Topics:
1. Limits
2. Gradients, Tangents and Derivatives
3. Differentiation from First Principles
4. Rules for Differentiation
5. The Chain Rule
6. The Product Rule
7. The Quotient Rule
8. Maxima and Minima
9. Curve Sketching
1 Limits
The limit of a function describes the behaviour of the function as the
variable approaches a particular value. We are not concerned with
the value of the function at the point of interest, but only what value
is being approached.
Definition
If the value of the function as we approach a particular value x = a is
the same whether we are approaching from the left or the right then
the limit exists. The function need not be continuous or even defined
at the point of interest. For the function below the limit of f ( x ) as x
approaches 2 is 4.
We say,
lim f ( x ) = 4.
x →2
Examples
1. Find the limit of the function f ( x ) = x + 2 as x approaches 2.
4
Solution:
The behaviour of f ( x ) as x → 2 is shown in the table:
x
1.95
1.995
99K 2 L99
2.005
2.05
f (x)
3.95
3.995
99K 4 L99
4.005
4.05
The table shows that as x approaches 2, f ( x ) approaches 4
Alternatively, because this function is defined at x = 2, we could
substitute x = 2 into the function:
f (2) = 2 + 2 = 4
Therefore
lim ( x + 2) = 4.
x →2
2. Find limh→0 (5x2 + 2xh + h) .
Solution:
The 2xh and h terms will approach zero as h approaches zero. The
limit can be found by substituting zero for h:
lim (5x2 + 2xh + h) = 5x2 + 2 × 0 + 0
h →0
= 5x2
2
3. Find limx→1 xx−−11 .
Solution: It is not possible to find the limit by substituting x = 1
as we would then have 0 in the denominator. We can find this limit, if
we simplify the expression first1 and then substitute.
In this case
You need to know the difference of
two squares,
1
a2 − b2 = ( a − b ) ( a + b ) .
2
( x −1)( x +1)
lim x −1 = lim
x −1
x →1 x −1
x →1
Applying this gives:
x 2 − 1 = ( x − 1) ( x + 1) .
= lim x + 1
x →1
= 2.
4. Find limh→0 sin (h) /h.
Solution:
It is not possible to find the limit by substituting h = 0. But consin(h)
sider the behaviour of f (h) = h as h → 0:
h
-0.5
-0.3
-0.2
-0.1
0
0.01
0.1
0.2
0.3
0.5
f (h)
0.959
0.985
0.993
0.998
?
0.99998
0.998
0.993
0.985
0.959
It appears that
sin(h)
= 1.
h
h →0
lim
5
Exercises
Determine the following limits:
1) limx→0 (4xh + 2)
2
2) limx→0 9−4x
2h
3) limh→0 xh−
h
4) limx→0
Answers
1) 2
2) 94
3) x − 2
4) 1
tan( x )
x
2 Gradients, Tangents and Derivatives
Gradient of a Curve
In this module we are concerned with finding a formula for the slope
or gradient of the tangent at any point on a given curve y= f ( x ).
The gradient at a point on a curve is defined as the gradient of the
tangent to the curve at that point.
y −y
The formula m= x22 − x1 may be used to find the gradient of a line
1
when two points on the line, (x1 , y1 ) and (x2 , y2 ), are known2 .
Consider the gradient of the curve defined by y = f ( x ) at the point
P (ie the gradient of the tangent line AB).
This gradient cannot be calculated as only one point (the point P)
on the line is known. But the point P has coordinates ( x, f ( x ))and
the point Q has coordinates ( x + h, f ( x + h)). The gradient of the
line PQ can be calculated and this can be used to approximate the
f ( x +h)− f ( x )
gradient of AB. The gradient of PQ =
. As the value of
h
h decreases (i.e Q becomes closer to the point P), the approximation
Special cases: horizontal and vertical
lines
A line parallel to the x-axis with
equation of the form y = k (k constant),
has a gradient of zero. As a line becomes closer to vertical its gradient gets
larger and larger. A line parallel to the
y-axis with equation of the form x =
c (c constant) has a gradient which is
undefined.
2
8
of the gradient is more accurate. The value of the gradient becomes
most accurate as h approaches zero.
The gradient formula for the curve y = f ( x ) is defined as the derivative function
f ( x + h) − f ( x )
, h6=0.
h
h →0
f 0 ( x ) = lim
The derivative function f 0 ( x ) gives the slope of the tangent to the
curve f ( x ) at any point x.
Example
The derivative function of f ( x ) = 3x is f 0 ( x ) = − x32 . Find the slope of
the tangent to the curve at x = 4
3
Solution: At x = 4, f 0 ( x ) = f 0 (4) = - 432 = − 16
. The slope of the
3
tangent at x= 4 is − 16
Exercises
1. If the derivative function for f ( x ) = x3 –x is f 0 ( x ) = 3x2 −1, find the
slope of the tangent to this curve at
a) x = 2
b) x = 0
c ) x = −9
2. If the derivative function of f ( x ) = sin(x) is f 0 ( x ) = cos(x) find the
gradient of f ( x ) = sin( x ) at
a) x = 0
b) x = π2
c) x = 3.5
( x + h )2 − x 2
and hence find the slope of the tangent
h
3. Determine limh→0
to the curve y = x2 at
a) x = 2
b) x = 0
Answers
1. a)11
2. a) 1
3. limh→0
a) 4
b) − 1
c) 242
b) 0
c) − 0.94
( x + h )2 − x 2
= 2x
h
b) 0
c) − 18
c ) x = −9
3 Differentiation from First Principles
The process of finding the derivative function using the definition
f 0 ( x ) = lim
h →0
f ( x + h) − f ( x )
, h 6 =0
h
is called differentiating from first principles.
Examples
1. If f ( x ) = x2 find the derivative of f ( x ) from first principles.
f ( x + h) − f ( x )
, h 6 =0
h
( x + h )2 − x 2
= lim
h
h →0
2
x + 2xh + h2 − x2
= lim
h
h →0
2
2xh + h
= lim
h
h →0
h(2x + h)
= lim
h
h →0
f 0 ( x ) = lim
h →0
[NB:( a + b)2 = a2 + 2ab + b2 ]
= lim 2x + h
h →0
= 2x
2. Determine, from first principles, the gradient function for the curve
f ( x ) = 2x2 − x and calculate its value at x = 3.
10
f ( x + h) − f ( x )
h 6 =0
h
2( x + h)2 − ( x + h) − (2x2 − x )
= lim
h
h →0
2x2 + 4xh + 2h2 − x − h − 2x2 + x )
= lim
h
h →0
4xh + 2h2 − h)
= lim
h
h →0
f 0 ( x ) = lim
h →0
= lim 4x + 2h − 1
h →0
= 4x − 1
The gradient function is f 0 ( x ) = 4x − 1. f 0 (3)= 4(3) − 1 = 11
3. Use differentiation from first principles to find the gradient function of y = 1x
f 0 ( x ) = lim
h →0
= lim
h →0
= lim
f ( x + h) − f ( x )
, h 6 =0
h
1
1
x+h − x
h
1
1
x+h
x
x+h × x − x × x+h
h
h →0
= lim
x
− x(xx++hh)
x ( x +h)
h
h →0
= lim
h →0
= lim
h →0
x −( x +h)
x ( x +h)
h
−h
x ( x +h)
h
−1
= lim
h →0 x ( x + h )
=
[NB: a common denominator is needed to add fractions]

NB:
−h
x ( x +h)
h
1
=
h

−h

x ( x + h)
−1
x2
Exercises
Find the derivative of the following functions using differentiation
from first principles and calculate the slope of the tangent to each
curve at x=1.
1) f ( x ) = 3x
2) f ( x ) = 5x2
3) f ( x ) = 2x2 − 1x
Answers
1) f 0 ( x ) = 3, m = 3
2) f 0 ( x ) = 10x, m = 5
3) f 0 ( x ) = 4x + x12 , m = 1
4 Rules for Differentiation
It is not always convenient to use differentiation from first principles
to find a derivative function. The "rules" shown below have been
established from first principles and can be used to find derivative
functions directly.
Alternative Notations
We will also introduce alternative notations for derivatives. When y is
a function of x, i.e. y = f ( x ), the derivative function may be written
dy
as y0 or dx or f 0 ( x ).
Operational Rules
• If g( x ) = k f ( x ), where k is a constant then g0 ( x ) = k f 0 ( x )
• If f ( x ) = k where k is a constant then f 0 ( x ) = 0
• If f ( x ) = g( x ) + h( x ) then f 0 ( x ) = g0 ( x ) + h0 ( x )
Derivative of a power of x:
If y = x n , then
dy
= nx n−1 .
dx
Examples
dy
1. If y = x7 , then dx = 7x7−1 = 7x6
dy
2. If y = x2020 , then dx = 2020x2020−1 = 2020x2019
√
1
3. If y = x = x 2 , then
1 1
dy
= x 2 −1
dx
2
1 1
= x− 2
2
1
= √
2 x
(Rewrite the expression in index form before differentiating)
(We usually give the answer in the same form as the question)
12
4. If y = x12 = x −2 then
dy
= −2x −3
dx
= −2x −2−1
2
=− 3
x
And using some of the operational rules:
dy
5. If y = x3 + 7x6 then dx = 3x2 + 42x5
dy
6. If y = 4x + 10 = 4x −1 + 10 then dx = −4x −2 + 0 = − x42
4
7. If f ( x ) = x√−x1 find f 0 ( x )
f (x) =
x4 − 1
√
x
x4
1
= √ −√
x
x
7
1
= x 2 − x− 2
7 5
1 3
f 0 (x) = x 2 + x− 2
2
2
(divide through by x1/2 )
Derivatives of Some Other Functions
The following table provides derivatives for some other functions.
Function
Derivative
f (x) = ex
f ( x ) = loge x
f ( x ) = sin( x )
f ( x ) = cos( x )
f ( x ) = tan( x )
f 0 (x) = ex
f 0 ( x ) = 1x
0
f ( x ) = cos( x )
f 0 (x ) = −sin( x )
f 0 ( x ) = sec2 ( x )
Examples
dy
1. y = sin( x ) + 3x2 ⇒ dx =cos( x ) + 6x
dy
2. y = ex - 5 ⇒ dx = ex
1
3. y = 10 cos ( x ) − 2x
= 10 cos ( x ) − 5/x = 10 cos ( x ) − 5x −1 .
dy
Therefore dx = −10 sin x + 5/x2 .
Exercises
1. Differentiate the following
13
a) y = x 7
√
f) y = 4 x
1
b) y = x 5
c) y = x −19
g) 5x6
h) y = 9x −5
2. Find the derivative of
a) sin ( x ) x − cos ( x )
d) 3 sin ( x ) − 2 + 1/x
d) y = x14
e) 53
√
i)
5
x
b)
e)
10 − loge ( x )
ex
7
6 −x
b) 5x14/5
c) −19x −20
j) 3x2 + 2x
c)
tan ( x ) −
√
x
Answers
1.
a) 7x6
f) 41 x
2.
− 43
a) cos ( x ) + sin ( x )
d) − x45
e) 0
g) 30x5
h) −45x −6
√
i) −x2 5
j) 6x + 2
b) − 1x
1
c) sec2 ( x ) − 2√
x
d) 3cos ( x ) − x12
e) e6 − 7x6
x
5 The Chain Rule
The "chain rule" is used to differentiate a function which is the
composition of two simpler functions. The derivatives of functions
4
such as y = sin x3 and f ( x ) = x2 − 1 can be found using the
chain rule.
The Chain Rule for Differentiation
If y = g(u) where u = h( x ), then
dy
dy
du
=
× .
dx
du
dx
Examples
1) Differentiate y = (2x − 1)4
Let u= 2x - 1 ⇒ y= u4
dy
3
Then du
dx =2 and du = 4u
dy
dy
du
=
×
dx
du
dx
= 4u3 . 2
= 8u3
= 8(2x − 1)3
[since u = 2x-1]
1
2) Find the derivative of y= √
3
2
5t +2t+1
y= 5t2 + 2t + 1
− 13
[change to index form for differentiation]
1
Let u = 5t2 + 2t + 1⇒ y = u− 3
dy
1 − 43
Then du
dt = 10t + 2 and du = − 3 u
16
dy
dy
du
=
×
dt
du
dt 1 −4
= − u 3 (10t + 2)
3
− 4
10t + 2 2
3
5t + 2t + 1
[after simplifying]
=−
3
3) Differentiate y= sin (5x )
Let u = 5x ⇒ y = sin(u)
dy
Then du
dx = 5 and du = cos( u )
dy
dy
du
=
×
dx
du
dx
= cos(u) × 5
= 5 cos(5x )
4) If f ( x )=cos3 x find f 0 ( x )
y = cos3 x = [cos( x )]3
Let u = cos( x ) ⇒ y=u3
dy
2
Then du
dx = − sin( x ) and du = 3u
f 0 (x) =
dy
dy
du
=
×
dx
du
dx
2
= 3u × [− sin( x )]
= 3 cos2 x × [− sin( x )]
= −3 sin x cos2 x
5) Differentiate y = (loge [4x ])3
NB:y is a composite of THREE functions
Let v = 4x and u = loge v ⇒ y = u3
dy
dv
1
2
Then dx
=4 and du
dv = v and du = 3u
dy
du
dv
dy
=
×
×
dx
du
dv
dx
1
= 3u2 × × 4
v
1
= 3 (loge v)2 ·
·4
4x
3
= (loge [4x ])2
x
Exercise
Find the derivatives of the following functions
17
1. y= tan 3x
2. f ( x )=loge 2x
3. y = sin
π
4 − 2x
4. y = cos2 x
5. f ( x ) = esin x
6. y = cos2 (10x )
Answers
1. y0 = 3 sec2 (3x )
2. f 0 ( x ) = 1x
3. y0 = −2 cos
π
4 − 2x
4. y0 =-2sin x cos x
5. f 0 ( x ) = esin x cos x
6. y’=-20sin(10x ) cos (10x )
6 Product Rule
The ‘product rule’ is used when we want to differentiate the product
of two functions. The derivatives of functions such as y = f ( x ) =
2x sin ( x ) and y = f ( x ) = xe x can be found using the product rule.
If y = f ( x ) = u( x ) · v( x ) then y0 = f 0 ( x ) = u( x )v0 ( x ) + u0 ( x )v( x )
This is often abbreviated toy0 = uv0 + u0 v
Examples
1. Find the derivative of f ( x ) = ( x + 3)6 (2x − 1)
Letu = ( x + 3)6 and letv = 2x − 1
⇒ u 0 = 6( x + 3)5 , v 0 = 2
Theny0 = uv0 + u0 v
= ( x + 3)6 .2 + 6( x + 3)5 (2x–1)
= 2( x + 3)6 + 6( x + 3)5 (2x–1)
...and this could (but does not have to be) simplified further...
= 2( x + 3)5 [( x + 3) + 3(2x–1)]
= 2( x + 3)5 (7x )
= 2( x + 3)5 (7x )
.
2. Differentiate e x sin (2x )
[by factorising]
20
Letu = e x and letv = sin (2x )
⇒ u0 = e x , v0 = 2 cos (2x )
0
0
[using the chain rule]
0
Theny = uv + u v
= e x · 2 cos (2x ) + e x sin (2x )
= 2e x cos (2x ) + e x sin (2x )
Exercises
1. Use the product rule to differentiate the following
a) y = ( x–2)(6x + 7)andsimplify as far as possible
b) f ( x ) = (2x2 + 4)( x5 + 4x2 –2) [do not simplify]
√
c)y = ( x − 1)( x2 + 1)
√
d)y = ( x3 –4x + x )(3x4 + 2)
2. Find the derivative of
a)y = e x tan x
b)y = x2 loge x
c) y = sinhx cos x
i
x
d)y = ex
Hint : 1x = x −1 ]
Answers
1.
2.
a)12x–5
b) (2x2 + 4)(5x4 + 8x ) + 4x ( x5 + 4x2 –2)
3
c) 25 x 2 –2x + 1 1
2x 2
√
3
3
4
d) 12x ( x –4x + x ) + (3x + 2) 3x2 –4 +
a) e x tan x + e x sec2 x
b)x + 2x loge x
c) cos2 x–sin2 x
x
x
d) ex − xe 2
1
1
2x 2
7 Quotient Rule
The quotient rule is used when we want to differentiate a function
which is the quotient of two simpler functions. Functions such as
2
y = f ( x ) = x21+ x , y = f ( x ) = sinx x and y = f ( x ) = xx++11 may be
differentiated using the quotient rule.
0
u( x )
v( x )u ( x ) − u( x )v0 ( x )
If f ( x ) =
then f 0 ( x ) =
v( x )
[v( x )]2
0
This is often abbreviated to y0 =
vu − uv0
.
v2
Examples
dy
1) If y = x12+−x3 , find dx
u = 1 + x and v = x2 − 3
⇒ u0 = 1 and v0 = 2x
0
vu − uv0
y =
v2
2
( x − 3)1 − (1 + x )2x
=
( x 2 − 3)2
0
=
x2 − 3 − 2x − 2x2
( x 2 − 3)2
=
− x2 − 2x − 3
( x 2 − 3)2
2
2) Differentiate logx x
e
u = x2 and v = loge x
⇒ u0 = 2x and v0 =
1
x
22
0
vu − uv0
v2
(loge x ).2x − x2 . 1x
=
(loge x )2
2x loge x − x
=
(loge x )2
y0 =
Exercise
Find the derivatives of the following functions
+1
1) f ( x ) = 2x
4x −3
2) f ( x ) = 3x23+1
√
3) y = 1−√x x
4) y =
ex
sin2 x
Answers
−10
(4x −3)2
2) f 0 ( x ) = −218x 2
(3x +1)
3) y0 = 1 1 1 2 [after simplifying]
2x 2 1− x 2
1) f 0 ( x ) =
4) y0 =
e x (sin x −2 cos x )
[after simplifying]
sin3 x
8 Maxima and Minima
The maximum or minimum points of a function occur where the
derivative is zero. We can therefore use calculus to solve problems
that involve maximizing or minimizing functions.
Examples
1. The distance s km, to the nearest km, of a fishing boat from port at any
time, t hours, is given by the formula s = 2 + 8t–2.5t2 . When is the boat
furthest from port and what is its distance from the port at that time?
s = 2 + 8t–2.5t2
ds
= 8 − 5t
dt
The maximum distance will occur when
ds
=0
dt
⇒ 8 − 5t = 0
5t = 8
t = 1.6
When t = 1.6 hours,
s = 2 + 8(1.6)–2.5(1.6)2
= 8.4
The boat is furthest from port after 1.6 hours and the distance at
that time is 8.4 km.
2. Find the maximum product of two numbers that have a sum of 10.
24
Let the numbers be a and b and the product P.
Then a + b = 10 ⇒ b = 10 − a
Also P = a × b
P = a × (10 − a)
P = 10a − a2
dP
= 10 − 2a
da
The maximum value of P will occur when
dP
=0
da
10 − 2a = 0
⇒
2a = 10
a=5
If a = 5then b = 10 − a
b=5
The numbers are both 5. Therefore P = 5 × 5 = 25.
Exercises
1. Find two positive numbers whose sum is 18 such that the sum of
their squares is a minimum.
2. Find the turning point of the parabola defined by y = f ( x ) =
5x2 − 30x + 17
3. A ball is thrown vertically upward. The height, h(t) m above the
ground is a function of time with the formula h(t) = 15t2 – 5t + 1.6
. Find the greatest height reached.
4. What is the maximum area that can be enclosed if a rectangle is
created with a piece of wire 48cm long?
5. The annual profit P made on a garment is related to the number n
that are produced by the formula P(n) = 300n – 7200 − 0.2n2 . How
many garments should be produced to maximize profit?
Answers
1. The two numbers are both 9.
2. (3, -28)
3. 12.85m
4. 144cm2
5. 750
9 Curve Sketching
To sketch a curve it is helpful to find
1. the x and y intercepts:
2. maximum and minimum stationary points
Stationary Points
A stationary point is a point on a graph of a function y = f ( x )
where the tangent to the curve is horizontal. At a stationary point
the derivative function f 0 ( x ) = 0.
The Sign Test
A maximum stationary point occurs at x = a if
f 0 ( x )>0 for x < aand f 0 ( x ) = 0 for x = a and f 0 ( x ) < 0 for x > a.
A minimum stationary point occurs at x = a if
f 0 ( x ) < 0 for x < a and f 0 ( x ) = 0 for x = a and f 0 ( x ) > 0 for
x > a.
26
Example 1
Find the turning point of the parabola defined by
y = x2 + 4x + 5
and state if it is a maximum or minimum.
Solution:
We have
f ( x ) = x2 + 4x + 5
and so
f 0 ( x ) = 2x + 4.
For a stationary point, f 0 ( x ) = 0 :
2x + 4 = 0
2x = −4
x = −2.
When x = −2, y = (−2)2 + 4(−2) + 5 = 1 and there is a stationary
point at (−2, 1).
Sign Test
A sign test can be used to determine whether the stationary point
is a minimum or a maximum by checking the slope of the tangent on
each side of the stationary point as shown in the figure below:
The sign test shows that at x = −2, the gradient changes from negative to positive hence the stationary point at (−2, 1) is a minimum.
27
Example 2
Sketch the graph of
y = x3 − x.
Solution strategy:
1. Find x and y intercepts.
2. Find stationary points
3. Determine if stationary points are minimums or maximums using
the sign test.
4. Sketch graph.
Find x −intercepts
x −intercepts occur whenever the value of the function is 0:
We have
f ( x ) = x3 − x
and so, for x −intercepts,
x3 − x = 0
x x2 − 1 = 0.
For this to be true, either x = 0 or x2 − 1 = 0. Hence the x −intercepts
are x = 0, x = −1, and x = 1.
Find y−intercepts
A y− intercept may occur at a point where the x − coordinate is 0.
For y = f ( x ) = x3 − x, we see the y−intercept occurs at (0, 0) .
Find Stationary points:
We have
f ( x ) = x3 − x
and so
f 0 ( x ) = 3x2 − 1.
For a stationary point f ( x ) = 0:
3x2 − 1 = 0
3x2 = 1
x2 =
1
3r
x=±
1
3
1
= ±√
3
≈ ±0.58 .
28
When x = 0.58,y = −0.38.
When x = −0.58,y = 0.38.
Sign tests
x = 0.58:
The sign test indicates there is a minimum point at (0.58, −0.38)
x = −0.58:
The sign test indicates there is a maximum point at (−0.58, 0.38)
Sketch Graph
First plot all the stationary points and intercepts as shown below.
Note that you should label all points of interest. Now you can
simply use the points to sketch the curve as shown below.
29
Exercise
Sketch the graphs of the following functions showing all intercepts
and turning points
1. y = x² − 4x
Answer:
2. y = x3 –2x² + x
Answer:
3. y = 6–x − x²
Answer:
30
4. y = ( x + 1)4
Answer: