h
3.7 POLYNOMIAL FUNCTION MODELS AND THE FIRST DERIVATIVE 233