5.

Rates of
change
• The gradient of a curve
• Calculating the gradient at a point on
a curve
• General statement of this ‘limiting chord’
process
• Notation
• Finding the equation of a tangent at
a point on y = ax n
• Differentiating f (x) ± g(x)
• Differentiating more general power
functions
• Miscellaneous exercise five
Situation One
Research scientists are testing a new, lightweight
alloy for its possible use in car, train and aeroplane
bodies. In one test a vehicle made of the alloy is

Shutterstock.com/Niv Koren
propelled along a straight horizontal railway in
such a way that the distance, y metres, that the
vehicle has travelled t seconds after it started
is given by
y = t3.
Unfortunately the vehicle starts to break up 8 seconds after it started.
The scientists knew that at that instant the vehicle was 512 metres from its starting point (512 = 83).
For those 8 seconds the average speed was given by:
distance travelled in the 8 seconds 512 m
=
time taken 8s
= 64 m/s = (230.4 km/h).
However the scientists want to know the speed the
vehicle was travelling at the instant that it started
to break up.
Try to determine the speed of the vehicle at this
instant (i.e. at t = 8).

Situation Two
It is the year 2035 and plans are well advanced for

Shutterstock.com/Nasa/Castleski
the building of a space station on the Moon. The
space station will be pressurised and will act as a
lunar laboratory and repair depot for space vehicles
servicing the various telecommunication and
surveillance satellites. The space station will obtain
its power from thousands of solar tiles on the roof.
Health and Safety experts are concerned that tiles dislodged from the roof could fall on astronauts
working outside the station and damage their space suits. They want tests to be carried out on Earth
to ensure that the space suits are strong enough to withstand the impact.
For such tests to be carried out, the speed any dislodged tiles will have at the instant they reach the
Moon’s surface is required. It is known that the tiles will fall from a height of 20 metres. Due to the
Moon’s gravitational pull any tile will have fallen y metres, t seconds after it is dislodged, where
y = 0.8t2.
• Calculate the value of t at the instant a dislodged tile hits the surface of the Moon.
(Take the time the tile is dislodged as t = 0.)
• Calculate the average speed of the tile during its fall.
• Calculate the speed of the tile at the instant it strikes the surface of the Moon.

ISBN 9780170390408 5. Rates of change 85

 The two situations on the previous page each involved finding the speed of an object at an instant,
i.e. the instantaneous speed. Perhaps with the first situation, having found the average speed for the
motion from t = 0 to t = 8 you then considered the average speed for the motion from t = 7 to t = 8,
and then perhaps you considered … etc.
Speed is the rate at which an object changes its position with respect to time. The two situations
required us to find the rate at which the distance variable (y) was changing with respect to the time
variable (t), at a particular instant.
Graphically, the rate of change of one variable, y, with respect to another variable, x, is the gradient of
the graph of the relationship. Therefore if either of the situations on the previous page had involved
linear functions it would have been an easy matter to determine the gradient by comparison with the
form y = mx + c, or in the situations given, y = mt + c. However neither y = t3 nor y = 0.8t2 are of this
form so, as we know, neither have straight line graphs. Therefore, if we wish to pursue this gradient
idea, we first need to think about what we might mean by the gradient of a curve.

The gradient of a curve

In the linear relationship shown graphed on the right, each unit y
increase in x sees a 3 unit increase in y. 8
y = 3x – 1
7
The straight line has a constant gradient, or slope, of 3. 6
5
We could say that the rate of change of y with respect to x is 3.
4
If the two variables are not linearly related the gradient, or 3
3
slope, is not constant. We must then refer to the gradient 2
1
at a particular point. 1

–2 –1 1 2 3 4 5 x
–1 (0, –1)
We define the gradient at some point P on the curve y = f (x) to –2
be the gradient of the tangent to the curve at the point P.

y y = f (x)
The tangent at P is the line that ‘just touches’ the curve at that
point (except if P is a point of inflection as we will see on the
next page). f (a) P
If we can determine the gradient of this tangent we know the
gradient of the curve at the point P and hence the rate of change
of y with respect to x at x = a.
a x

Consider the graph shown on the right and in particular consider y

the positive or negative nature of the gradient of this curve at the
points P, Q, R, S and T. Q

P R T

S
x

86 MATHEMATICS METHODS  Unit 2 ISBN 9780170390408

The diagram on the right now has the tangents to the curve y
at points P, Q, R, S and T drawn.
Q
From this we can notice the following. R
P T
At points P and T the gradient is positive.
S
(Uphill for increasing x.) x
At point R the gradient is negative.
(Downhill for increasing x.)
At points Q and S the gradient is zero. (The tangent is horizontal.)
Between points P and Q we say the function is increasing (a positive gradient).
Between points Q and S we say the function is decreasing (a negative gradient).
Between points S and T the function is increasing (a positive gradient).
The Preliminary work section at the beginning of this book reminded us of some other vocabulary used
to describe some key features of the graphs of functions. The following dot points expand on some of
this vocabulary, again referring to the graph shown above.
• In the graph the point Q is a maximum turning point and point S is a minimum turning point.
The gradient of the curve is zero at these two points. (The tangents at these points are horizontal.)
• We can also refer to point Q as a local maximum point (sometimes referred to as a relative
maximum point). There may be points on the graph that are higher than Q but in the locality
of point Q it is the highest point.
• Looking at the section of the graph displayed the highest point overall is at the right hand end.
This would be the global maximum for the section shown.
• Similarly point S is a local minimum point (or relative minimum point) and, for the section of
graph shown, S is also the global minimum.
• From the extreme left of the display to point R the curve is concave down.
• From the point R to the extreme right of the display the curve is concave up.
• Point R, where the concavity changes, is a point of inflection. (Notice that at this point the tangent
line actually cuts the curve.)
• If the graph were to continue its upward path as x increases then as x gets large positively (we say
‘as x tends to infinity’ and we write ‘x → ∞’) then y also gets large positively.
We write: As x → +∞ then y → +∞.
And similarly: As x → −∞ then y → −∞.
• Local maximum and local minimum points are sometimes referred to as turning points. At all such
points the gradient is zero.
• Maximum points, minimum points and points of horizontal inflection are sometimes referred to as
stationary points.

Turning Points
or

Minimum Maximum Horizontal

point point inflection

Stationary Points

ISBN 9780170390408 5. Rates of change 87

 Exercise 5A
1 For the function on the right, points A and I are end f (x)
F
points, points B, D, F and H are stationary points and B
G
C, E, G and H are points of inflection.
C E
A H
a Between which points is the function increasing, x
i.e. a positive gradient? (Give your answer in the D I
form J → K, M → N, etc.)
b Between which points is the function decreasing?
c At which points is the gradient zero?

2 For each of the statements I → X state the letters of those graphs A → H for which the statement
is true.
I The gradient is zero at least once.
II The gradient is always positive.
III The gradient is always negative.
IV The gradient is never negative.
V The gradient is constant.
VI The gradient is zero exactly twice.
VII For all negative x values the gradient is positive.
VIII The gradient is positive as x gets very large positively. (I.e. x → ∞.)
IX The gradient is positive as x gets very large negatively. (x → −∞.)
X The gradient is negative when x = 0.

A B
y C
y y

x x x

D E
y F
y y

x x x

G H
y y

x x
Line is
momentarily
horizontal.

88 MATHEMATICS METHODS  Unit 2 ISBN 9780170390408

 3 For the graph below state
a which six of the points A → P are places where the gradient is zero
b which six of the points A → P are places where the gradient is positive
c which four of the points A → P are places where the gradient is negative.

K
C y

L P
B D J
O
E
M
A F N
G I x
H

4 The diagram below shows the graph of y = x2 with the tangents to the curve drawn at the point
(1, 1) and the point (2, 4).

y y = x2
9
8
7 Tangent at point (2, 4)
6
5
4 Tangent at point (1, 1)
3
2
1

x
–3 –2 –1 1 2 3
–1

a Use the graph to suggest the gradient of y = x2 at the point (1, 1).
b Use the graph to suggest the gradient of y = x2 at the point (2, 4).
c Use the graph to suggest the gradient of y = x2 at the point (0, 0).
d Suggest the gradient of y = x2 at the point on the curve where x = −1.
e Suggest the gradient of y = x2 at the point on the curve where x = −2.
f Suggest the gradient of y = x2 + 3 at the point where x = 1.
g Suggest the gradient of y = (x − 2)2 at the point where x = 3.

ISBN 9780170390408 5. Rates of change 89

 5 Sketch the graph of a function that satisfies all of the conditions stated below.
(You do not need to determine the equation of such a function.)
• The function cuts the x-axis at (1, 0) and (4, 0) and nowhere else.
• The gradient of the function is zero for x = 2.5.
• For x < 2.5 the gradient is always positive.
• For x > 2.5 the gradient is always negative.

6 Sketch the graph of a function that satisfies all of the conditions stated below.
(You do not need to determine the equation of such a function.)
• The function cuts the x-axis at (0, 0) and nowhere else.
• The gradient of the function is zero for x = 2.
• For x < 2 and for x > 2 the gradient is always positive.

7 Sketch the graph of a function that satisfies all of the conditions stated below.
(You do not need to determine the equation of such a function.)
• The function cuts the x-axis at (−2, 0), (1, 0), (6, 0) and nowhere else.
• The gradient of the function is zero for x = −1, x = 3 and x = 5.
• For x < −1 and for x > 5 the gradient is always positive.
• For −1 < x < 3 and 3 < x < 5 the gradient is always negative.

WS
Calculating the gradient at a point on a curve
Now that we know what we mean by the gradient of a curve, how do we determine its value at various
Rates of change:
gradients of secants points on a curve? Well one way would be to draw the tangent to the curve at those points and estimate
its gradient, as in one of the questions of the previous exercise. However, drawing the tangent accurately
is difficult and deciding exactly which straight line is the tangent at a particular point involves a certain
amount of guesswork. So how do we calculate the gradient at a particular point accurately?
To answer this question let us return to the idea mentioned after the two situations at the beginning of
this chapter. It was suggested there that to determine the rate of change of the curve y = t3, at the point
where t = 8, you perhaps considered the rates of change of intervals closer and closer to t = 8. Let us try
this approach to determine the gradient of y = x2 at various points on the curve.
Consider the graph of y = x2. y

The tangent drawn through (0, 0) will be the x-axis and this has a gradient
y = x2
of zero.
Thus the gradient of y = x2 at x = 0 is zero but what will be the
gradient of y = x2 at x = 1, 2, 3, 4, 5, …?
x
2
For y = x x 0 1 2 3 4 5 …
gradient 0 ? ? ? ? ? ???

90 MATHEMATICS METHODS  Unit 2 ISBN 9780170390408

To determine the gradient at x = 1 we first determine the y
gradient of the chord PQ where P is the point (1, 1) and y = x2 10

Q is some other point on the curve. We then move Q to Q

8
positions Q1, Q2, Q3, …, each position being closer to P
Q1
than the previous position, and determine the gradient of 6
the chord in each case. As Q gets closer and closer to P Q2
then so the gradient of PQ will be a better and better 4
approximation of the gradient of the tangent at P. Q3
2
P (1, 1)

x
–3 –2 –1 1 2 3
–1

This process is shown tabulated below:

Point P Point Q Gradient of chord PQ

9−1
(1, 1) (3, 9) =4
3−1

(1, 1) (2, 4) 4 −1
=3
2−1

(1, 1) (1.5, 2.25) 2.25 − 1

= 2.5
1.5 − 1

(1, 1) (1.1, 1.21) 1.21 − 1

= 2.1
1.1 − 1

(1, 1) (1.05, 1.1025) 1.1025 − 1

= 2.05
1.05 − 1

(1, 1) (1.01, 1.0201) 1.0201 − 1

= 2.01
1.01 − 1

(1, 1) (1.001, 1.002 001) 1.002001 − 1

= 2.001
1.001 − 1

As Q approaches P the gradient of PQ approaches 2.

We say that the limit of the gradient of PQ, as Q approaches P, appears to be 2.
This suggests that the gradient of y = x2 at x = 1 is 2.
Thus we now have:

For y = x2 x 0 1 2 3 4 5 …
gradient 0 2 ? ? ? ? ???

The first two questions of the next exercise involve determining more of the unknowns in this table.

ISBN 9780170390408 5. Rates of change 91

 Exercise 5B
1 Complete the following table to find the gradient of y = x2 at the point P(2, 4).

Point P Point Q Gradient of chord PQ

16 − 4
(2, 4) (4, 16) =?
4−2

?−?
(2, 4) (3, 9)  = ?
?−?

(2, 4) (2.5, ???) ?

(2, 4) (2.1, ???) ?
(2, 4) (2.01, ???) ?
(2, 4) (2.001, ???) ?
(2, 4) (2.0001, ???) ?

Thus the gradient of y = x2 at x = 2 is ???.

2 Repeat the ‘limiting chord’ process used in Question 1 to determine the gradient of y = x2 at (3, 9),
(4, 16) and (5, 25) and hence, together with your answer from number 1, copy and complete the
following table.

For y = x2 x 0 1 2 3 4 5
gradient 0 2 ? ? ? ?

Use your table to suggest a rule for determining the gradient of y = x2 at some point (a, a2).

3 Repeat the ‘limiting chord’ process to determine the gradient of y = 3x2 at (2, 12), (3, 27) and
(4, 48) and hence copy and complete the table below.

For y = 3x2 x 0 1 2 3 4 5
gradient 0 6 ? ? ? 30

Use your table to suggest a rule for determining the gradient of y = 3x2 at some point (a, 3a2).
Shutterstock.com/Guitar Photographer

92 MATHEMATICS METHODS  Unit 2 ISBN 9780170390408

General statement of this ‘limiting chord’ process WS

Let us again consider the process we go through to find the

Finding derivatives
gradient at a particular point, P, on a curve y = f  (x). Q from first principles
f (x + h)
We choose some other point, Q, on the curve whose
x-coordinate is a little more than that of point P. Suppose P WS

has an x-coordinate of x and Q has an x-coordinate of (x + h). P

f (x) Limits
h
The corresponding y-coordinates of P and Q will then be f (x)
and f (x + h).
x x+h
f ( x + h) − f ( x )
Thus the gradient of PQ =  .
h
This gives us the average rate of change of the function from P to Q.
For example the average rate of change of the function y = x2 from P (3, 9) to Q(4, 16) is given
16 − 9
by  = 7.
4 −3
We then bring Q closer and closer to P, i.e. we allow h to tend to zero, and we determine the limiting
value of the gradient of PQ.
f ( x + h) − f ( x )
i.e. Gradient at P = limit of   as h tends to zero.
h
We write this as:

f ( x + h) − f ( x )
Gradient at P(x, f  (x)) = lim
h→0 h

This gives us the instantaneous rate of change of the function at P using an algebraic approach,
rather than having to create tables as we did earlier.
For example the instantaneous rate of change of the function y = x2 at the point P(3, 9) is given by

f (3+ h ) − f (3) (3+ h ) − (3)

2 2
lim = lim
h→0 h h→0 h
9 + 6h + h 2 − 9
= lim
h→0 h
6h + h 2
= lim
h→0 h
= lim (6 + h )
h→0

=6 because h → 0 then so (6 + h ) → 6

Does this agree with the answer you obtained numerically in Question 2 of the previous exercise?

ISBN 9780170390408 5. Rates of change 93

 This algebraic method for determining the gradient at a point on f (x) = x2 is certainly a quicker
process than creating the tables that we did earlier. However, rather than using this quicker algebraic
method each time we want to determine the instantaneous rate of change at some particular point on
y = x2, we could instead apply the technique once for the general point (x, x2), obtain a formula for the
gradient, and then apply this formula each time.
Consider the general point P(x, x2) lying on the function f (x) = x2.
Applying the general result obtained previously:

( x + h )2 − x 2
Gradient at P ( x , x 2 ) = lim
h →0 h
x 2 + 2x h + h 2 − x 2
= lim
h →0 h
2x h + h 2
= lim
h →0 h
= lim (2x + h )
h →0

= 2x because as h → 0 then so (2x + h ) → 2x .

Thus for the curve y = x2 the gradient formula or gradient function is 2x.
Does this agree with your answers and suggested rule for Exercise 5B Question 2?
This process of determining the gradient formula or gradient function of a curve or function is called
DIFFERENTIATION (part of the branch of mathematics known as CALCULUS).
If we differentiate y = x2 with respect to the variable x, we obtain the gradient function 2x.
We say that 2x is the derivative of x2.
Similarly:
• If we differentiate y = t2 with respect to the variable t we obtain the gradient function 2t.
• If we differentiate z = y2 with respect to the variable y we obtain the gradient function 2y.
• If we differentiate v = z2 with respect to the variable z we obtain the gradient function 2z, etc.
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94 MATHEMATICS METHODS  Unit 2 ISBN 9780170390408

Exercise 5C
On the previous page the result
f ( x + h) − f ( x )
Gradient at P(x, f (x)) = lim
h→0 h

was used to determine the gradient function of y = x2 as 2x.

Use this same procedure to prove the following results.

1 The gradient function of y = 4x2 is 8x.

2 The gradient function of y = 2x3 is 6x2.

3 The gradient function of y = x4 is 4x3.

The results given in Exercise 5C, and that you should have found in Exercise 5B, suggest that if
y = axn then the gradient function is anxn − 1.
In words this general statement can be remembered as:
‘multiply by the power and decrease the power by one’
This ‘suggested’ general statement is indeed true but can we prove it? Well, to do so we simply
have to go back to the basic principle that

f ( x + h) − f ( x )
Gradient at P(x, f (x)) = lim
h→0 h
and apply it to the function y = axn.

However, before turning the page and seeing it done for you, try it yourself first.
One result that you may find useful is the binomial expansion, a result you were reminded of in the
Preliminary work section at the beginning of this book:
(p + q) n = p n + n C1 p n − 1q1 + n C 2  p n − 2q2 + n C3p n − 3q3 + … + n C n p0q n
An alternative approach would be to use another result that was mentioned in the Preliminary work:
pn − qn = (p − q)(pn − 1 + pn − 2q + pn − 3q2 + pn − 4q3 + … pqn − 2 + qn − 1)
Have a go!

ISBN 9780170390408 5. Rates of change 95

 Consider some general point P(x, axn) on f (x) = axn.

f ( x + h) − f ( x )
The gradient at P ( x , ax n ) = lim
h→0 h
a ( x + h ) − ax n
n
= lim [1]
h→0 h

= lim
a ( ( x + h) n
− xn )
h→0 h

= lim
(
a x + C1x n − 1h + nC 2 x n − 2h 2 +  + h n − x n
n n
)
h→0 h

= lim
a ( n
C1x n −1 n
h + C2x n−2 2
h + + hn )
h→0 h
(
= lim a C1x
h→0
n n −1
+ a C 2 x n − 2h + a nC3x n − 3h 2  + ah n − 1
n
)
= a nC1x n − 1
= anx n −1

The reader is left to show that the same result can be arrived at by applying the rule given at the
bottom of the previous page to equation [1] above.

Notation
In the expression
f ( x + h) − f ( x ) f (x + h)
Q
lim ,
h→0 h
‘h’ is a small increment in the variable x and
P
f (x)
[f (x + h) − f (x)] h

is the corresponding small increment in the variable y.

x x+h
Denoting this small increment in x as δx, where ‘δ’ is a Greek
letter pronounced ‘delta’, and the small increment in y as δy,
we have:
f ( x + h) − f ( x )
Gradient function = lim
h→0 h
δy
= lim
δ x →0 δ x

dy
which we write as (pronounced ‘dee y by dee x’).
dx

96 MATHEMATICS METHODS  Unit 2 ISBN 9780170390408

 dy
Thus if y = x2, then , the gradient function, is 2x.
dx
dy
If y = x3, then , the gradient function, is 3x2.
dx
dy
If y = x4, then , the gradient function, is 4x3.
dx

dy
If y = axn then , the gradient function, is anxn − 1.
dx

This can also be written as

d
dx
d
( y ) = ax n
dx
( )
= anx n − 1.
We say ‘dee by dee x’ of axn is anxn − 1. I.e. the derivative of axn is anxn − 1.

EXAMPLE 1
Determine the gradient function for each of the following.
a y = 3x2 b y = 7x3 c y = 2x5
d y = 3x e y=7

Solution
y = 3x2 then
a If y = 7x3 then
b If y = 2x5 then
c If
dy dy dy
= 3(2)x2 − 1 = 7(3)x3 − 1 = 2(5)x5 − 1
dx dx dx
= 6x = 21x2 = 10x4

d If y = 3x (i.e. 3x1) then y = 7 (i.e. 7x0) then

e If
dy dy
= 3(1)x1 − 1 = 7(0)x0 − 1
dx dx The derivatives can also be
=3 = 0 obtained from some calculators.
(as expected because (as we would expect
y = 3x is a straight line because y = 7 is a
horizontal line) d
with gradient 3) (3x2)
dx
6·x
d
(7x3)
dx
21·x2
d
(2x5)
dx
10·x4
d
(3x)
dx
3
d
(7)
dx
0

ISBN 9780170390408 5. Rates of change 97

 EXAMPLE 2
Determine the gradient of the curve y = 3x4 at the point (2, 48).
Solution
If y = 3x4
dy
then = 12x3.
dx
At (2, 48), x = 2.
d
(3·x4x=2)
dy dx
Thus = 12 (2)3 96
dx
= 96.
The gradient of y = 3x4 at (2, 48) is 96.

TECHNOLOGY

Get to know the capability of your calculator with regard to finding the derivative of a function
and of finding the value of the derivative for a specific x value. However make sure that if the
course requires it you can also determine derivatives, and gradients at a point, yourself, without
access to a calculator.

Note: For the moment we are differentiating functions of the form y = axn for n a non-negative
integer. Later in this chapter we will consider more general polynomial functions which,
as the reader should know, are of the form
f  (x) = an xn + an – 1xn − 1 + an – 2 xn − 2 + … + a2 x2 + a1x + a0
where n is a non-negative integer and an, an – 1, an – 2, … are all numbers, called the coefficients
of xn, xn − 1, xn − 2 etc.
The highest power of x is the order of the polynomial.
Thus linear functions, y = mx + c, are polynomials of order 1,
quadratic functions, y = ax2 + bx + c, are polynomials of order 2,
cubic functions, y = ax3 + bx2 + cx + d, are polynomials of order 3, and so on.
Shutterstock.com/Matt Gore

98 MATHEMATICS METHODS  Unit 2 ISBN 9780170390408

 EXAMPLE 3
Find the coordinates of any points on the curve y = x3 where the gradient is 12.
Solution
dy
If y = x3 then = 3x2
dx
Thus we require points for which 3x2 = 12,
i.e. x2 = 4
giving x = 2 or x = −2
3
If x = 2, y = 2 and if x = −2, y = (−2)3
= 8
= −8
Thus y = x3 has a gradient of 12 at (2, 8) and (−2, −8).

Note:
dy df d
• If y = f (x) then the derivative of y with respect to x can be written as , or  f (x).
dx dx dx
(This last version is pronounced: ‘Dee by dee x of eff of x’.)
• A shorthand notation using a ‘dash’ is sometimes used for differentiation with respect to x.
dy
Thus if y = f (x) we can write as f  ′(x) or simply y ′ or f  ′.
dx

EXAMPLE 4
Determine f  ′(x) for
a f (x) = 7x5 b f (x) = 20 c f (x) = 6x9

Solution
a If f (x) = 7x5, b If f (x) = 20, c If f (x) = 6x9,
then f  ′(x) = 35x4 then f  ′(x) = 0 then f  ′(x) = 54x8
Shutterstock.com/Gray wall studio

ISBN 9780170390408 5. Rates of change 99

 WS
Finding the equation of a tangent at a point on
Slopes of curves
y = axn
EXAMPLE 5
Find the equation of the tangent to y = 0.5x3 at the point (2, 4).

Solution
Find the derivative either algebraically or by calculator.
d
dy (0.5x3)x=2
3
If y = 0.5x then = 1.5x2 dx
6
dx
dy
Thus at (2, 4), = 1.5(2)2
dx
=6
Thus the gradient of the curve y = 0.5x3, at the point (2, 4), is 6.
Thus the gradient of the tangent to y = 0.5x3, at the point (2, 4), is also 6.
The tangent is a straight line and has an equation of the form y = 6x + c.
But (2, 4) lies on this tangent \ 4 = 6(2) + c
giving c = −8
The tangent to y = 0.5x3 at the point (2, 4) has equation y = 6x − 8.

Some calculators and internet programs are able to determine the equation of a tangent at a point on a
curve directly, given the appropriate instructions. Whilst you are encouraged to explore this capability of
such programs make sure you can carry out the process shown in the above example yourself.

Though we have been concentrating on finding the gradients at points on various curves it is important
to remember that the gradient tells us the rate at which one variable is changing with respect to
another. Rates of change are important in everyday life.
Differentiation can be used to find:
• the rate at which a vehicle is changing its position with respect to time, i.e. the vehicle’s speed.
• the rate of change in the population of a country.
• the rate of change in the number of people suffering a disease.
• the rate of change in the value of one currency with respect to another.
• the rate of change in the total profit we get from a particular item with respect to the unit cost of
that item.
Etc.

100 MATHEMATICS METHODS  Unit 2 ISBN 9780170390408

In the remainder of this chapter we will concentrate on improving our ability to differentiate various
functions. In the next chapter we will apply these skills to some real-life rate of change situations.

Exercise 5D
dy
Determine the gradient function for each of the following.
dx

1 y = x2 2 y = x3 3 y = x 4 y = x4

5 y = 3 6 y = 6x2 7 y = 6x4 8 y = 7x

9 y = 16x 10 y = 2x7 11 y = 7x2 12 y = 9x

x2 2x 6 3x 6 2x 7
13 y = 14 y = 15 y = 16 y =
10 3 2 7

Differentiate each of the following with respect to x.

17 4x2 18 5x4 19 8x3 20 9

21 x7 22 4x6 23 9x2 24 5x

Determine f  ′(x) for each of the following.

25 f (x) = 5 26 f (x) = 6x3 27 f (x) = 8x4 28 f (x) = 3x5

29 f (x) = x6 30 f (x) = 6x7 31 f (x) = 4x4 32 f (x) = 10x

Determine the gradient of each of the following at the given point.

33 y = 2x2 at the point (3, 18)

34 y = 4x3 at the point (1, 4)

35 y = 4x3 at the point (−1, −4)

36 y = x5 at the point (2, 32)

37 y = 7x at the point (2, 14)

38 y = 5x2 at the point (−2, 20)

39 y = 0.25x2 at the point (4, 4)

x2
40 y = at the point (2, 0.8)
5
iStock.com/JohnnyGreig

ISBN 9780170390408 5. Rates of change 101

 Find the coordinates of the point(s) on the following curves where the gradient is as stated.

41 y = x4, gradient 4. 42 y = x3, gradient 3.

43 y = 3x2, gradient 9. 44 y = 2x3, gradient 1.5.

45 y = x6, gradient 6. 46 y = x6, gradient −6.

Find the equation of the tangent to the following curves at the indicated point.

47 y = 2x3 at the point (1, 2) 48 y = 3x2 at the point (−1, 3)

49 y = 5x2 at the point (2, 20) 50 y = 5x2 at the point (−2, 20)

x4 x3
51 y = at the point (2, 8) 52 y = at the point (6, 36)
2 6

53 If f (x) = 3x3, find

a f  (2) b f  (−1) c f  ′(x) d f  ′(2)

54 If f (x) = 1.5x2, find

a f  (2) b f  (4) c f  ′(x) d f  ′(2)

55 For y = 2x3, determine:

a by how much y changes when x changes from x = 2 to x = 5.
b the average rate of change in y, per unit change in x, when x changes from x = 2 to x = 5.
c the instantaneous rate of change of y, with respect to x, when x = 2.
d the instantaneous rate of change of y, with respect to x, when x = 5.

56 The straight line y = 8x + 16 cuts the curve y = 8x2 at two points. Find the coordinates of each
point and the gradient of the curve at each one.

57 The straight line y = 4x cuts the curve y = x3 at three points. Find the coordinates of each point
and the gradient of the curve at each one.

58 The tangent to the curve y = ax4 at the point (3, b) has a gradient of 2. Find the values of a and b.

59 The tangent to the curve y = ax3 at the point (−1, b) is perpendicular to the line y = 2x + 3.
Find the values of a and b.
(As mentioned in the Preliminary work: If two lines are perpendicular the product of their
gradients is −1.)

102 MATHEMATICS METHODS  Unit 2 ISBN 9780170390408

Differentiating f   (x) ± g (x) WS
dy
We know that if y = 3x then =3 Derivatives of
dx polynomials

dy
and if y = x2 then = 2x.
dx WS

2
It then seems reasonable to suggest that if y = 3x + x Basic differentiation

dy
then = 3 + 2x.
dx
WS
To check whether this seemingly reasonable suggestion is true, we differentiate 3x + x2 from first
Derivative of a sum
f ( x + h) − f ( x ) of terms
principles, i.e. by determining  lim .
h→0 h
Let f  (x) = 3x + x2, WS

then f  (x + h) = 3(x + h) + (x + h)2. Derivatives of linear

products
f ( x + h) − f ( x ) [3( x + h ) + ( x + h )2 ] − (3x + x 2 )
Thus lim = lim
h→0 h h→0 h
WS
3x + 3h + x 2 + 2x h + h 2 − 3x − x 2
= lim
h→0 h Tangents to a curve

= lim (3 + 2x + h )
h→0

= 3 + 2x
Though the above example only considers the particular function (3x + x2), it is in fact true that

If y = f  (x) ± g (x)
dy
then = f  ′(x) ± g ′(x)
dx

Note: • (For interest.)

If some operation, which we will call Z, applied to the functions f and g is such that
Z(  f  + g) = Z(  f  ) + Z( g)
and Z(k × f  ) = k × Z(  f  )  for k a constant
then Z is said to be a linear operator.
d dy
Hence the facts that
dx
( ky ) = k dx   for k a constant

d
and
dx
( y1 + y2 ) = dy 1 dy2
+
dx dx
confirm what may be referred to as the ‘linearity property’ of derivatives.
• (For interest.)
The use of δx and δy to represent small increments in x and y respectively, and the use of
dy
the term , is referred to as ‘Leibniz notation’ in honour of the German mathematician
dx
Gottfried Leibniz (1646−1716).

ISBN 9780170390408 5. Rates of change 103

 EXAMPLE 6
Find the gradient of y = x2 − 3x at the point (5, 10).

Solution
By calculator (typical display below): Algebraically:
If y = x2 − 3x,
d 2 dy
(x -3·x)x=5 then = 2x − 3.
dx
7 dx
Therefore, at the point (5, 10),
dy
= 2(5) − 3
dx
= 7.

The gradient of y = x2 − 3x at (5, 10) is 7.

Exercise 5E
dy
Find the gradient function for each of the following.
dx
1 y = x2 + 3x 2 y = x3 − 4x + 7 3 y = 6x2 − 7x3 + 4

4 y = 3x4 + 2x3 − 5x 5 y = 6 + 7x + x2 6 y = 6x2 − 3x

7 y = 4x2 + 7x − 1 8 y = 5x3 − 4x2 + 8 9 y = 5x4 − 3x + 11

10 y = 2x2 + 7x + 1 11 y = 5 − 3x2 + 7x 12 y = 1 + x + x2 + x3 + x4

13 y = 5 − 4x + 3x2 − 2x3 + x4

14 Find the gradient of y = x3 − 3x2 at the point (1, −2).

15 Find the gradient of y = 17 + 2x3 at the point (−2, 1).

16 Find the gradient of y = x3 − x2 − 8 at the point (3, 10).

17 Find the gradient of y = 1 + 3x − 2x3 + x4 at the point (2, 7).

18 Find the equation of the tangent to y = x2 + 3x at the point (2, 10).

19 Find the equation of the tangent to y = 2x2 − 7x at the point (5, 15).

20 Find the equation of the tangent to y = x3 − 5x2 + 14 at the point (4, −2).

21 Find the equation of the tangent to y = 5x4 − 4x5 at the point (1, 1).

22 Find the coordinates of any point on the curve y = x3 + 6x2 − 10x + 1 where the gradient is 5.

23 The curve y = x2 − 2x − 15 cuts the x-axis in two places. Find the coordinates of each of these
points and determine the gradient of the curve at each one.

104 MATHEMATICS METHODS  Unit 2 ISBN 9780170390408

 24 Find the coordinates of any point on the curve y = x2 − 7x where the gradient is the same as that of
3y = 9x − 1.

25 Find the coordinates of any point on the curve y = x3 + 3x2 − 7x − 1 where the gradient is the same
as that of y = 2x + 3.

Differentiating more general power functions

Functions of the form y = axn are called power functions and in this chapter we have considered such
functions for non-negative integer values of n. We have differentiated such functions using the fact that
dy
if y = axn then = anxn − 1.
dx
Polynomial functions, which are linear combinations of such power functions, could then be
differentiated using the fact that
dy
if y = f  (x) ± g(x) then = f  ′(x) ± g ′(x).
dx
Now let us remove the restriction that n must be a non-negative integer and consider more general
1
power functions y = axn, for example y = x and y = .
x
Though not proved here, it is the case that for negative and fractional values of n, the same rule applies, i.e,

dy
If y = axn then = anxn − 1.
dx
1
Thus if y = i.e. y = x−1,
x
dy
then = −1x−2
dx
1
= − 2, d  1
x  
1 dx  x  −1

and if y = x i.e. y = x 2, x2
1
d
dx
( )x
dy 1 − 2 1
then = x 
dx 2 2⋅ x
1
= .
2 x
It then follows that:
5
If y = 3x2 − 2x + 7 +  3 x 2 −
x2
2
i.e. y = 3x2 − 2x + 7 + x 3 − 5x−2
1
dy 2 −
= 6x − 2 + x 3 + 10x−3
dx 3
2 10
= 6x − 2 + 3 + 3
3 x x

ISBN 9780170390408 5. Rates of change 105

 EXAMPLE 7
16
Find the gradient of y = x2 + at the point (4, 20).
x

Solution
By calculator (typical display below). Algebraically:
16
If y = x2 +
d  2 16  x
 x +  x=4
dx
7 x
= x2 + 16x−1

dy
then = 2x − 16x−2
dx

Therefore, at the point (4, 20),

dy
= 2(4) − 16(4)−2
dx
=7
16
The gradient of y = x2 + at (4, 20) is 7.
x

EXAMPLE 8

Find the equation of the tangent to y = 12 x at the point (4, 24).

Solution
Either algebraically:        or by calculator:
1 1
dy −
If y = 12x 2 =
then 6x 2 d
dx dx
( )
12 x x=4
dy 1 3
= 6(4 ) 2
−
Thus at (4, 24), 
dx
6
=
4
=3

We determine that the gradient of the curve y = 12 x , at the point (4, 24), is 3.
Thus the gradient of the tangent to y = 12 x , at the point (4, 24), is also 3.
The tangent, being a straight line, will have equation of the form y = 3x + c.
But (4, 24) lies on this tangent \ 24 = 3(4) + c
giving c = 12
The required tangent has equation y = 3x + 12.

106 MATHEMATICS METHODS  Unit 2 ISBN 9780170390408

Exercise 5F
(Whilst you are encouraged to use your calculator to obtain expressions for the derivative, and to
determine its value at particular points on a curve, it is suggested that you do most of the following
questions algebraically to ensure that you can follow the basic processes without a calculator.)
dy
Determine the gradient function for each of the following.
dx

1 3
1 y = x 2 y = 3 y =
x x
1 1
4 y = 6 x 2 5 y = 6 x 3 6 y = x 3

1 1
7 y = 23 x 8 y = 9 y =
x3 x4

2 5
10 y = 11 y = 12 y = x2 + x
x3 x4

1 1
13 y = 3x2 − 4 x 14 y = x + 15 y = x2 −
x x2

3 1 1
16 y = x + 17 y = x2 + x + 1 + +
x x x2

Determine f  ′(x) for each of the following.

2
18 f (x) =
x

3
19 f (x) =
x
6
20 f (x) = 3
x
1
21 f (x) = 3
x
4
22 Find the gradient of y = − x2 at the point (2, −2).
x

1  1
23 Find the gradient of y = at the point  −2,  .
x 2  4 

1
24 Find the gradient of y = 1 − at the point (4, 0.75).
x

2
25 Find the gradient of y = 3x3 − at the point (1, 1).
x

ISBN 9780170390408 5. Rates of change 107

 26 Find the gradient of y =  3 ( x 4 )  at the point (8, 16).

2
27 Find the gradient of y = 6 3 x +  at the point (1, 8).
x3
2 16
28 Find the gradient of y = + x2 + 2 at the point (2, 9).
x x
1 1
29 Find the coordinates of the point(s) on the curve y = where the gradient is equal to − .
x 4

30 Find the coordinates of the point(s) on the curve y = x where the gradient is equal to 1.

31 Find the coordinates of any point on the curve y = x2 − 108 x where the gradient is zero.

32 Find the equation of the tangent to the curve y = x at the point (4, 2).

1
33 Find the equation of the tangent to the curve y = at the point (1, 1).
x
1
34 Find the equation of the tangent to the curve y = at the point (2, 0.25).
x2
1
35 Find the coordinates of any point on the curve y = 2x − where the gradient is the same as that of
x
16y = 41x + 6.

36 (Challenge)
Use the first principles definition

f ( x + h) − f ( x )
Gradient at P(x, f (x)) =  lim
h→0 h
1 dy 1
to show that if y= , then = − 2
x dx x

dy 1
and if y = x, then = .
dx 2 x

108 MATHEMATICS METHODS  Unit 2 ISBN 9780170390408

Miscellaneous exercise five
This miscellaneous exercise may include questions involving the work of this chapter, the
work of any previous chapters, and the ideas mentioned in the Preliminary work section at
the beginning of the book.

1 Find the value of n in each of the following.

a 5 × 5 × 5 × 5 = 5n b 24 = n c 2n = 8
d 63 × 64 = 6n e 26 × 8 = 2n f 32 × 3n = 36
g 100 × 10n = 106 h 16 × 8 = 2n i 4 × 16 = 4n
j 85 ÷ 8n = 82 k 15n = 1 l 32 × 3n × 3 = 37
m 59 ÷ 53 × 5n = 58 n 59 ÷ (53 × 5n) = 52 o 8 × 8 × 8 = 2n

2 a Find the average rate of change of the function y = x2 from the point P(4, 16) to the point
Q(5, 25).
b Find the instantaneous rate of change of the function y = x2 at the point with coordinates
(8, 64).

3 a Find the average rate of change of the function y = x3 from the point on the curve
where x = 1 to the point on the curve where x = 3.
b Find the instantaneous rate of change of the function y = −2x3 at the point on the curve with
coordinates (−2, 16).

4 If we consider our ‘ancestors’ to be our

parents, grandparents, great grandparents,
great great grandparents etc. then going

Shutterstock.com/Monkey Business Images

back one generation we each have two
ancestors in that generation, going back
two generations we each have four
ancestors in that generation, etc. (assuming
no repetition of ancestors). How many
ancestors does a person have
a in the tenth generation back?
b in the thirtieth generation back?
How many ancestors does a person have altogether if we sum the ancestors from
c the first generation back to the tenth generation back?
d the first generation back to the thirtieth generation back?

5 Differentiate each of the following with respect to x.

1
a 5 − x3 b 5x2 − 6 x c 5x2 + 6 +
x2
1
6 Find the gradient of y = at the point (0.5, 2).
x

ISBN 9780170390408 5. Rates of change 109

 7 For each of the tables shown below determine whether the relationship that exists between x and y is:
• linear, (rule can be written in the form y = mx + c),
• quadratic, (rule can be written in the form y = ax2 + bx + c),
• cubic, (rule can be written in the form y = ax3 + bx2 + cx + d ),
• exponential, (rule can be written in the form y = a × bx ),
k
• reciprocal, (rule can be written in the form y = ),
x
or none of the above five types.
For those that are one of the above five types determine the algebraic rule for the relationship in
the form ‘y = ???’.
(‘undef’ indicates that the function is undefined for that value of x.)
a x −4 −3 −2 −1 0 1 2 3 4
y 1.5 2 3 6 undef −6 −3 −2 −1.5

b x −4 −3 −2 −1 0 1 2 3 4
y 17 10 5 2 1 2 5 10 17

c x −4 −3 −2 −1 0 1 2 3 4
y −7 −4 −1 2 5 8 11 14 17

d x −4 −3 −2 −1 0 1 2 3 4
y 0.0016 0.008 0.04 0.2 1 5 25 125 625

e x −4 −3 −2 −1 0 1 2 3 4
y 12 6 2 0 0 2 6 12 20

Hint: To obtain the rule for part e consider the y values as:
−4 × −3 −3 × −2 −2 × −1 −1 × 0 0×1 1×2 2×3 3×4 4×5

f x −4 −3 −2 −1 0 1 2 3 4
y 0.0001 0.001 0.01 0.1 1 10 100 1000 10 000

g x −4 −3 −2 −1 0 1 2 3 4
y 0.25 0.5 1 2 4 8 16 32 64

h x −4 −3 −2 −1 0 1 2 3 4
y 6 8 12 24 undef −24 −12 −8 −6

i x −4 −3 −2 −1 0 1 2 3 4
y −56 0 20 16 0 −16 −20 0 56

110 MATHEMATICS METHODS  Unit 2 ISBN 9780170390408

 8 A triangle has its three angles in arithmetic progression. If the smallest angle is 10° find the size
of the other two angles.

9 A particular sequence is geometric with a common ratio of 5 and a fourth term equal to 100.
Define the sequence by stating the first term, T1, and giving Tn + 1 in terms of Tn.

10 Given that a = 2 × 107 and b = 4 × 104 evaluate each of the following, without the assistance of
a calculator, giving your answers in standard form (scientific notation).
a a × b b b × a c a3 d b 2 e b ÷ a f a÷b

11 For each of the following sequences:

Sequence 1: T1 = 5 and Tn + 1 = 3Tn + 2
Sequence 2: T1 = 0.125 and Tn + 1 = Tn × 2
Sequence 3: T1 = −5 and Tn + 1 = Tn + 10
a State the first five terms.
b State whether the sequence is arithmetic, geometric or neither of these.
c State the sum of the first five terms.
d State the eighteenth term. (Use a calculator or spreadsheet.)
e State the sum of the first 18 terms. (Use a calculator or spreadsheet.)

12 Find the equation of the tangent to the curve y = 2x3 − x + 3

a at the point (1, 4),
b at any points on the curve where the gradient is 23.

13 What do each of the following displays tell us about the rate of change of f (x) = x3 + 3x2 + 4?
a b
3 2 3 2
Define f(x)=x +3x +4 Define f(x)=x +3x +4
f ( 6 ) − f (1)  f ( 5 + h) − f ( 5 ) 
lim  
64 5 h→ 0  h
105

dy
14 One of the graphs A to D shown below has = x(x + 3). Which one?
dx
A y B y C y D y

x x x x

ISBN 9780170390408 5. Rates of change 111

 dy
15 A curve is such that = x(x + 6)(x − 6).
dx
a At how many places on the curve is the gradient zero?
b For x → ∞ is the gradient positive or is it negative?
c For x → −∞ is the gradient positive or is it negative?

16 Figure 1 below shows a child’s feeding bowl and figure 2 shows the same bowl with the shape
of the interior shown.

Figure 1         Figure 2

An unfortunate ant has found its way into the bowl and is at the bottom, hoping to get out.
However the bowl’s surface is very slippery so the ant may not be successful.
y
8

6
2 3
y = 6x – 2x
4 25 125

x
2 4 6 8 10 12

The above graph shows that the route the ant must follow, from the bottom of the bowl to
6 x 2 2x 3
the top, can be accurately modelled by part of the curve y = − .
25 125
The ant starts his (her?) climb to the top but, due to the slippery surface, will slip when the
144
gradient of the slope is .
125
Clearly showing the use of calculus and algebra, show
that this gradient occurs twice in the section of the curve
shown in the graph, stating the x-coordinate of each of Shutterstock.com/paulrommer

these points and the y-coordinate of the lower point.

Check the x-coordinates just determined using a calculator
that has the ability to differentiate functions.

112 MATHEMATICS METHODS  Unit 2 ISBN 9780170390408