MATH 162: Calculus

Moderator: Kwame Piesie

July 18, 2023

Functions

A function is a rule that assigns to an element of a set X, one

and only one element in a set Y . The set X is called the
domain of the function and Y is called the co-domain of the
function. Symbolically, a function f is written as f : X −→ Y .
For each element x ∈ X, f (x) ∈ Y is the image of x

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There are instances where some element in the co-domain, Y
may not serve as images for elements in the domain X. If
every element in Y is an image for some element in X then Y
is called the range of the function.
The range of a function is the set of all images. The range of
a function is a subset of the co-domain.
A function f : A −→ B can also be defined as a set of ordered
pairs; {(a, b) : a ∈ A and b ∈ B}

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Example
Let f : A −→ B be defined by f (x) = 2x + 3. Evaluate the
following

1. f (0) 2. f (1) 3. f (−3) 4. f (a)

Example
A function is defined by f : x −→ 3x−1
x−3
on {−2, −1, 0, 1, 2, }.

1. Find the range of f

2. What values of x make the function undefined?

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Determination of the Domain of a Function

To determine the domain of a function, we exclude all possible

values of x which makes the function undefined. i.e.

1. The values of x which make the denominator of a rational

function zero.
2. The values of x which produce a negative sign under an
even indexed radical

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Example
State the largest possible domain of the following functions.

1. f (x) = √1−2x
2x−1
2. f : x −→ x+5
x−3
3. f (x) = 1 − 2x
√
4. g : x −→ 2x − 1
5. g : x −→ √2x−1
x
√
6. g(x) = 1 − x2
√
7. f (x) = 5x − 2
√
8. f (x) = x−2
x2 −4

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Limit

The two main branches of calculus; differential calculus and

integral calculus, depend on the limit concept.
To help grasp the idea on which calculus is based, we first
provide a practical introduction of limit of a function.

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The Meaning of x −→ a

For a given function f , x −→ a gives the behaviour of f (x)

for values of x around a. Consider the behaviour of
f (x) = 2x + 3 around 2.

x 1 1.9 1.99 1.999 1.9999 2 2.0001 2.001 2.01 2.1 3

f(x) 5 6.8 6.98 6.998 6.9998 7.0002 7.002 7.02 7.2 9

From the table above, we see clearly that f (x) starts at 5 and
increases progressively to 6.9998 as we move from left to right
approaching 2. Again, f (x) starts from 9 and reduces
progressively to 7.0002 as we move from right to left
approaching 2.

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Therefore as we allow x to approach 2, the values of f (x)
approach 7.
We say that "as x approaches 2, f (x) approaches 7" or "as x
tends to 2, f (x) tends to 7.
The mathematical notation for this is expressed as
lim (2x + 3) = 7 or 2x + 3 −→ 7 as x −→ 2.
x−→2

The number 7 is called the limit of f (x) as x −→ 2.

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Examples
Evaluate each of the following limits.
x2 −1
1. lim
x−→1 x−1
2. lim (2x + 3)
x−→4
n 2 o
3. lim xx+1 −1
x−→1
n 2 o
4. lim xx−2 −4
x−→2
n 2 o
5. lim x −7x+10
x2 −4
x−→4
 3 
6. lim xx−1 −1
x−→1
n o
7. lim 3− 9−x
√
x
x−→9
n 2
o
8. lim 4−9−x√
x2 +7
x−→3

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Left-hand and Right-hand Limits

lim f (x) indicates "the limit of f (x) as x approaches a

x−→a−
from the left of a". This is called left-hand limit.
Also, lim+ f (x) indicates " the limit of f (x) as x approaches
x−→a
a from the right of a. This is called the right-hand limit.
Definition
For any function f , lim f (x) = L if and only if
x−→a
lim− f (x) = lim+ f (x) = L.
x−→a x−→a

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Remark
1. If the left-hand limit does not equal the right-hand limit,
then there is no limit. Under such circumstances, we say
that the limit does not exist.
2. The existence of lim f (x) does not depend on whether
x−→a
f (a) is defined.
3. The existence of lim f (x) does not depend on the value
x−→a
f (a) if f (a) is defined.

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Limit Theorems

If a, c and n are real numbers then

1. lim c = c
x−→a
2. lim x = a
x−→a
3. lim c f (x) = c lim f (x)
x−→a x−→a
4. lim {f (x) + g(x)} = lim f (x) + lim g(x)
x−→a x−→a x−→a
5. lim {f (x) − g(x)} = lim f (x) − lim g(x)
x−→a x−→a x−→a
6. lim {f (x) · g(x)} = lim f (x) · lim g(x)
x−→a x−→a x−→a
n o lim f (x)
x−→a
7. lim f (x)
g(x)
; lim {g(x)} =
= ̸ 0
x−→a lim g(x) x−→a
x−→a
n on
8. lim {f (x)} = lim f (x)
n
x−→a x−→a

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Examples
Compute each of the following limits.

1. lim (x + 3)(x2 − 5)
x−→2

1 − x, x ≤ 2
2. lim f (x) when f (x) =
x−→2 4, x>2
|x|
3. lim
x−→0 x

x + 1, x ̸= 1
4. lim f (x) when f (x) =
x−→1 π, x=1

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5. Determine the existence or otherwise of lim f (x) when
x−→4
√x − 4, if x > 4


f (x) =
8 − 2x, if x < 4

 x−1 , x ̸= 1
6. Evaluate lim f (x) when f (x) = x −1
2

x−→1 2, x=1

7 Show that lim |x| = 0.
x−→0

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 1
Limit of the Form lim n
x−→0 x

As x gets closer to 0, xn also gets closer to 0, and 1

xn
becomes
very large. Consider f (x) = x12 as x −→ 0.

x -1 -0.5 -0.2 -0.1 -0.05 -0.01 -0.001 0 0.001 0.01 0.05 0.1 0.2 0.5 1
f(x) 1 4 25 100 400 10000 1000000 1000000 10000 400 100 25 4 1

As clearly shown above, f (x) can be made arbitrarily large by

taking x close enough to 0. Then the values of f (x) do not
approach a number, so lim x1n does not exist. We use the
x−→0
notation lim x1n = ∞ to describe the behaviour of f (x) as
x−→0
x −→ 0.

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Remark
lim 1n = ∞ does not mean we are regarding ∞ as a number,
x−→0 x
or the limit exists. It simply expresses the particular way in
which the limit does not exist. f (x) = x12 can be made as
large as we like by taking x close to zero.
In general, we write symbolically lim f (x) = ∞ to indicate
x−→a
that the values of f (x) become larger and larger (or increases
without bound) as x becomes closer and closer to a.

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Limits of the Form lim f (x)
x−→∞

The limit of a function f (x) as x approaches infinity (∞) or

becomes very large is denoted as lim f (x).
x−→∞

1. lim 1
n = 0 if the index, n is positive.
x−→∞ x
2. lim xn does not exist if the index n is positive.
x−→∞

3. lim f (x) can be found by first dividing

the numerator
x−→∞ g(x)
and the denominator by the variable with the highest
index of x.

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Example
Evaluate the following limits.
 2 
1. lim 5x 2
−1
x−→∞ 2x +1
2. lim x−4 7x


x−→∞
 3 
3. lim 2x x−7x+3
3 +2
x−→∞
4. lim x+2 3x


x−→∞

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Limit of Simple Trigonometric Functions

Consider the function f (x) = sin x

x
, where x is in radian.

x 2 1.5 1.0 0.5 0.3 0.2 0.1 0.01 0.001 0

f(x) 0.45 0.66 0.84 0.96 0.99 0.99 0.998 0.999 0.999 ?

As x −→ 0, sin x
x
−→ 1, thus lim sin x
x
= 1.
x−→0

We note the following.

1. lim cos x = 1
x−→0
2. lim sin x = 0
x−→0
3. lim tan x = 0
x−→0
4. lim tanx x = 1
x−→0
5. lim cosx x = 0
x−→0
6. lim sin x
x
=1
x−→0
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Examples
1. Show that lim tan x
=1
x−→0 x
2. Evaluate lim sin 2x
x−→0 x
3. Compute lim sec x
x−→0

Example
Evaluate the following limits.

1. lim (cos x + sin x)

x−→0
2. lim 2 tan x−x


x−→0 sin 2x
√ 
3. lim 2−sin x
cos x
x−→0
4. lim sin 5x


x−→0 x

5. lim cot 2x
cot x


x−→0

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Calculus

Calculus is the mathematical study of change. It has two

major branches.

1. Differential calculus, which deals with the rate of change

and slope of curves.
2. Integral calculus, which is concerned with the
accumulation of quantities and the areas under curves.

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The Fundamental Theorem of Calculus

This theorem states that if a function f is continuous on an

interval (a, b), and if F is a function whose derivative is f on
Z b
the interval (a, b), then f (x) dx = F (b) − F (a).
a
Furthermore, for any x in the interval (a, b),
Z x
d
f (t) dt = f (x)
dx a
.
The fundamental theorem of calculus simply says that
differentiation and integration are inverse/reverse operations.

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Continuity of Real-Valued Functions

A function f is said to be continuous if its graph has no breaks

in it such as holes, gaps, or jumps. This means the graph of
the function can be drawn without lifting the pencil off the
paper. If the graph of the function has a hole, a gap, or a jump
at x = a, we say that the function is discontinuous at x = a.
Definition
A function f is said to be continuous at a point x = a if the
following conditions are satisfied.
1. f (a) is defined or f is defined at a.
2. lim f (x) exists.
x−→a
3. f (a) = lim f (a).
x−→a

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Definition
A function f (x) is continuous on an interval (a, b) if it is
continuous for all values of x in the interval (a, b).
Examples
√
1. Show that f (x) = 1 − 1 − x2 is continuous on the
interval [−1, 1]
x2 −4
2. verify the continuity of f (x) = x−2
at x = 2.

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Remark
1. Every polynomial function is continuous for all real x.
2. A rational function is continuous for all real x except
those values of x for which the denominator equals zero.
3. Every exponential function is continuous for all real x.
4. The natural logarithmic function, ln |x| is continuous for
all real x in its domain.
5. A radical function, or rational exponent function is
continuous for all real x in its domain.

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Example
Determine the continuity of the given function at the point
indicated.

1. f (x) = 3x − 2; at x = −2
2. f (x) = 2x + 6; at x = −3
3. h(x) = x2 − x − 6;at x = 3
4. h(x) = 2x3 x2 − 5; at x = 2
5. g(x) = x+1
x−3
; at x =3
6. g(x) = x−5
x+2
; at x = −2
2
7. f (x) = x −25
x−5
; at x=5
x2 −25
8. f (x) = x−5
; at x=0

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 
x + 2, x≤1
9. f (x) = at x = 1
x2 + 3, x > 1

x − 3, x≤0
10. f (x) = at x = 0
2
x + x − 3, x > 0

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Remark
If f and g are continuous at a, and if c is a constant, then the
following functions are also continuous at a.

1. f + g
2. f − g
3. cf
4. f g
5. f
g
; g(a) ̸= 0

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Example
Determine the intervals within which each of the following
functions is continuous.
1. 2x+1
f (x) = (x−1)(2x+1)
2. f (x) = (1.12)x
√
3. f (x) = x − 7
4. f (x) = 7x2 − 3.2x + 0.5
5. x+1
f (x) = (x+1)(x−3)
6. f (x) = 4x3 − 2x2 + 1.3x − 5
√
7. f (x) = 2x + 3
8. f (x) = e−0.25x
9. f (x) = −2.1 + ln x + 3
10. f (x) = 3x−7
2x+1
11. 2x−3
f (x) = (2x−3)(x+4)

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The Derivative of a Function f (x)

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In Fig 1 figure above, A and B are two points on the curve
y = f (x). AB is a chord joining A and B. Let m denote the
slope of the chord AB.
Then m = ∆x ∆y
= f (x+h)−f
(x+h)−x
(x)
. As ∆x approaches 0, ∆x
∆y

approaches a limiting value and the gradient of the chord

approaches the gradient of the tangent at A. Thus the slope
of the curve at A can be written as

∆y f (x + h) − f (x)
lim = lim
∆x−→0 ∆x h−→0 (x + h) − x
.
If this limit exists then it is called the derivative of f at x. It is
denoted by dx dy
or (f ′ (x)). This process of finding the
derivative of f is called differentiating from first principles.

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Remark
1. f ′ (a) = lim f (x)−f (a)
x−a
x−→a

2. f ′ (x) = lim f (x+h)−f (x)

h
h−→0

Remark
1. The derivative of y with respect to x, denoted by dy
dx
, is
rate of change of y as a result of a change in x.
2. It is the instantaneous rate of change of y at x.
3. It is also the gradient function of the curve y = f (x) at a
point along the curve.

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Example
Find from first principles the derivative of each of the
following.
1. f (x) = x2
2. f (x) = x2 + x
3. f (x) = 5x
4. f (x) = 8
5. f (x) = x12
Example
Differentiate the following functions from first principles.
1. f (x) = 2x3 + 3
2. f (x) = 2x2 + 4x − 3
3. f (x) = 3x2 + 1
√
4. f (x) = x
5. Kwame
f (x) = (2x + 3)2
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Rules of Differentiation

1. For any constant k, if f (x) = k then

d
f ′ (x) = (k) = 0
dx
2. If f (x) = xn , where n is any real number then

f ′ (x) = nxn−1

3. If y = k g(x), where k is any constant, then dx

dy
= k g ′ (x)
4. If h(x) = f (x) ± g(x), where f and g are differentiable
functions of x, then

h′ (x) = f ′ (x) ± g ′ (x)

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Product Rule of Differentiation

If h(x) = f (x) g(x), where f (x) and g(x) are differentiable

functions of x, then

h′ (x) = f ′ (x) g(x) + f (x) g ′ (x)

. Examples
1. Differentiate h(x) = 3x3 (x4 + 2).
2. Find the derivative of the following functions.
a. f (x) = (2x2 + 4x + 5)(5x − 4)
√
b. y = x(3x3 − 4x2 + 8)
c. y = (6x4/3 + 2x)(3x5/3 + 4x − 1)

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Quotient Rule of Differentiation

If h(x) = fg(x)
(x)
where g(x) ̸= 0, and f and g are differentiable
functions of x then,

f ′ (x) g(x) − f (x) g ′ (x)

h′ (x) = provided g(x) ̸= 0
{g(x)}2

Example
Find the derivative of the following functions;
(x−3)2
1. f (x) = x+3
x−2 6. y = x2 +1
4
2. y = xx2−3x +1 7. y = √3x−1
x2 +1
3. y = (2x+1)(3x−2)
x+1
8. y = x
1+x2
x2 −1
4. y = 1−x2
1+x
9. y = x3 +1
x2 −1
5. Kwame
y =Piesie
2x2 +1 37 / 172
Non-Differentiable Functions

In spite of all the rules of differentiation, some functions are

non-differentiable i.e. some functions cannot  be differentiated
x, x≥0
at certain values. Consider f (x) = |x| =
−x, x < 0
f (x+h)−f (x)
f ′ (x) = lim h
h−→0
f (0+h)−f (0)
f ′ (0) = lim h
h−→0
f (h)−f (0)
= lim h
h−→0
|h|−0
= lim h
h−→0
|h|
= lim
h−→0 h

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 |h| |h|
Now, lim− h
= −1 and lim+ h
=1
h−→0 h−→0
|h| |h|
Since lim− h
̸= lim+ , lim |h|
h h−→0 h
does not exist. Hence
h−→0 h−→0
f ′ (0) does not exist. Therefore f (x) = |x| is not differentiable
at x = 0.
Again, consider f (x) = x1/3 . f ′ (x) = 13 x−2/3 = 1
3x2/3
. Clearly
f ′ (x) = x1/3 is not differentiable at x = 0.
The two examples above shows that a function is not
differentiable at a point x = a if the function is discontinuous
at x = a.

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Theorem: (Differentiability Implies Continuity)

If f is differentiable at x = a, then f is continuous at x = a

Proof
f (x)−f (a)
f ′ (a) = lim x−a
x−→a

lim f (x) − f (a)

x−→a
f ′ (a) =
lim (x − a)
x−→a

f ′ (a) lim (x − a) = lim (f (x) − f (a))

x−→a x−→a
′
=⇒ f (a) · 0 = lim f (x) − lim f (a)
x−→a x−→a

=⇒ 0 = lim f (x) − f (a)

x−→a

=⇒ lim f (x) = f (a)

x−→a

Hence f is continuous at a if f is differentiable at a.

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Remark
Differentiability always implies continuity but continuity does
not necessarily imply differentiability

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The Chain Rule

This rule is used to differentiate composite functions that have

the form h(x) = f (g(x)). If y = f (u) and u = g(x) then
y = f (g(x)). Therefore

dy dy du
= ·
dx du dx

provided that dudy

and du
dx
exists
Example: Find the derivatives of the following functions using
the chain rule.

1. f (x) = (5x3 + 3x)4 4. f (x) = 5

(2x−3)2
2. f (x) = (x2 + 1)15 5. g(x) = (x + 3)5
√
3. f (x) = 3 2x − 4 6. y = (2x2 − 5x)4
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The Generalised Power Rule

If u is a differentiable function of x and n is any real number

with f (x) = {u(x)}n , then

f ′ (x) = n{u(x)}n−1 · u′ (x)

. In simple terms, if y = (ax + b)n , then

dy d
= n(ax + b)n−1 · (ax + b)
dx dx
Example
Find the derivatives of the following functions

1. y = (x2 + 2)5 3. y = (2x − 5)−4

2. y = (3x4 − 5)7 4. y = (3x + 4)5
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Derivatives of Logarithmic Functions

The function f (x) = loge x, where x > 0 is a variable and

e = 2.718, is a constant called the natural logarithm. The
natural logarithm function is commonly written as
f (x) = ln x, x ≥ 0 or f (x) = ln |x|

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Properties of Natural Logarithm

If x > 0 and y > 0 then;

1. ln xy = ln x + ln y
2. ln xy = ln x − ln y
3. ln xy = y ln x
4. ln 1 = 0
5. ln e = 1
6. eln x = x
7. ln ex = x

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If f (x) = ln x with x > 0 then
f ′ (x) = lim ln (x+h)−ln (x)
h
and thus d
dx
(ln x) = x1 .
h−→0

In general if
y = ln f (x)

where f is a differentiable function of x then

dy f ′ (x)
=
dx f (x)

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Example
Compute the derivatives of the following functions.
1. f (x) = 2 ln x
2. f (x) = ln x3
3. f (x) = 7 − 4 ln x
4. f (x) = ln (x4 − 2x)
√
5. y = ln 6x − 1
6. y = (ln x)3
Example
Determine the derivative of the following functions.
1. y = 4x3 (ln x)
2. y = (ln x)2
3x5
3. y = ln x
4. y = 12x3 (ln x2 )
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Derivative of the General Logarithmic Function

The general logarithmic function is the function

f (x) = logb x

, where b > 0, b ̸= 1. The function f (x) = logb x can be

written as f (x) = ln x
ln b
via the change of base formula. Hence
f (x) becomes f (x) = ln1b x1
′

In general, if
f (x) = logb {g(x)}

then
′ 1 g ′ (x)
f (x) = ·
ln b g(x)

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Example
Differentiate the following functions.

1. y = log6 x
2. y = log (x3 + 9)
3. y = log5 x
4. y = 2x5 logb x
5. y = log5 (3x + 9)
6. y = log10 x+3
x2 +1
3
7. y = log2 x2x−1

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Derivative of the Exponential Function

The function
h(x) = ef (x)

, where e = 2.718 is a constant is called the exponential

function. If f is a differentiable function of x, then

h′ (x) = ef (x) · f ′ (x)

. Examples
Differentiate the following functions with respect to x.

1. y = x2 ex 4. h(x) = e3x−3
x2 1 6
2. y = ex
5. g(x) = e6x− 2 x
3. f (x) = loge ex
Kwame Piesie 50 / 172
Derivative of the General Exponential Function

The general exponential function is any function of the form

f (x) = ax

, where a > 0 and a ̸= 1.

To differentiate this function, we resort to the natural log
differentiation. i.e. we take log of both sides of y = ax and
differentiate it implicitly. i.e. if
ln y = x ln a then 1 dy
y dx
= ln a, thus dy
dx
= y ln a
Therefore
dy
= ax ln a
dx

Kwame Piesie 51 / 172

Examples
Determine the derivative of the following functions.

1. y = 10x
2. y = ( 43 )x
3. y = 10x+3
4. y = 5x
2

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In general, if
y = af (x)

where f is a differentiable function of x then

dy
= af (x) · f ′ (x) ln a
dx
. Proof
Let y = af (x)
Then ln y = f (x) ln a
1 dy
=⇒ y dx
= f ′ (x) ln a
dy
=⇒ dx
= y f ′ (x) ln a
dy
∴ dx
= af (x) · f ′ (x) ln a

Kwame Piesie 53 / 172

Example
Find the derivative of these functions.

1. y = 59x−5
2. y = 3ln x+5
3. y = 29−x
√
4. y = 4 x

5. y = 4x−3
6. y = 5x
2

Kwame Piesie 54 / 172

Derivative of Trigonometric Functions

1. d
dx
(sin x) = cos x
2. d
dx
(cos x) = − sin x
3. d
dx
(tan x) = sec2 x
4. d
dx
(sec x) = sec x tan x
5. d
dx
(cosec x) = −cosec x cot x
6. d
dx
(cot x) = −cosec2 x

In general, we have the following.

1. if y = sin f (x) then dy

dx
= f ′ (x) cos f (x)
2. if y = cos f (x) then dy
dx
= −f ′ (x) sin f (x)

Kwame Piesie 55 / 172

Example
Differentiate the following with respect to x.

1. y = sin x2
2. y = sin (x2 − 3x + 1)
3. y = cos (2x3 + x2 − 1)
4. y = cos (4x3 )
5. y = sin (3x2 + 1)
6. y = tan 6x2

Kwame Piesie 56 / 172

Derivative of Powers of Trigonometric functions

In general, if
y = cosn x

then
dy d
= n cosn−1 x · (cos x)
dx dx
.
Similarly, if y = sinn x, then

dy d
= n sinn−1 x · (sin x)
dx dx

Kwame Piesie 57 / 172

Find the derivative of the following functions with respect to x

1. y = tan2 x
2. y = 2 cos2 (x2 + 1)
3. y = sin3 x
4. y = 3 tan4 x
5. y = x sin 3x
6. y = x3 cos 2x
7. y = cos x sin3 x
8. y = 2x3 cos 3x
9. y = x4 cos x
10. y = cos (x2 − x)

Kwame Piesie 58 / 172

Implicit Differentiation

An implicit function in two variables x and y is a function

where neither variable can be expressed explicitly in terms of
the other. For example, the equation x2 + y 2 = 3xy is an
implicit function. Thus far, all the functions that we have
studied have the form y = f (x). Algebraically, this means that
we can solve for y explicitly in terms of x such as
y = x3 + 3x2 − 1. However, not all equations are expressed in
this form. For example, for the equation,
xy 4 + y 2 − xy + 3x2 − 5 = 0, solving for y explicitly will be
difficult. How then do we find the derivative dxdy
in this case?
The key is to use implicit differentiation. To differentiate a
function implicitly, we follow the steps below.

Kwame Piesie 59 / 172

1. Differentiate the function term by term to give an
equation in dx
dy
, x and y.
2. Make dy
dx
the subject of the equation.
3. The following are worth noting.
d dy
a. dx (y) = dx
d dy
b. dx (y 2 ) = 2y dx
d dy
c. dx (2y 3 ) = 6y 2 dx
dy
d. In general dx d
(f (y)) = f ′ (y) dx , where f ′ (y) is the
derivative of f (y) with respect to y.
4. Terms such as 3x2 y, xy, x dx
dy
should be treated as
x2 2x
products, and y , y should be treated a quotients.

Kwame Piesie 60 / 172

Example
Use implicit differentiation to determine dy
dx
for the following
functions

1. y 3 − 4x2 − 7 = 0
2. x1/3 + y 1/3 − 10 = 0
3. 4ey = x2
4. 2x + 1 =
p
2 − y2
5. x2 y 2 − 7x3 − 5 = 0
6. x2 + y 2 = 25
7. 3xy + 6y 2 = 15
8. y + xy + y 2 = 2

Kwame Piesie 61 / 172

Higher Order Derivatives

So far we have considered the rate of change of y with respect

to x when y = f (x). This is called the first derivative. The
result of differentiating dx
dy
with respect to x is called the
second derivative of y with respect to x. This is denoted by

d dy d2 y
( )= 2
dx dx dx
If the second derivative is also differentiated with respect to x
d3 y
the third derivative dx 3 is obtained and so on.

Remark
d2 y dy 3
If y = f (x) then dy
dx
= f ′ (x), dx2
= f ′′ (x), dx3
= f ′′′ (x) etc.

Kwame Piesie 62 / 172

Example
dy d2 y d3 y
Find dx , dx2 and dx3
for each of the following functions.

1. y = 3x4 + 2x3 + 2
x2
2. y = x6 + 2x3 − 3x + 4x + 1
2

3. y = x4 − x3 + 4x − 1
4. y = 3x3 (x2 + 4)2
5. y = −5 cos 3x
6. y = x
1+x2

Kwame Piesie 63 / 172

The Derivative of a Function with Parametric
Representation

Consider the function y = f (x) with parametric representation

x = f (t) and y = f (t), where t is the parameter. In such
cases the gradient function dx
dy
is given in terms of the variable
parameter using the chain rule of differentiation i.e.

dy dy dt
= ·
dx dt dx
. Now by virtue of the fact that x is a function of t, we can
only find dx
dt
and write dx
dt
as

dt 1
= dx
dx dt

Kwame Piesie 64 / 172

Example
1. If x = t2 and y = t3 , find dx
dy
in terms of t and show that
the above parametric representation is of the curve
y = x3/2 .
2. Find dx
dy
in terms of t when x = at2 and y = 2at, where t
is a parameter and a is a constant, Find the equation
with this parametric representation.
3. Find dy
dx
in terms of t in each of the following cases
t t2
a. x = 1−t , y= 1−t
2t 3t
b. x = t+2 , y= t+3
4. Find the curve whose parametric representation is given in
each case in (3) above.

Kwame Piesie 65 / 172

Examples

1. A cylindrical container of thin metal is to hold 10cm3 of

water. If an engineer wishes to use the least material of
metal for the construction of the cylinder, show that
A = 2πr2 + 20 r
, where r is the radius of the base area, h
is the height of the cylinder and A is the surface area.
Hence show that the area of the metal needed is minimum
when the height of the cylinder is twice the base radius.
2. A point P whose x-coordinate is α is on the line
y = 3x − 7. If Q is the point (4, 1), show that
|P Q|2 = 10α2 − 56α + 180, and find the value of α
which makes |P Q|2 minimum.

Kwame Piesie 66 / 172

3. A sealed cylindrical can is of radius r and height h. The
area of its total surface is A and its volume is V
a. Find an expression for A in terms of r and h.
b. If A = 24π, find an expression for h in terms of r.
Hence find an expression for V in terms of r.
c. Find the value of r which maximises V .
4. A cylinder is such that the sum of its height and the
circumference of its base is 5cm. Express the volume V
in terms of the radius of the base r. What is the
maximum value of the cylinder?

Kwame Piesie 67 / 172

Applications of Differentiation

1. The gradient Function: If y = f (x) then dx dy

or f ′ (x) is
called the gradient function. The gradient of a curve at a
point P (a, b) is obtained by substituting the values a and b
into the expression for dx
dy
. This value dx
dy
|(a,b) is the same as
the gradient of the tangent to the curve at the point P (a, b)

Kwame Piesie 68 / 172

Example
Find the gradient of the curve at the indicated point.

1. f (x) = (x2 + 3)2 at x = 1

2
2. y = (x − 3)(x2 + 2) at, x = 1
3. x2 + y 2 = 9 at x = 1
4. y = x2 − 3 at x = 1
5. y = 3x2 − 2 at x = 3
√
6. y = x at x = 2
7. y = (x − 3)(x + 4) at x = −1
8. y = 1
x
at x = 3

Kwame Piesie 69 / 172

Examples
1. Find the coordinates of the point on the curve y = 2
x2
where its gradient is 21
2. Find the values of x for which the gradient function of
the curve y = 14 x4 + 23 x3 − 12 x2 − 2x + 1 is zero
3. Find the coordinate of the point on the given curve where
its gradient, m has the given value.
(a) y = x2 − x + 3; m = 1
(b) y = x3 + x2 ; m = 1
(c) y = 5 + 3x − 2x2 ; m = −3
(d) y = (x + 3)(x − 5); m = 0
(e) y = (x + 1)(x − 1); m = 2

Kwame Piesie 70 / 172

2. Equations of Tangents and Normals to a Curve:

a. If y = f (x) then dx
dy
|(a,b) is the gradient of the tangent to
the curve at P (a, b). If m = dx dy
|(a,b) then the equation of
the tangent to the curve at P (a, b) is

y − b = m(x − a).

b. The normal to a curve at the point of contact is the

straight line perpendicular to a tangent to the curve at
that point of tangency. If m1 is the gradient of the
tangent to the curve and m2 is the gradient of the normal
then
m1 · m2 = −1

Kwame Piesie 71 / 172

Example
Given that y = x3 − 4x2 + 5x − 2 and that P is a point on the
curve at x = 3,

(a) find dy
dx
,
(b) find the coordinates of P .
(c) calculate the gradient of a tangent at P
(d) find the equation of the tangent at P
(e) find the equation of the normal at P
(f) find the value of x for which the curve has gradient 5

Kwame Piesie 72 / 172

Example
Find the equation of the tangent and the normal to the curve
y = x3 − 3x + 2 at the point P (2, 4).
Example
Find the coordinates of the points on the curve
y = x3 − 4x2 − 3x − 2 where the tangent is parallel to the
x-axis.
Example
The gradient function of a curve at a point P (x, y) is
dy
dx
= 4x2 − 3x + 1. If the curve passes through the point
A(1, 0), find the equation of the tangent to the curve at A.

Kwame Piesie 73 / 172

Example
Given that y = (x − 2)(x2 − 3x − 5), find dxdy
. Determine the
gradient of the tangent to the curve at the point where it cuts
the x − axis.
Example
Find the gradient and the equation of a tangent and a normal
to the curve y = 5x3 − 7x2 + 3x + 2 at (1, 3).
Example
A tangent to the curve y = x3 + kx, where k is a constant, at
x = 1 passes through the points A(−1, 6) and B(2, −15).
Find the value(s) of k.
Example
x2 y2
Find the gradient of 25
+ 4
= 1 at P (3, 85 ).

Kwame Piesie 74 / 172

3. Rates of Change: Suppose we have some quantity y which
depends on x and also varies with x. The rate of change of y
with respect to x is given by dx
dy
. For instance, if V is the
volume of a sphere and r is its radius then dV
dr
is the rate of
change of V with respect to r. If the quantity y is increasing
with respect to x then dx
dy
> 0 and if y is decreasing with to x
then dx < 0.
dy

Remark
Rates of change are by convention with respect to time.

Kwame Piesie 75 / 172

Example
A circular ink blot has a radius 2cm. At what rate is the area
increasing with respect to the radius?
Example
A spherical balloon is being inflated and the volume is
increasing at the rate of 15cm3 /sec. At what rate is the
radius increasing when it is 10cm?

Kwame Piesie 76 / 172

Example
A spherical balloon is blown up so that its volume increases at
the rate of 2cm3 /s. Find the rate of increase of the radius
when the volume of the balloon is 50cm3 .
[Hint : Vs = 43 πr3 , SA = 4πr2 ]
Example
A spherical bubble is decreasing in volume at a rate of
2cm3 /s. Find the rate at which the surface area is diminishing
when the radius is 3cm. [Hint : VS = 43 πr3 , SA = 4πr2 ]

Kwame Piesie 77 / 172

Examples
1. A spherical balloon is inflated by a gas being pumped in
at a constant rate of 200cm3 /s. What is the rate of
increase of the surface area of the balloon when the
radius is 100cm?
2. An elastic spherical balloon is being blown up so that the
radius is increasing at a rate of 1cm/s. Calculate the rate
at which the volume of the balloon increasing when the
radius is 5cm
3. The side of a cube is increasing at a rate of 60m/s. Find
the rate of increase of the volume when the length of the
side is 9cm
4. The volume of a cube is increasing at a rate of 25 cm3 s−1 .
Find the rate of change of the side of the base when its
length is 2cm
Kwame Piesie 78 / 172
4. Small Changes: From previous discussions, we learnt that

∆y dy
lim = .
∆x−→0 ∆x dx

This means that ∆y

∆x
≃ dy
dx
when ∆x is so small. Hence

dy
∆y ≃ · ∆x.
dx
This approximation can be used to estimate the small change
∆y in y if dx
dy
can be found when the small change ∆x in x is
given. Thus the change in y is given by

dy
∆y = · ∆x.
dx

Kwame Piesie 79 / 172

We can also use this approximation to estimate percentage
changes. In general, if x is increased by p%, then
p
∆x = ∗x
100
and the approximate percentage increase in y is

∆y
∗ 100
y
.

Kwame Piesie 80 / 172

Examples
1. The surface area of a sphere is 4πr2 . If the radius of the
sphere is increased from 10cm to 10.1cm, what is the
approximate increase in the surface area?
2. The radius r of a circle is 5cm. Find the increase in the
area A of the circle when the radius expands by 0.01cm
3. The side of a cube increases by 5%. Find the
corresponding percentage increase in the volume
4. Find the approximate value for the square root of 16.01
√
[Let y = x, x = 16, ∆x = 0.1]
5. The radius of a sphere is increased by 2.5%. Find the
percentage increase in the volume of the sphere
√
6. If y = x, find the approximate increase in y if x is
increased from 4 to 4.01

Kwame Piesie 81 / 172

5. Maxima and Minima:
1. Stationary Points: A point on a curve at where dx dy
= 0 is
called a stationary point and the value of the function at
this point is called the stationary value. At the stationary
point, the tangent to the curve is parallel to the x-axis.
To find the stationary point, we put dxdy
= 0 and solve for
the resulting equation. Stationary points are also called
turning points.

Kwame Piesie 82 / 172

2. Procedure for testing and distinguishing between
stationary points:
dy d y 2
a. Given y = f (x), find dx and dx2
dy
b. Put dx = 0 and solve the resulting equation for the
turning value.
c. Substitute the value of x into the original equation
y = f (x) to find the corresponding y coordinate value.
This establishes the coordinate of the stationary point
d2 y
d. Substitute the value(s) of x found in step (b.) into dx 2.
d2 y
If dx2
< 0 then the point is a maximum point. Finally, If
d2 y
dx2
> 0 then the point is a minimum point.

Kwame Piesie 83 / 172

Examples
1. A curve is defined by f (x) = x3 − 6x2 − 15x − 1. Find;
a. the derivative of f
b. the gradient of the curve at the point where x = 1
c. the maximum and the minimum points
d. the maximum and the minimum values of (f)
2. Find the minimum/minimum point of y = x2 − 3x + 5.
3. Find the stationary points of y = 31 x3 − 2x2 + 3x and
identify their nature.
4. The curve y = x2 − ax + b has a turning point at P (1, 3).
Find the values of a and b.
5. The function f (x) = ax2 + bx + c has gradient function
4x + 2 and stationary value of 1. Find the values of a, b
and c.

Kwame Piesie 84 / 172

Examples

1. A rectangular sheet of metal 60cm by 50cm has squares

each of side x cm cut off from its corners. The remainder
is then formed into a cuboid of volume V cm3 . Find the
value of x for which the volume is maximum. Hence find
the maximum volume.
2. The length of the sides of a rectangular sheet of metal are
8cm and 3cm. A square of side x cm is cut off from each
corner of the sheet and the remaining piece is folded to
make an open box.
(a) Show that the volume, V of the box is given by
V = 4x3 − 22x2 + 24x cm3 .
(b) Find the value of x for which the volume is maximum
and find the maximum volume.

Kwame Piesie 85 / 172

3. A rectangular box with a square base x cm and height
h cm is to be constructed from a cardboard,
a. If the box holds a volume of 200cm3 , find a formula for
h in terms of x.
b. Show that the area of the cardboard needed to make the
box is given by 2x2 + 800x
c. Find, to 1 d.p the least area of cardboard needed to
make the box.
4. A rectangle has perimeter 28m. What is its maximum
area?
5. 100cm of fencing is to be used to make a rectangular
enclosure. Find the greatest possible area that can be
enclosed.

Kwame Piesie 86 / 172

6. A ball is thrown vertically upwards from the ground level
and its height after t seconds is h = (50t − 16t2 ) m.
Find,
a. The greatest height reached.
b. The time taken to reach the greatest height
7. Determine the area of the largest piece of rectangular
ground that can be enclosed by 100m of fencing, if part
of an existing straight wall is used as one side.
8. A cylindrical can has a radius r and a height h. The sum
of the radius and the height is 2m. Find an expression for
the volume, V of the cylinder in terms of r only. Hence
find the maximum volume

Kwame Piesie 87 / 172

6. Increasing and Decreasing Functions: The derivative of a
function y = f (x); dx
dy
gives the gradient function of the curve.
If dx > 0 then the function is increasing and decreasing when
dy

dy
dx
< 0.

Example
Find the range of values of x for which the following functions
are increasing.

1. x2 − 2x − 5
2. 2x3 + 3x2 − 12x + 5
3. x3 − 3x2 + 3x + 2

Kwame Piesie 88 / 172

Example
Find the interval on which f (x) = x3 + x2 − 5x − 5 is

1. Increasing
2. Decreasing

Example
Find the range of values of x for which the function
f (x) = x3 − 12x + 5 is

1. Increasing
2. Decreasing

Kwame Piesie 89 / 172

7. Curve Sketching: Let y = f (x) be a function that
represents a curve we wish to sketch. The steps below are
used to sketch any curve whose equation is given.
Step 1: Find the intercepts on the x and y axes

1. For intercept on x-axis, put y = 0 and solve for x

2. For intercept on y-axis, put x = 0 and solve for y

Step 2: Find the turning points. At the turning points put

dy
dx
= f ′ (x) = 0 and solve the resulting equation for x.
Substitute these value(s) of x into the original equation
y = f (x) to find the corresponding y-coordinate(s). This
establishes the coordinates of the turning point(s).

Kwame Piesie 90 / 172

Step 3: Test whether the turning points are
maximum/minimum. Use the following conditions.
2
1. If dx2 > 0 at the turning points then the turning points
d y

are minimum.
2
2. If dx2 < 0 at the turning points then the turning points
d y

are maximum.
2
3. If dx2 = 0 at the turning points then the turning points
d y

are inflection points.

Kwame Piesie 91 / 172

Example
1. Sketch the curve y = x3 − 7x2 + 15x − 9 indicating
clearly its points of intersection with the coordinate axes
and its turning points.
2. Sketch the following curves.
a. y = x(x2 − 16)
b. y = x2 − 4x − 5
c. y = x2 − 6x
d. y = 5x − x2
3. A function f is defined by f (x) = 2x3 + 3x2 .
a. Find the turning points and distinguish between them.
b. Sketch the curve of f (x).
x3 x2
4. A function f is defined by f (x) = 3
− 2
− 6x + 5
3
a. Find the turning points and distinguish between them.
b. Sketch the curve of f (x).

Kwame Piesie 92 / 172

Applications of Differentiation to Linear Kinematics

Kinematics is the study of displacements, velocity, and

acceleration. Consider a particle in motion in a straight line
such that its displacement from a fixed point is x meters after
time t seconds. Then displacement is a function of time i.e.
x = f (t). The S.I unit of displacement is cm, m or km
Velocity: It is the rate of change of displacement with time
i.e. v = dxdt
. The S.I unit of velocity is cms−1 , ms−1 , or
kmh−1 . If v = 0 then the particle is momentarily at rest. If
v < 0 then the particle is moving in the opposite direction to
its initial direction.

Kwame Piesie 93 / 172

Acceleration: Acceleration is the rate of change of velocity
with time i.e. a = dv
dt
. The S.I unit of acceleration is cms−2 ,
ms−2 , or kmh−2 . If a = 0 then the velocity of the particle is
constant. If a > 0, the particle is accelerating and it is
decelerating/retarding when a < 0.
Average Velocity: If a particle moves a distance x1 in time
t1 and another distance x2 in time t2 , where t2 > t1 , then the
average velocity of the particle within the time interval t1 and
t2 is given by;

x2 − x1 ∆x
Average Velocity = =
t2 − t1 ∆t

Kwame Piesie 94 / 172

Average Acceleration: If the velocity of a particle at t1 is v1
and at t2 is v2 , where t2 > t1 , then the average acceleration of
the particle within the time interval t1 and t2 is given by;

v2 − v1 ∆v
Average Acceleration = =
t2 − t1 ∆t

Kwame Piesie 95 / 172

Examples

1. A particle moves in a straight line such that its distance x

meters from a fixed point 0 after t seconds is given by
x(t) = t3 − 2t2 + t + 10. Calculate
(a) its average velocity after 2 seconds,
(b) its average velocity during the 3rd second,
(c) the average acceleration from t = 1 to t = 3.
(d) When and where is the particle momentarily at rest?
2. An object is thrown into the air such that its height, h
meters after t seconds is h(t) = 36t − 4t2 . Find
a. the velocity at t = 2,
b. the time taken to reach the maximum height.

Kwame Piesie 96 / 172

3. An object projected vertically upwards satisfies the
equation h(t) = 27t − 3t2 , where h meters is the height
after t seconds.
a. Find the time it takes to reach the highest point.
b. How high does it go?
4. A ball thrown vertically upwards at time t has height
h(t) = 32t − 8t2 .
(a) Find the velocity of the ball at t = 1 second
(b) Find the max height it reached.
5. A particle p moves in a straight line so that its distance
from a fixed point 0 after t seconds is
x(t) = 13 t3 − 23 t2 + 2t. Show that the particle is at rest at
two different times and find these times. Find the
acceleration of the particle at these times and interpret
your results.

Kwame Piesie 97 / 172

6. A particle moves in a straight line so that its distance x
from a fixed point 0 after t seconds where t ≥ 0 is given
by x(t) = 9t2 − 2t3 .
(a) Find v and a at t = 3.
(b) Find x when t = 4 and show that the motion is towards
the fixed point 0.
7. A particle projected from a fixed point with velocity
10ms−1 , moves in a straight line such that its velocity
after t seconds is v(t) = bt2 + 7t + c, where b and c are
constants. 1 second after projection, its acceleration is
1ms−1 . Find the value of b and c.

Kwame Piesie 98 / 172

8. A particle moves along a straight line such that its
distance X from a fixed point after t seconds is
X(t) = − 13 t3 + t2 − 75
(a) Find the time when the particle is momentarily at rest.
(b) Find the average acceleration of the particle between
t = 0 and t = 10.
9. A particle p moves along the x-axis such that its distance
x from a fixed point 0 is given by x(t) = t3 − 6t2 + 9t.
Calculate
(a) the times at which the particle is stationary.
(b) the distance of p at these times.
(c) the acceleration of p at these times.
(d) the velocity of the particle when a = 0.

Kwame Piesie 99 / 172

10. A missile fired from the ground level rises x meters
vertically upwards in t seconds and x(t) = 100t − 252
t2 .
Find
(a) the initial velocity of the missile.
(b) the time when the height of the missile is maximum.
(c) the maximum height reached.
(d) the velocity with which the missile strikes the ground.
11. A particle p moves in a straight line such that
displacement from a fixed point A is x(t) = (3t2 + 4)
(a) Find the velocity of the particle at t = 2.
(b) Find the initial displacement of p from A.
12. A particle starts from rest and travels in a straight line so
that its acceleration is (t + 3)ms−2 at time t. Find the
distance traveled in the interval of the time from t = 1 to
t = 3.

Kwame Piesie 100 / 172

Application of Differentiation to Business and Eco-
nomics

Marginal Analysis: Many decisions made by managers in

business involve analyzing the effect on the dependent variable
when a small change is made to a specific independent value.
For example, a firm may wish to consider changing the price of
an item and examining how this change affects the revenue or
profit of the product. Marginal analysis can be defined as the
study of the amount of change in the dependent variable that
results from a change in an independent variable. A unit
change means a change of one single unit. This change in the
dependent variable is a direct application of the derivative.

Kwame Piesie 101 / 172

The price-demand function: The price-demand function, P
is the price P (x) at which exactly x units of the product is
bought.
The cost function: The cost of producing x units of the
product with variable cost a and fixed cost b is given by the
cost function C(x) = ax + b i.e.

T C = V C + F C.

It should be noted that since variable costs are often expressed

as a function, C(x) may be a higher-order polynomial function.
The total revenue function: The total revenue function
generated by producing and selling x units of the product at a
price P is given by R(x) = xP (x).

Kwame Piesie 102 / 172

The profit function: The profit function π generated after
producing and selling x units of the product is given by
π(x) = R(x) − C(x) i.e. Profit=T R − T C

Marginal Cost Function: The marginal cost function

M C(x) is the approximate cost of producing one additional
unit at a production level x. It is the derivative of total cost
i.e. M C(x) = dx
d
C(x)

Marginal Revenue Function: The marginal revenue function

M R(x) is the approximate gain in revenue by producing one
additional unit at a production level. It is the derivative of the
total revenue i.e. M R(x) = dxd
(R(x)).

Kwame Piesie 103 / 172

Marginal Profit Function: The marginal profit function
M P (x) is the approximate gain in profit by producing one
additional unit at a production level x. It is the derivative of
profit function i.e. M P (x) = dx
d
(π(x))

Average Cost Function: The average cost function is the

per unit cost of producing x items. It is given by
AC(x) = C(x)x
= N o. ofTunits
otal cost
produced

Kwame Piesie 104 / 172

Average Revenue Function: This is the per unit revenue
generated by producing and selling x items. It is given by
AR(x) = R(x)
x
= NTo.otal revenue
of units sold

Average Profit Function: The average profit function is the

per unit profit of producing and selling x items. It is given by
AP (x) = π(x)
x
= N o.T otal prof it
of units sold

Kwame Piesie 105 / 172

Conditions for Minimising Average Cost

Now, AC(x) = C(x)

x
d d
d
x dx (C(x)) − C(x) dx (x)
=⇒ dx
(AC) = 2
x
′
d xC (x) − C(x)
=⇒ dx (AC) =
x2
For minimum average cost dxd
(AC) = 0
xC ′ (x) − C(x)
∴ =0
x2
=⇒ xC ′ (x) − C(x) = 0
C(x)
=⇒ C ′ (x) = x
..........(1)
Equation (1) says that average cost is minimum when
marginal cost is equal to average cost
Kwame Piesie 106 / 172
Example
A company estimates that the cost of producing x items of a
commodity is given by C(x) = 2680 + 2x + 0.001x2 .

1. Find the cost of producing 100 items.

2. Find the average cost of producing 200 items.
3. Find the marginal cost of producing 300 items.
4. At what production level will the average cost be
minimum? What is the minimum average cost?

Kwame Piesie 107 / 172

Conditions for Maximum Profit

If x units are sold then total profit is π(x) = R(x) − C(x).

=⇒ π ′ (x) = R′ (x) − C ′ (x) ..........(∗)

For maximum product π ′ (x) = 0

=⇒ R′ (x) = C ′ (x) ..........(2)

Equation (2) says that for profit to be maximum, marginal

revenue should be equal to marginal cost.

Kwame Piesie 108 / 172

From (∗), π ′′ (x) = R′′ (x) − C ′′ (x).
But for max profit, π ′′ (x) < 0.

Therefore R′′ (x) − C ′′ (x) < 0

=⇒ R′′ (x) < C ′′ (x) ..........(3)

(3) says that the rate of increase of marginal revenue should
be less than rate of increase of marginal cost for profit to be
maximum.

Kwame Piesie 109 / 172

Examples

1. Determine the production level that will maximize the

profit for a company with respective cost and dd function
C(x) = 84 + 1.26x − 0.01x2 + 0.00001x3 and
P (x) = 3.5 − 0.01x.
2. The average cost of producing x units of a commodity is
AC(x)=21.4-0.002x. Find the marginal cost at a
production level of 1000 units and interpret your answer.

Kwame Piesie 110 / 172

3. For each cost function given below, find;
a. the cost, average cost and the marginal cost at a
production level of 1000 units.
b. the production level that will minimize the average cost.
c. The minimum average cost.
i. C(x) = 40000 + 300x + x2
ii. C(x) = 25000 + 120x + 0.1x2
iii. C(x) = 16000 + 200x + 4x3/2
iv. C(x) = 10000 + 340x − 0.3x2 + 0.0001x3
4. For the given cost and dd function, find the production
level that will maximize profit.
a. C(x) = 680 + 4x + 0.01x2 ; P (x) = 12
x
b. C(x) = 680 + 4x + 0.01x2 ; P (x) = 12 − 500
c. C(x) = 1450 + 36x − x2 + 0.001x3 ; P (x) = 60 − 0.01x
d. C(x) = 16000 + 500 − 1.61x2 + 0.004x3 ;
P (x) = 1700 − 7x

Kwame Piesie 111 / 172

Example
DBS Industry has determined that its cost of producing x IBR
brand of roofing sheets can be modeled by the function
C(x) = 2x2 + 15x + 1500, 0 ≤ x ≤ 200, where x is the
number of IBR roofing sheets produced each week and C(x)
represents the weekly cost in Ghana cedis. The company also
determines that the price -dd function for the roofing sheets is
P (x) = −0.3x + 460.

1. Determine the profit function π for the roofing sheets.

2. Determine the marginal profit function.
3. Compute M P (60) and M P (145) and interpret the results

Kwame Piesie 112 / 172

Example
The revenue function for the production of x boxes of roofing
nails per week is R(x) = 0.3x2 + 460x. Compute R(110) and
M R(110) and interpret them.
Example
Odike shoe manufacturing company knows that, for its
executive shoes, the daily cost function can be modeled by the
√
function C(x) = 700 x + 500, 0 ≤ x ≤ 500, where x is the
number of pairs of shoes produced daily and C(x) is the daily
cost in cedis.

1. Determine the average cost function, AC(x).

2. Evaluate and interpret C(400) and AC(400).

Kwame Piesie 113 / 172

Example
Determine the marginal cost function M C(x) and marginal
profit function M π(x) for each of the following;

1. C(x) = 5x + 500; P (x) = 6

2. C(x) = 12x + 4500; P (x) = 15
x2
3. C(x) = 100
+ 7x + 1000; x
P (x) = − 20 + 15
4. C(x) = 1
100
x2 + 12 x + 8; P (x) = 1 − x
200

Kwame Piesie 114 / 172

Integration

Until now, we have focused almost exclusively on how to

determine the derivative of a function y = f (x) and interpret
the resulting rate of change function. But what happens if we
are given the rate of change function and we wish to find the
original function? In this section, we will study the basis of
integration, a process that reverses the process of
differentiation and finds the original function when its
derivative is given.

Kwame Piesie 115 / 172

The Indefinite Integral

If F ′ (x) = f (x), then

Z
f (x) = F (x) + c.

In this notation,
Z
• is the integral sign
• f (x) is the integrand.
• F (x) is the integral of f (x).
• c is a constant of integration.
• dx tells us the variable of integration.

Kwame Piesie 116 / 172

Rules of Integration

1. The Power Rule: For any Zreal number n, where n ̸= −1 ,

the indefinite integral of xn is xn dx = n+1
1
xn+1 + c

Example
Compute the following integrals.
Z
1. x8 dx
√
Z
2. x dx
Z
3. x−5 dx
Z
4. x1/3 dx

Kwame Piesie 117 / 172

2. Constant Rule: If k is any constant,
Z thenZ the indefinite
integral of k with respect to x is k dx = k dx = kx + c

Example
Evaluate the following integrals.
Z
1. (2x − 5) dx
Z
2. 3x2 dx
Z
3. 2x4 dx

Kwame Piesie 118 / 172

3. Sum and Difference Rule: For integrable functions f (x)
and
Z g(x) we have Z Z
(f (x) ± g(x)) dx = f (x) dx ± g(x) dx

Example
Find the following integrals
Z
1. (x2 + 3) dx
Z
2. (x1/3 + 5) dx
Z
2x3 −3x
3. 4x
dx
√
Z
4. (x3 + x12 + x) dx
Z
5. (2x2 − x12 + x) dx

Kwame Piesie 119 / 172

Definite Integrals

Integrals containing an arbitrary constant c in their results are

called indefinite integrals since their precise value cannot be
determined without further information.
Definite integrals are those with limit of integration. The
definite integral of f (x) from a to b is given by
Z b
f (x) dx = [F (x)]ba = F (b) − F (a),
a

whenever F ′ (x) = f (x).

Kwame Piesie 120 / 172

Example
Evaluate the following integrals with respect to x
Z 3
1. (3x2 + 2x + 4) dx
Z1 2
(x+1)(x2 −2x+2)
2. x2
dx
1
Z 2
3. (x3 − x2 + 1) dx
−2
Z 2
4. (x2 − 2)(2x − 1) dx
−2

Example
Z a
If 2(x + 2) dx = 15, where a > 0, find the value(s) of a.
−1

Kwame Piesie 121 / 172

Integration of Simple Trigonometric Functions
Z
1. cos x dx = sin x + c
Z
2. sin x dx = − cos x + c
Z
3. tan x dx = ln sec x + c
Z
4. sec2 x dx = tan x + c
Z
5. cosec2 x dx = − cot x + c
Z
6. sec x tan x dx = sec x + c
Z
7. cosecx cot x dx = −cosecx + c

Kwame Piesie 122 / 172

Integration by Substitution

Functions which are to be integrated are not always in the

standard form which can easily be integrated. However, it is
possible to change them into a form which can be integrated
by using an algebraic substitution.
Example
Evaluate the following integrals
√
Z Z
1. x 1 − x dx 2 5. x2 (1 − x3 ) dx
Z Z
2. 3x(4x + 3) dx
2 5
6. (2 + 7x)7 dx
Z Z
3. √ 2x
4x2 −1
dx 7. 1
(4x+5)3
dx
Z Z
4. x(x − 3) dx
2 4
8. √ 1
1−2x
dx
Kwame Piesie 123 / 172
Example
Compute the following definite integrals
Z 2
1. √ 3x
2x2 +1
dx
0
Z 3
2.
p
5x (2x2 + 7) dx
Z1 1
√
3. x2 x3 + 1 dx
Z0 2
4. x(x2 + 1)3 dx
Z0 2
5. 3
(2y+3)1/3
dy
−1
Z 10
6. √ 1
2y+5
dy
2

Kwame Piesie 124 / 172

Integration by Partial Fraction

Definition: A rational function is any function of the form

P (x)
,
Q(x)

where P (x) and Q(x) are polynomials in x and Q(x) ̸= 0. For

example,
x+1
f (x) = 2
x −1
and
x2 − 2x − 8
g(x) =
x+4
are rational functions.

Kwame Piesie 125 / 172

If the degree of P (x) is less than that of Q(x) then Q(x)
P (x)
is a
proper rational function. Example f (x) = x2 −1 is a proper
x+1

rational function.
If the degree of P (x) is greater than or equal to the degree of
Q(x) then Q(x)
P (x)
is an improper rational function. Example
x2 −2x−8
g(x) = x+4
is an improper rational function.

Kwame Piesie 126 / 172

Resolution of Rational Functions into Partial Frac-
tions

A rational function which may be expressed as a sum of

separate fractions is said to be resolved into partial fractions.
For example,

1 3 4x − 5
+ = 2 .
x−2 x+1 x −x−2

The reverse process of splitting x24x−5

−x−2
into its component
fractions x−2 + x+1 is called resolving into partial fractions
1 3

Kwame Piesie 127 / 172

Type I - Denominator with Linear Factors

Example
Express (x+3)(x−3)
6
as a partial fraction.

6 A B
= + .
(x + 3)(x − 3) (x + 3) (x − 3)

Example
Resolve the following rational functions into partial fractions.
1. 3x+2
x2 +x−2
2. 11−3x
x2 −2x−3
3. 9
2x2 +x
4. 4x−2
x2 +2x

Kwame Piesie 128 / 172

Type II - Denominator with Quadratic Factors

Example
11−4x−x2
Resolve the rational function (x+2)(x2 +1)
into partial fractions.

11 − 4x − x2 A Bx + C
2
= +
(x + 2)(x + 1) (x + 2) (X 2 + 1)
Examples
Express each of the following in partial fractions.

1. 3x+1
(x−1)(x2 +1)
5. 13x
(x−3)(x2 +x+1)
7x2 +5x+13 x2 +9x+6
2. (x2 +2)(x+1)
6. (x2 +1)(x+1)

3. x2 +1 7. x
(x−1)(x2 +1)
(x2 +2)(x−1)
11−4x−x2
4. 1
x(x2 +1)
8. (x+2)(x2 +1)

Kwame Piesie 129 / 172

Type III - Denominator with Repeated Factor

Example
Express (x+2)(x−1)
1
2 as a partial fraction.

1 A B C
= + +
(x + 2)(x − 1)2 (x + 2) (x − 1) (x − 1)2

Example
Express these rational functions as a partial fraction.

1. 1
(x+2)(x−1)2
4. 5x+4
(1−x)(x+2)2

2. 5x2 −2x−19 5. 11x+12

(x+2)(x−3)2
(x+3)(x−1)2
x2 −1
3. 2
x2 (x−1)
6. x2 (2x+1)

Kwame Piesie 130 / 172

Type IV - Improper Fractions

Example
2 +1
Resolve x2x−3x+2 into partial fractions.

x2 + 1 3x − 1
=1+ 2
x − 3x + 2
2 x − 3x + 2
Example
Resolve these rational functions into partial fractions.
2x2 +1 x2 −2
1. (x−1)(x+2)
4. (x+3)(x−1)
x2 +1 x2 +3
2. x2 −3x+2
5. (x−2)(x+2)
x3 +3
3. (x+1)(x−1)

Kwame Piesie 131 / 172

Integrals that yield Logarithmic Functions

We recall that dxd

(ln x) = x1 , and in general if f is a
differentiable function of x then dx d 1
ln (f (x)) = f (x) · f ′ (x).
Therefore Z ′
f (x)
dx = ln |f (x)| + c
f (x)

Also,
Z the power rule for integration states that
1
xn dx = n+1 · xn+1 + c, provided n ̸= −1.
Z
To find the integral of x when n = −1 we have x−1 dx = x1
n
Z
since dx ln x = x . We then conclude that x1 dx = ln |x| + c
d 1

Kwame Piesie 132 / 172

Examples
Compute the following integrals
Z −2
1. 1
x
dx
−6
Z 5
2. 6
x
dx
Z1

3. 1
x+2
dx
Z
4. 4x
x2 +5
dx
Z
5. 1
2x+3
dx
Z
6. 2x
3x2 −2
dx
Z
x3
7. x4 +10
dx

Kwame Piesie 133 / 172

Example
Compute the following integrals
Z
1. 5
(x−2)(x+3)
dx
Z
2. 4
x2 −4
dx
Z
3. 2x−1
(x+1)2
dx
Z
4. x
4−x2
dx
Z 3
5. 5+x
(1−x)(5+x2 )
dx
2
Z 3
6. x−4
(x+2)(x−1)
dx
2

Kwame Piesie 134 / 172

Example
Evaluate the following integrals, correct to 3 sf
Z 5
1. 2
x2 −1
dx
3
Z 0
2. 2
(x−1)(1+x2 )
dx
−1
Z 3
3. x−9
x(x−1)(x+3)
dx
2

Kwame Piesie 135 / 172

Example
Find the following integrals
Z
1. x
1+x2
dx
Z
2. x
1+x
dx
Z
3. x
1−x2
dx
Z
4. x−2
x2 −4x−5
dx
Z
5. 7x+2
3x3 +x2
dx
Z
6. x
16−x2
dx

Kwame Piesie 136 / 172

Integration of Exponential Functions

Z
We recall that d
dx
(ex ) = e . Thus
x
ex dx = ex + c.
In general, if f (x) is a differentiable function of x which yields
a constant upon differentiation then
Z
1
ef (x) dx = · ef (x) + c.
f ′ (x)
Z
Also ekx dx = k1 ekx + c.

Kwame Piesie 137 / 172

Examples
Z Z
1. e2x dx 4.
2 +2x
(2x + 1)e2x dx
Z Z
2. (2 − e2x ) dx 5. e2x+3 dx
Z Z
3. 10xex dx 6.
2 3
x2 ex dx

Kwame Piesie 138 / 172

Application of Integration

1. Average Value of a Function: We recall that the

average/mean of n numbers x1 , x2 , x3 , ..., xn is given by
n
1X x1 , x2 , x3 , ..., xn
x= xi = .
n n
i=1

Now the average value of a continuous function f , on the

closed interval [a, b] is given by
Z b
1
f (x) dx .
b−a a

Kwame Piesie 139 / 172

Example
Determine the average value of f (x) = x3 − 2x2 − 5x + 1 on
the interval [−2, 4]

Example
The annual per capita consumption of alcoholic beverages on
KNUST campus can be modeled by the function

f (x) = 43e−0.012x , 1 ≤ x ≤ 11,

where x is measured in years. Determine the average per

capita consumption of alcoholic beverages on campus.

Kwame Piesie 140 / 172

2. Area under a Curve: Suppose y = f (x) is the equation
of a curve and that it is required to find the area enclosed by
the curve, the x-axis, and the lines x = a and x = b. This is
given by
Z b
A= f (x) dx .
a
The area is positive if it is above the x-axis, and negative if it
is below the x-axis

Kwame Piesie 141 / 172

Examples

1. Determine the area enclosed by the curve y = 2x + 3, the

x-axis and the ordinates x = 1 and x = 4.
2. Find the area enclosed by the curve y = 3x + 4, the
x-axis and the lines x = 1 and x = 4.
3. Determine the area bounded by the curve y = x2 + 5, the
x-axis and the lines x = 0 and x = 3.
4. Sketch y = x2 − 4x − 5 and find the area bounded by the
curve and the x-axis.
5. Determine the area enclosed between the curve
y = x(x − 1)(x − 2) and the x-axis.

Kwame Piesie 142 / 172

3. The Area between two Curves: The area enclosed
between two curves y = f1 (x) and y = f2 (x) is given by
Z b Z b
A= f2 (x) dx − f1 (x) dx
a a

Examples
1. Sketch the curves y = x2 + 1 and y = 7 − x on the same
axis and find the area enclosed between the curves.
2. Determine the coordinate of the points of intersection of
the curves y = x2 and y 2 = 8x. hence sketch the curves
y = x2 and y 2 = 8x on the same axis and calculate the
area enclosed by the two curves.
3. On the same axis, sketch and label clearly the curves
y = x2 − 5 and y = 5 − x2 . Find the area of the finite
region enclosed by the two curves.
Kwame Piesie 143 / 172
4. Finding the equation of a curve when the gradient
function is given:
When the gradient function dx
dy
of a curve y = f (x) is given,
we can find the equation of the curve by integration. Thus if

dy
= h(x)
dx
Z Z
then dy = h(x) dx. Therefore dy = h(x) dx and
Z
y = h(x) dx +c.

Kwame Piesie 144 / 172

Examples
1. A curve passes through the point (1, 0). If the gradient
function at any point P (x, y) is 3x2 − 1, find the
equation of the curve.
2. The gradient function of a curve at any point P (x, y) is
4x2 − 3x + 1. If the curve passes through the point
A(1, 0), find the equation of the curve.
3. The gradient function of a curve at any point is given by
(2x − 3x2 ). If the curve passes through the point (1, 2),
find the equation of the curve.
4. A curve passes through the point (1, −11) and its
gradient function at any point is ax2 + b, where a and b
are constants. A tangent to the curve at the point
(2, −16) is parallel to the x-axis. Find;
a. the values of a and b,
b. the equation of the curve.
Kwame Piesie 145 / 172
Consumers’ and Producers’ Surplus

1. Consumers Surplus: It is the benefit enjoyed by

consumers who are willing and able to offer prices above the
equilibrium price. This benefit is shown as the shaded area in
the diagram below.

Kwame Piesie 146 / 172

Let p0 and q0 be the equilibrium price and the quantity
demanded respectively. If the demand function is p = f (q),
then consumers surplus (CS), is given by
Z q0
C.S = f (q) dq −p0 q0
0

Kwame Piesie 147 / 172

Examples
1. The demand function for a product is p = 100 − 0.5q,
where p is the price and q is the quantity demanded. If
the supply function is p = 10 + 10q
, determine the
consumers surplus
2. In each of the cases below, determine the total
benefit/gain to consumers who are willing to pay more
than the equilibrium price.

a. pd = 20 − 0.8q d. pd = 400 − q 2
ps = 4 + 1.2q ps = 200q + 100
b. pd = 900 − q 2 e. q = 100(10 − p)
ps = 100 + q 2 q = 80(p − 1)
50 √
c. pd = q+5 f. q = 100 − p
q
ps = 10 + 4.5 q = p2 − 10

Kwame Piesie 148 / 172

2. Producers Surplus: There is an economic gain to
producers who are willing to supply their products at prices
lower than the equilibrium price p0 . The total gain to
producers is the shaded area shown in the diagram below

Kwame Piesie 149 / 172

Let p0 and q0 be the equilibrium price and the quantity
supplied respectively. The shaded area depicts the total gain
to producers. If the supply function is p = f (q), then
producers surplus (PS), is given by
Z q0
P.S = p0 q0 − f (q) dq
0

Example
The demand function for a product q is p = q+2
90
and the supply
equation is p = q + 1. Determine the producer’s surplus.

Kwame Piesie 150 / 172

Examples

1. Given the demand function p = 42 − 5q − q 2 and

assuming that the equilibrium price is GH¢6. Find the
consumers’ surplus.
2. Find the consumers’ surplus when equilibrium price and
quantity are respectively GH¢32.5 and 25 and
p = 45 − 0.5q
3. Given that p = (q + 3)2 , find the producers’ surplus at
p0 = 81 and q0 = 6.
4. Given the demand function p = 25 − q 2 and the supply
function p = 2q + 1, assuming that there is pure
competition in the economy, find
(a) the consumers’ surplus,
(b) producers’ surplus.

Kwame Piesie 151 / 172

Exercise

1. Given the demand function p = 113 − 3q 2 and the supply

function p = (q + 1)2 under pure competition, find the
consumers’ and the producers’ surplus.
2. Under monopoly, the quantity sold and market prices are
determined by the demand function. If the demand
function for a profit maximizing monopolist is
P = 274 − q 2 and M C = 4 + 3q. find the consumers’
surplus.

Kwame Piesie 152 / 172

Remark
For profit maximization, M C = M R
Proof
Let π denote the profit function. Then π = R − C, where
R = total revenue and C = total cost
Now from π = R − C,
dπ d d
dq
= dq (R) − dq (C)
But for maximum profit dπ
dq
=0
d d
∴ dQ
(R) − dq
(C) =0
dR dC
=⇒ dq
= dq

But dR
dq
= M R and dC
dq
= MC
∴ M R = M C as required.

Kwame Piesie 153 / 172

Differential Equations

A differential equation is an equation that contains an

unknown function and one or more of its derivatives. The
order of a differential equation is the order of the highest
derivative that occurs in the equation.
d2 y d2 y
Examples are dy
dx
= xy, dx2
+ dy
dx
= 4x2 , dx2
= −ax
The general solution of a differential equation is a function
that satisfies the differential equation together with a constant
of integration. This is called the general solution. If conditions
that enable us to determine the value of the constant of
integration are given then the solution is called a particular
solution. The conditions are called initial condition

Kwame Piesie 154 / 172

Separable Differential Equations

A differential equation of the form

dy
= f (x) · g(y)
dx
or
dy f (x)
= , g(y) ̸= 0
dx g(y)
is separable if everything involving x can be moved to one side
of the equation and everything involving y can be moved to
a
other side of the equation. For example dx dy
= xyb is a first-order
separable differential equation. The method of solving such a
differential equation is called separation of variables.

Kwame Piesie 155 / 172

Example
Find the general solution of the following differential equations

1. dy
dx
= 3x2 5. x2 dx
dy
= y(y − 1)
x2
2. dy
dx
= y3
6. dy
dx
= x
y
3. dy
dx
= 2x 7. dy
dx
= y
x
4. dy
dx
= xy

Example
Determine the particular solution for the following differential
equations

1. dy
dx
= 3xy; x = 0, y = 3
2. dy
dx
= xy − 3x; at y(0) = 1

Kwame Piesie 156 / 172

Example
The annual rate of increase in the number of employees
between 1950 and 1985 at the NewMedia Company can be
modeled by the differential equation

dy
= 0.3t0.6 , 0 ≤ t ≤ 35,
dt
where t represents the number of years and y represents the
number of employees in 100s

1. If the company had 1800 employees in 1965, find the

employee growth function for the company
2. Use your answer in (1) to estimate the number of
employees at the company in 1983.

Kwame Piesie 157 / 172

Example
The StrikeNow Company determines that the marginal cost for
their MaxiPak of lighters is given by the differential equation

dC
= −0.02x + 6, 0 ≤ x ≤ 300,
dx
where x is the number of MaxiPaks produced and C is the
cost in dollars.

1. Determine the general solution of the cost function C

2. Find the particular cost function, if the cost of producing
10 Paks is 400.

Kwame Piesie 158 / 172

Exercise

Determine the general solution of the following differential

equations.

1. dy
dx
= −2
2. dy
dx
=e
3. dy
dx
= 1 − 3x
√
4. dy
dx
= x+2
5. dy
dx
= 5
x2
6. dy
dx
= 10x4 − x2 + 3

Kwame Piesie 159 / 172

Exercise

Determine the particular solution of the following differential

equations.

1. dy
dx
= 12x; x = 1, y = 8
2. dy
dx
2
= 3x + 4x; y(1) = 6
3. dy
dx
= xe ; x2
y(0) = 2
4. dy
dx
= x3 − 2x; y(0) = −2

Kwame Piesie 160 / 172

Partial Differentiation

So far we have taken a look at functions of one independent

variable and their related rates. If y = f (x) then dx
dy
is the
total derivative of y with respect to x. Now if z = f (x, y)
then the derivative of z with respect to either x or y is called
the partial derivative of z. An example of the form z = f (x, y)
represents a function of two independent variables. The
domain of z = f (x, y) is the set of all ordered pairs (x, y) for
which z is defined.

Kwame Piesie 161 / 172

Examples

1. Consider f (x, y) = 2x2 − 3y 3 . Determine the following

a. f (2, 1) b. f (−1, 2) c. f (1, −2) d. f (2, 3)

2. The cost of producing x tennis balls and y volleyballs is

given by C(x, y) = 0.97x + 0.89y
a. Determine the domain of C(x, y).
b. Evaluate C(250, 700) and interpret it.

Kwame Piesie 162 / 172

Definition of Partial Derivatives

1. Let f (x, y) be a function of two variables. The partial

derivative of f (x, y) with respect to x is defined as

∂f f (x + h, y) − f (x, y)
= lim .
∂ x h−→0 h

The partial derivative of f (x, y) with respect to x is found by

treating y as a constant and performing our ordinary
differentiation techniques with respect to x. The notation for
this partial derivative is fx or ∂f
∂x
.
2. The partial derivative of f (x, y) with respect to y is defined
as the limit
∂f f (x, y + h) − f (x, y)
= lim
Kwame Piesie ∂ x h−→0 h 163 / 172
This is obtained by treating x as a constant and performing
our ordinary differentiation techniques with respect to y. The
notation for this partial derivative is fy or ∂f
∂y
.

Kwame Piesie 164 / 172

Examples

1. Determine fx (x, y) and fy (x, y) for f (x, y) = xy − y 2 .

2. Compute the partial derivatives for
f (x, y) = x2 y + y 3 x − 2xy = y.
3. Determine fx and fy for each of the following.
a. f (x, y) = 5x2 − 6y 3
b. f (x, y) = 2xy − y 2 + 1
c. f (x, y) = x3 y 2
d. f (x, y) = x2 y 4
a. f (x, y) = xy

Kwame Piesie 165 / 172

Applications of the Partial Derivatives

1. Analysis of Marginal Revenue

Example; A company sells two products: military boots and
life jackets. Let x and y be the number of military boots sold
per month and the number of life jackets sold per month
respectively. Suppose that p1 = 1200 − 18x − y and
p2 = 523 − 1.5x − 0.15y are the demand function for the
military boots and life jackets respectively. Determine

1. the revenue function R(x, y).

2. Rx (x, y) and Ry (x, y) and interpret it,
3. evaluate and interpret Rx (50, 25).

Kwame Piesie 166 / 172

Example

1. Donyma Steel Complex spends x thousand cedis each

week on newspaper advertising and y thousand cedis each
week on radio advertising. The company has weekly sales,
in terms of thousands of cedis, given by
S(x, y) = 2x2 + y. Determine
a. Sx (x, y) and Sy (x, y).
b. Sx (3, 5) and Sy (3, 5) and interpret each.
2. If S(x, y) = 3x + 2y 3 , compute Sx (2, 6) and Sy (2, 6) and
interpret each.
3. For f (x, y) = x2 + y 2 − 8x + 2y + 7, determine the value
of x and y such that fx = 0 and fy = 0 simultaneously.

Kwame Piesie 167 / 172

2. Marginal Productivity of Labour and Capital
If L represents the units of labour and K represents the units
of capital, then the total production Q is given by the
Coff-Douglas production function.

Q(L, K) = ALa K b ,

where A, a and b are constants.

Example
If f (x, y) = 4x0.75 y 0.25 , determine the realistic domain of
f (x, y) and evaluate f (x, y) when x = 50 and y = 60 and
interpret it.

Kwame Piesie 168 / 172

Definition
For any function Q(x, y) = Axa y b where A, a and b are
constants and x and y represent units of labour and capital
respectively, Qx gives the approximate change in productivity
per unit change in labour. It is called the marginal productivity
of labour, and Qy gives the approximate change in the
production per unit change in capital. Qy is called the
marginal productivity of capital

Kwame Piesie 169 / 172

Example
The production function for a company is given by
f (x, y) = 1.64x0.64 y 0.4 Q, where x is the number of labour
hours in hundreds and y is capital in thousands, and Q is the
quantity in 1000s of gallons.

1. Compute fx and fg .
2. Find fx (100, 18) and interpret it.
3. Find fy (110, 20) and interpret it.

Kwame Piesie 170 / 172

Remark
For some amounts invested in labour and in capital, we wish
to know whether production would increase more rapidly if
additional resources were invested in labour or in capital. We
can approximate the change in productivity by any of the
following
1. The change in productivity with respect to labour =
fx ∆x
2. The change in productivity with respect to capital =
fy ∆y
Example
Suppose that a production function is given by

f (x, y) = 4.23x0.37 y 0.66 ,

Kwame Piesie 171 / 172

where x=labour (million) and y=capital (million). Currently,
x = 5 and y = 1. Would production increase more by
spending additional GH¢1m on labour or GH¢500000 on
capital equipment?

Kwame Piesie 172 / 172