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FIRST YEAR CALCULUS
W W L CHEN
c


W W L Chen, 1982, 2008
...

It is available free to all individuals, on the understanding that it is not to be used for financial gain,
and may be downloaded and/or photocopied, with or without permission from the author
...


Chapter 3
INTRODUCTION TO DERIVATIVES

3
...
Introduction
We begin by looking at a simple example
...
For 10 hours,
its distance from the point of origin is noted at hourly intervals and recorded
...
Then the average speed of the car between the 3-hour mark and the 8-hour
mark will be given by
change in distance over the time interval
s(8) − s(3)
620 − 190
=
=
= 86
length of the time interval
8−3
8−3
kilometres per hour
...

Then the table above is not of much use
...
We take the position s(3) of
the car at the 3-hour mark
...
Then we calculate the average speed
s(3 + ∆t) − s(3)
∆t
of the car over this small time interval
...
We are therefore looking at some quantity, if it exists at all, like
s(3 + ∆t) − s(3)

...
Suppose that P (a, b) is a point on the curve y = f (x)
...
We draw the line joining
the points P (a, b) and Q(x, y), and obtain the picture below
...

x−a
x−a

Now let us keep the point P (a, b) fixed, and move the point Q(x, y) along the curve towards the point P
...

y

y = f (x)

P (a, b)

x
We are interested in the slope of this tangent line
...

dx x=a
In this case, we say that the function y = f (x) is differentiable at the point x = a
...
Sometimes, when we move the point Q(x, y) along the curve y = f (x) towards the point
P (a, b), the line P Q does not become the tangent1 to the curve y = f (x) at the point P (a, b)
...
An example of such a
situation is given in the picture below
...

In this chapter, we assume that the reader has some idea of the notion of a limit of a function f (x) as
x → a
...
The three parts are
respectively called the sum, product and quotient rules for limits
...
Suppose that the functions f (x) → L and g(x) → M as x → a
...

In the remainder of this first section, we recall some well known facts concerning derivatives
...
Indeed, we do not prove
any statements in this chapter, as we have chosen not to develop the theory of limits at this point
...
Here we begin by looking at a concrete example
...
1
...
Consider the function f (x) = x2
...
Let us consider the point (2, 4) on the curve, and denote this
point by P
...
We may do
the following
...
Then the line through P
and Q has slope
x2 − 4

...
Then as Q approaches P ,
this line through P and Q approaches the tangent at P , while its slope approaches the value
x2 − 4

...

Chapter 3 : Introduction to Derivatives

page 3 of 20

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First Year Calculus

W W L Chen, 1982, 2008

We can obviously repeat the same argument with any arbitrary function f (x), and investigate whether
there is a tangent at the point (a, f (a))
...
We say that a function f (x) is differentiable at x = a if the limit
lim

x→a

f (x) − f (a)
x−a

(1)

exists
...

Example 3
...
2
...
For every a ∈ R, we have
f (x) − f (a)
=0→0
x−a
as x → a
...

Example 3
...
3
...
For every a ∈ R, we have
f (x) − f (a)
=1→1
x−a
as x → a
...

Example 3
...
4
...
For every a ∈ R, we have
xn − an
f (x) − f (a)
=
= xn−1 + xn−2 a + xn−3 a2 +
...
It follows that f 0 (a) = nan−1 for every a ∈ R
...
1
...
Consider the function f (x) =
f (x) − f (a)
=
x−a

√

√

x
...
It follows that f 0 (a) = 1/2 a for every positive a ∈ R
...
1
...
Consider the function f (x) = sin x
...
It follows that f 0 (a) = cos a for every a ∈ R
...
1
...
Consider the function f (x) = cos x
...
It follows that f 0 (a) = − sin a for every a ∈ R
...
1
...
We state, without proof, the following
important result
...
Suppose that n ∈ R is fixed and non-zero
...

Example 3
...
8
...
We have
f 0 (a) =

35 −1/36
36 a

for every positive a ∈ R
...
In Chapter 8, we shall establish the
following result
...
Suppose that the functions f (x) and g(x) are differentiable at x = a
...

g
g 2 (a)
Example 3
...
9
...
We know that
tan x =

sin x

...

2
cos a
cos2 a

Example 3
...
10
...
We know that
csc x =

1

...

sin2 a

Example 3
...
11
...

x2 + 3

We can write f (x) = g(x)/h(x), where g(x) = x3 sin x and h(x) = x2 + 3
...
It follows that
f 0 (a) =

h(a)g 0 (a) − g(a)h0 (a)
(a2 + 3)(a3 cos a + 3a2 sin a) − 2a4 sin a
=

...
We shall further write
y = f (x)
Chapter 3 : Introduction to Derivatives

and

dy
= f 0 (x)
...

An important technique in differentiation is through the use of composite functions
...

Example 3
...
12
...
To calculate the derivative dy/dx, note that we can first of all
write y = x6 + 2x3 + 1, and then differentiate to obtain
dy
= 6x5 + 6x2 = 6x2 (x3 + 1)
...
We can write y = u2 , where u = x3 + 1
...

dx

Note that
dy du
= 6ux2 = 6x2 (x3 + 1)
...

dx
du dx
Perhaps this is a coincidence
...
If we write u = f (x) = x3 + 1 and
y = g(u) = u2 , then (g ◦ f )(x) = g(f (x)) = g(x3 + 1) = (x3 + 1)2
...
As we vary x, the value u = f (x) changes at the rate of du/dx
...
It is therefore not unreasonable to expect the change in x causes a change in y at the rate
(dy/du)(du/dx)
...

PROPOSITION 3C
...
Then y is a differentiable function of x
...

dx
du dx
Example 3
...
13
...
Then we can write y = u5/2 , where u = 1 + x2 , so that
dy
5
= u3/2
du
2

and

du
= 2x
...

dx
du dx
Chapter 3 : Introduction to Derivatives

page 6 of 20

c


First Year Calculus

W W L Chen, 1982, 2008

Example 3
...
14
...
Then we can write y = sin u, where u = x5 + 3x, so that
dy
= cos u
du

and

du
= 5x4 + 3
...

dx
du dx
Example 3
...
15
...
Then we can write y = sin u, where u = v 1/2 and v = x2 + 1,
so that
dy
= cos u
du

and

du
1
= 1/2
dv
2v

and

dv
= 2x
...

dx
du dv dx
v
(x2 + 1)1/2

3
...
Stationary Points and Second Derivatives
We have indicated earlier that the derivative f 0 (a) of a function f (x) at a point x = a may be interpreted
as the slope of the tangent at the point (a, f (a)) to the curve representing the function f (x)
...
We now want to find a suitable interpretation for the
case f 0 (a) = 0
...

However, this is not quite correct
...

Definition
...

is said to have a local minimum at x = a if there is an open interval I containing
a and such that f (x) ≥ f (a) for every x ∈ I
...


Example 3
...
1
...
Since f 0 (x) = 2x
for every x ∈ R, the only stationary point is at x = 0
...
It follows that there is a local minimum at x = 0
...
2
...
Try to make a rough sketch of the graph of the function f (x) = x3
...
On the other hand, note that for every x < 0,
we have f (x) = x3 < 0 = f (0), whereas for every x > 0, we have f (x) = x3 > 0 = f (0)
...

To detect a local minimum or local maximum, the (first) derivative of a function f (x) is a useful
tool
...
On the other hand, if a
continuous function decreases before a point x = a and increases after x = a, it is reasonable to accept
the point x = a as representing a local minimum
...

Indeed, we shall establish the following result in Chapter 8
...
Suppose that I is an open interval containing a
...

(a) If f 0 (x) > 0 for every x < a in I and f 0 (x) < 0 for every x > a in I, then the function f (x) has a
local maximum at x = a
...

Example 3
...
3
...
Since f 0 (x) = 2x for every x ∈ R, it is clear
that f 0 (x) < 0 for every x < 0 and f 0 (x) > 0 for every x > 0
...

Example 3
...
4
...
Since
f 0 (x) = 6x2 − 18x + 12 = 6(x2 − 3x + 2) = 6(x − 1)(x − 2)
for every x ∈ R, it is clear that the only stationary points are at x = 1 and x = 2
...

It is easy to see that

 > 0 if x ∈ (0, 1),
f 0 (x) < 0 if x ∈ (1, 2),

> 0 if x ∈ (2, 3)
...

Example 3
...
5
...
It is not
difficult to see that f (x) is continuous everywhere
...


It follows that f (x) has a local maximum at x = 0
...

If the first derivative measures the rate of change of a function, then the second derivative measures
the rate of change of the first derivative
...

Suppose now that we have a function f (x) differentiable in an open interval containing a
...
Then we cannot have f 0 (a) > 0, otherwise f (x) is increasing
at x = a
...
It follows that we must
have f 0 (a) = 0
...
If we move a little to
the right from x = a, then clearly f (x) decreases, so that the tangent to the curve now has a negative
slope
...
It follows that if we move from a little to the
left of x = a to a little to the right of x = a, the slope of the tangent changes from positive to negative
...
This means that the second derivative must be negative
...

PROPOSITION 3E
...
Suppose further
that the function f (x) is differentiable at every x ∈ I, and that f 0 (a) = 0
...

(b) If f 00 (a) > 0, then the function f (x) has a local minimum at x = a
...
2
...
Let us return to the function f (x) = x2
...
On the other hand, we have f 00 (x) = 2 for every x ∈ R, so that
f 00 (0) > 0
...

Example 3
...
7
...
Since
f 0 (x) = 6x2 − 18x + 12 = 6(x2 − 3x + 2) = 6(x − 1)(x − 2)
for every x ∈ R, it is clear that the only stationary points are at x = 1 and x = 2
...
It follows that f (x) has a
local maximum at x = 1 and a local minimum at x = 2
...
We say that a function f (x) has a point of inflection at x = a if f 00 (a) = 0
...
2
...
Recall our discussion of the function f (x) = x3 in Example 3
...
2
...
Furthermore, we have f 0 (0) = 0
and f 00 (0) = 0
...
In fact, the function has a point of inflection here
...


3
...
Curve Sketching
In this section, we study a few important aspects of curve sketching
...
Not every step is applicable to every function
...

We use the convention y = f (x)
...
SYMMETRY
...
The two
most basic aspects of symmetry are even functions and odd functions
...

(1) A function f such that f (−x) = f (x) is called an even function
...

It is easy to see that the graphs of even functions are symmetric across the vertical axis, whereas the
graphs of odd functions are symmetric across the origin
...
3
...
Try to draw the graph of f (x) = x2 + 3
...

Example 3
...
2
...
This is an odd function
...
3
...
Try to draw the graph of f (x) = sin x
...

Example 3
...
4
...
This is an even function
...
PERIODICITY
...
We may therefore be
able to draw part of the graph, and obtain the rest by repetition
...
3
...
The function f (x) = sin x has period 2π
...
Then
complete the graph by repetition
...
LOCATING A FEW POINTS OF THE GRAPH
...
INTERCEPTS
...
This may be a simple exercise in some cases, but extremely difficult in other cases
...
It follows that the graph of the function f (x) intersects the y-axis at the point (0, f (0)),
provided that the function is defined at x = 0
...
3
...
Consider the function f (x) = x2 + 2x − 3
...

To see where the graph intersects the x-axis, we note that this happens precisely when f (x) = 0
...

Example 3
...
7
...
To see where the graph intersects the
x-axis, we have to solve the equation x2 + 2x − 3 = 0
...
It follows that
the roots are −3 and 1
...

Example 3
...
8
...

Of course, it is not absolutely crucial to locate all the points where the graph of the function intersects
the coordinate axes
...
ASYMPTOTES
...

Example 3
...
9
...
It is easy to see that the graph gets
rather close to the coordinate axes
...
It is easy
to see that the graph gets rather close to the lines x = 0 and y = 1
...

We have a horizontal asymptote y = L if
lim f (x) = L

x→+∞

or

lim f (x) = L
...


x→a−

While determining the asymptotes, we must also determine which side (possible both) of the asymptote
the graph lies
...
3
...
Try to draw the graph of f (x) = 5 + x−3
...
Also,
f (x) → 5 from above as x → +∞, and f (x) → 5 from below as x → −∞
...


x→0−

page 10 of 20

c


First Year Calculus

W W L Chen, 1982, 2008

However, there is insufficient information yet to complete the graph
...
Does the curve go up and down?
Example 3
...
11
...
This has horizontal asymptote
y = 0
...
On the other
hand, we have
lim f (x) = +∞,

x→1−

lim f (x) = −∞,

x→1+

lim f (x) = −∞

x→2−

and

lim f (x) = +∞
...
For example, we have yet to fully
understand the behaviour of the function when 1 < x < 2
...
However, it is still useful to investigate
the behaviour when x → −∞ and when x → +∞
...
3
...
Consider again the function f (x) = x3 − 3x − 1
...
Such information is important
...
STATIONARY POINTS
...
We determine where the function is increasing, where it is decreasing, and locate all
the stationary points, local maxima, local minima and points of inflection by using our knowledge of
Section 3
...

Example 3
...
13
...
Simple calculation
gives f 0 (x) = −3x−4
...
Next, note that f 0 (x) < 0 whenever
x 6= 0
...
We now
supplement our earlier effort with this extra information
...
3
...
Let us continue our investigation of the function f (x) = 1/(x − 1)(x − 2)
...

(x − 1)3 (x − 2)3

It follows that there is a stationary point at x = 3/2
...
Note that f (3/2) = −4, so that the local maximum is at the point (3/2, −4)
...

It follows that the function is increasing in the open intervals (−∞, 1) and (1, 3/2), and decreasing in the
open intervals (3/2, 2) and (2, +∞)
...

STEP 7
...
If we use the second derivative, we may also be able to
see how the tangent to the curve varies
...
On the other hand, if f 00 (a) < 0, then f 0 (x) is decreasing at
x = a, so that the slope of the curve is decreasing
...
3
...
Let us return to the function f (x) = 5 + x−3 one last time
...
It follows that
n
< 0 if x < 0,
f 00 (x)
> 0 if x > 0
...
With this
extra information, we should get a reasonably good sketch of the graph
...
3
...
Let us return to the function f (x) = 1/(x − 1)(x − 2) one last time
...


This means that the slope of the tangent increases when x < 1 and when x > 2, and decreases when
1 < x < 2
...

Example 3
...
17
...
Clearly f (x) is not even, odd or periodic, so
that Steps 1 and 2 do not apply
...
For Step 4, we note that the graph intersects the y-axis at the point
(0, 0), and that for the intersection points with the x-axis, we need to solve the equation x4 − 2x3 = 0,
with roots x = 0 and x = 2
...
For Step 5,
note that f (x) → +∞ as x → +∞ or as x → −∞, so there are no horizontal asymptotes
...
Next, let us
consider stationary points
...
It
follows that we have stationary points at x = 0 and x = 3/2
...
It
follows that the function has a point of inflection at (0, 0) and a local minimum at (3/2, −27/16)
...
Finally, note that

> 0 if x < 0,



 = 0 if x = 0,
f 00 (x) = 12x(x − 1) < 0 if 0 < x < 1,



 = 0 if x = 1,
> 0 if x > 1
...
It also shows that there is a point of inflection at (1, −1)
...


3
...
Linearization of Error and Approximation of Derivative
To motivate this section, we consider two examples
...
4
...
Consider again the function f (x) = x2
...
Clearly the value f (x) changes
to f (x + ∆x), giving rise to the error
∆f = f (x + ∆x) − f (x) = (x + ∆x)2 − x2 = 2x∆x + (∆x)2 ,
and the relative error
∆f
f (x + ∆x) − f (x)
=
= 2x + ∆x
...

∆x

Note that the first of these suggests that ∆f is essentially directly proportional to ∆x, and the second
shows that the relative error is an approximation of the derivative
...
4
...
Consider next the function f (x) = x3
...
Clearly the value f (x) changes
to f (x + ∆x), giving rise to the error
∆f = f (x + ∆x) − f (x) = (x + ∆x)3 − x3 = 3x2 ∆x + 3x(∆x)2 + (∆x)3 ,
and the relative error
f (x + ∆x) − f (x)
∆f
=
= 3x2 + 3x∆x + (∆x)2
...

∆x

Note again that the first of these suggests that ∆f is essentially directly proportional to ∆x, and the
second shows that the relative error is an approximation of the derivative
...

conditions, demonstrating
establish in Chapter 8 the

essentially directly proportional to ∆x, and with the derivative f 0 (x) as the
This estimate holds for all functions f (x) that satisfy mild differentiability
that the derivative is useful in the study of properties of a function
...


PROPOSITION 3F
...
Suppose further that f 0 (a) exists for every
a ∈ (A, B)
...


y = f (x)

A

B

To understand the Mean value theorem, it is easiest to rewrite the conclusion as
f (B) − f (A)
= f 0 (c)
...
It follows
that the theorem merely says that the tangent to the curve is sometimes parallel to this line
...

PROPOSITION 3G
...
Suppose further that f 0 (a) exists for every a ∈ (A, B)
...

(b) If f 0 (a) > 0 for every a ∈ (A, B), then f (x) is strictly increasing in [A, B]
...

Proof
...
Applying the Mean value theorem to the function f (x) in the
closed interval [x1 , x2 ], we have
f (x2 ) − f (x1 ) = (x2 − x1 )f 0 (c)
for some c ∈ [x1 , x2 ] ⊆ [A, B]
...

Example 3
...
3
...
It follows that if
0 6∈ (A, B), then we can apply Proposition 3G(c) immediately to conclude that f (x) is strictly decreasing
in [A, B]
...
However, if A ≤ x1 < 0 < x2 ≤ B, then we clearly
have f (x1 ) > f (0) > f (x2 )
...

Example 3
...
4
...
It follows
from Proposition 3G(b) that f (x) is strictly increasing in the closed interval [−π/2, π/2]
...
4
...
Consider the function f (x) = 6x + 5 cos x
...

It follows from Proposition 3G(b) that f (x) is strictly increasing in any closed interval
...
5
...
To study the limit
lim

x→a

f (x)
,
g(x)

(2)

we cannot simply write down the quotient f (a)/g(a), since this is indeterminate
...
However, a simple technique is given by the following very useful result, which we shall
state without proof
...
(L’HOPITAL’S
RULE) Suppose that f (x) → f (a) = 0 and g(x) → g(a) = 0 as
x → a
...

Chapter 3 : Introduction to Derivatives

page 14 of 20

c


First Year Calculus

W W L Chen, 1982, 2008

Example 3
...
1
...
Then f (x) → 0 and g(x) → 0 as x → 0
...
It follows from the quotient rule
that f 0 (x)/g 0 (x) → 1 as x → 0
...

Example 3
...
2
...

x4
12

To do this, let f (x) = 2 cos x + x sin x − 2 and g(x) = x4
...

Consider f 0 (x) = x cos x − sin x and g 0 (x) = 4x3 instead
...
Consider f 00 (x) = −x sin x and g 00 (x) = 12x2 instead, and note that
1 sin x
f 00 (x)
=−

...

00
x→0 g (x)
12
lim

Applying l’Hˆ
opital’s rule once, we have
f 0 (x)
f 00 (x)
1
=
lim
=−
...

x→0 g(x)
x→0 g (x)
12
lim

Remark
...
Similar techniques apply for resolving indeterminate limits like ∞/∞, and also in limiting
situations like x → a+, x → a−, x → +∞ and x → ∞
...
5
...
We shall show that
x2
= 1
...
We have f (x) → +∞ and g(x) → +∞ as x → +∞
...
Again, we have f 0 (x) → +∞ and g 0 (x) → +∞ as
x → +∞
...

x→+∞ g 00 (x)
lim

Applying l’Hˆ
opital’s rule once, we have
f 0 (x)
f 00 (x)
=
lim
= 1
...

x→+∞ g(x)
x→+∞ g (x)
lim

Example 3
...
4
...

x→0+ cot x
lim

To do this, let f (x) = x + x−1 and g(x) = cot x
...

Consider f 0 (x) = 1 − x−2 and g 0 (x) = − csc2 x instead
...
Note, however, that
f 0 (x)
1 − x−2
sin2 x
− sin2 x
...


Applying l’Hˆ
opital’s rule and using (6) and (7), we have
f (x)
f 0 (x)
= lim 0
= 1
...


3
...
Implicit Differentiation
So far, all our functions have been given by some formula which gives explicitly the value f (x) for every
x in the domain
...
The
question is to find the derivative of this function
...

The advanced reader may choose instead to proceed immediately to the next section
...
6
...
Consider the function y = f (x) described by the curve 9x2 + y 2 = 25 − x, with y > 0
...
Alternatively, we can
use the Chain rule in the following way
...
Then
dw
dw dy
dy
=
= 2y
...

dx
dx
Since
d
d
dw
dy
(9x2 + y 2 ) =
(9x2 ) +
= 18x + 2y
dx
dx
dx
dx

and

d
(25 − x) = −1,
dx

we have
18x + 2y
Chapter 3 : Introduction to Derivatives

dy
= −1
...
6
...
Consider the function y = f (x) described by the equation
2x2 y + cos y = x3
...
However,
d
d 3
(2x2 y + cos y) =
(x )
...

dx

Example 3
...
3
...


(9)

subject to the constraint

Differentiating (8) and (9) with respect to x, we obtain respectively
dz
dy
=1+2
dx
dx
and
2x + 2y

dy
= 0
...

dx

(11)

Combining (10) and (11) and eliminating dy/dx, we obtain
2x − y = 0
...
Obviously, z = 10 when x = 2, while z = −10 when x = −2
...


Chapter 3 : Introduction to Derivatives

page 17 of 20

c


First Year Calculus

W W L Chen, 1982, 2008

Problems for Chapter 3
1
...

[Hint: Use the identity α3 − β 3 = (α − β)(α2 + αβ + β 2 )
...
Suppose that f (x) and g(x) are twice differentiable at x = a
...

3
...

a) Show that f (x) is differentiable at x = a for every non-zero a ∈ R
...
]
b) Comment in view of Proposition 3D
...
Differentiate each of the following functions:
p
a) f (x) = sin−1 x + 1
x
c) f (x) =
−1
sin x
5
...

x2 + 4

dy

...

a) Find

6
...

7
...

a) Locate the four stationary points by studying f 0 (x)
...

c) By evaluating the second derivative at the four stationary points, show that f (x) has local maxima
at two of these points and local minima at the other two
...

x→+∞

x→−∞

g) Sketch the graph, clearly marking the intercepts, maxima, minima and points of inflection
...

x2 − 1
a) Find the (largest) domain of y = f (x) as a real valued function, all the stationary points and
determine their nature
...

c) Sketch the curve and find the range of y = f (x)
...
Given y = f (x) =

9
...
Sketch each of the following curves, clearly marking the maxima and minima:
x
x+1
x2 − 3
a) y = 2
b) y = 2
c) y =
x +1
x −9
x−2
Chapter 3 : Introduction to Derivatives

page 18 of 20

c


First Year Calculus

x

...

Find the composite functions (f ◦ g)(x) = f (g(x)) and (g ◦ f )(x) = g(f (x))
...

Find all the asymptotes of g(x)
...
Is g(x) a monotonic function in the interval (−1, 1); in other words,
is g(x) always increasing or always decreasing in the interval (−1, 1)? Give your reasons
...


11
...
Consider the function f (x) =
a) Show that

x2 + 3x − 3
, continuous everywhere except at x = 1
...

if x < 1,
if x > 1
...

Explain why f (x) has no points of inflection
...

We have f (0) > 0 and f (−1) < 0
...
Use the Mean value theorem to explain why there is no other real number
c < 0 such that f (c) = 0
...
You may use the following additional information:
f (x) → −∞ as x → 1− or x → −∞,
f (x) → +∞ as x → 1+ or x → +∞
...
The function y = f (x) is given implicitly by the equation 2x2 + y 3 = 9
...

a) Find the first derivative
dx
b) Find the coordinates of the point(s) on the curve where the tangent(s) is (are) horizontal
...

dx2

14
...
The function y = f (x) is defined implicitly by sin x + y 3 = 8
...
The function y = f (x) is given implicitly by x3 y + y 3 = 9
...

dx

dy
at the point (2, 1)
...
Consider the ellipse (x + 1)2 + 2(y − 1)2 = 6
...

b) Determine the slope of the tangent at the point (1, 0)
...
Use L’Hopital’s rule to find each of the following:
tan x − x
x − sin x
b) lim
a) lim
3
x→0
x→0
x
x3

Chapter 3 : Introduction to Derivatives

c) lim

x→1 x3

x3 − 3x + 2
+ x2 − 5x + 3

page 20 of 20



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