Calculus For Economists

Unit 3-Derivative
UNIT 3. DIFFERENTIAL CALCULUS-DERIVATIVE

3.0 AIMS AND OBJECTIVES

 Find the rate of change of an endogenous variable with respect to the exogenous variable.

 The mathematical concept of derivative is directly related to the notion of rate of change.

 Will concentrate on functions with a single independent variable.

 An extension of this will be discussed in unit five.

3.1 INTRODUCTION

 Will learn the derivative of different types of functions.

 Will learn the sum, difference, product & quotient rules.

 Other rules like the chain rule and inverse rules will be introduced.

 Rules of differentiation for exponential and logarithmic functions are also discussed.

3. 2 DEFINITION OF THE DERIVATIVE

 Define derivative in terms of the limit concept.

 The concept of derivative is the basis of calculus.

 Derivative tells us many things about the function and has a variety of applications.

 Example: in the analysis of rates of change, curve sketching, optimization problems etc.,

we require the concept of derivative.

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The general form of the function
The general form of the function is y = f (x
(x).
For a function of the form y = f (x
(x), we can find the average rate of change of y when
change in x, ∆x
∆x, is very small.
Let x changes from x0 to x0 + ∆x
∆x, the value of the function y = f (x
(x) changes from f (x0) to f
(x0 + ∆x
∆x).

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 The change in y per unit change in x can be represented by the difference quotient

 Measures the average rate of change of y

 Is also called a difference quotient
 It can be interpreted as the slope of a secant

Example 1: Given y = x2, find the average rate of change of y.

Solution f(x0) = x02 and f(x0 + ∆x

∆x) = (x ∆x)2
(x0 + ∆x
The next step is to form the difference quotient

 Let x0 = 5 and ∆x
∆x = 2; then the average rate of change of y will be 2(5) + 2 = 12.
 This means that, on the average,
average, as x changes from 5 to 7, the change in Y is 12 units per
unit change in x.
 In economics, we are frequently interested in the rate of change of y when ∆x
∆x is very small

 In the above example, if ∆x

∆x is very small, it tends to zero, then the difference quotient

can be reduced to 2x
2x0
 That is, as ∆x
∆x approaches zero, 2x
2x0 + ∆x
∆x will approach to 2x
2x0

 Symbolically, as , or it can be expressed by the equation

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 (x) = x2, provided that the limit of
This is the derivative of the function y = f (x exists as

∆x approaches zero.

Therefore, given a function Y = f (x), the derivative of the function f at x, written

f1(x) or dy/dx, is defined as

, provided the limit exists or let's use h instead of ∆x

, provided the limit exists

Notations: f1(x) is read as "f

"f prime of x"
dy/
dy/dx is read as "the derivative of y with respect to x"

Therefore, for a function y = f (x

(x), the derivative may be denoted by

dy/dx = f1(x) =
Note that
 A derivative is a function (a derived function).
 The original function is called primitive function.
function.
 The difference quotient measures the average rate of change where as the derivative
measures both the slope and the instantaneous rate of change of the original function f(x)
at a given point.
 Steps on how to compute the derivative of f(x), f1(x).

1. From the difference quotient (Note that ∆x

∆x = h)
2. Simplify the difference quotient algebraically to eliminate the factor h from the
denominator.
3. Let ∆x
∆x or h approaches zero in the simplified difference quotient.
if this limit exists

Example 2. Given f (x 2x + 3, find f1(x)

(x) = 2x

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Solution:
f 1( x ) =

Now take the limit of 2.

(x) = 2x2 + 3x + 5, find f1(x)

Example 3: Given f (x

Solution:

Form the difference quotient

This is the simplified expression

Next evaluate the limit of the simplified expression as h approaches zero.

i.e. f1(x) = (4x

(4x + 2h
2h + 3) = 4x
4x + 3

Check Your Progress Exercise – 1

Use the above steps to find the derivative of the following functions.
a) f(x) = 3x
3x + 5 b) f(x) = x2 + 1
c) f(x) = 1/x
1/x 4x2 – 5x
d) f(x) = 4x 5x + 1

3.3 GEOMETRIC INTERPRETATION OF THE DERIVATIVE

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 Will see the derivative as a measure of the slope of the line tangent to the curve at a given
point.

Figure 3.1 The Graph of f(x) = x2

 The derivative f1(x) expresses the slope of the tangent to the curve y = f(x) as a function of
the x coordinate of the point of tangency.
 To say that f is differentiable at a point means geometrically that the curve y = f(x) has at
that point a unique tangent line which is not parallel to the y – axis.
 Therefore, in terms of the derivative, the slope of the function y = f (x
(x) curve at point A
corresponds to the particular derivative value f1(x).

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3.4 DIFFERENTIABILITY AND CONTINUITY

 Use the concept of limit to see the relationship b/n continuity and differentiability.
 A function is said to be differentiable at x = c if it has a derivative when x = c.
 If it does not have a derivative at c, we say it is not differentiable at c.
 To be differentiable at a point, a function must
I. be continuous at that point and
II. have a unique tangent at that point
 But continuity is not a sufficient condition for differentiability.
 In fig. 3.4 (a) and (b) the functions are continuous at x = 0 but they are not differentiable.
 The graph of a differentiable function cannot have a sharp point, a cusp or a break.

f(x) f(x)

x
x

f (x ) = / x / f(x) = x2/3
a) - Corner: no tangent at x = 0 b) -cusp: vertical tangent at x = 0
- f(x) is continuous at x = 0
- Not differentiable at x = 0

f(
f(x) =

c) -The function is undefined at x = 0 d) –Discontinuity at x =0

- Not continuous at x = 0
- Not differentiable at x = 0

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Example 1:
1:
 Show that the function f(x) = /x
/x/ is continuous at x = 0 but not differentiable at x = 0
Solution:
Step 1: Check continuity first.
i) f(0) is defined

ii) f(x) exists ( = 0. Why the value of the function to the

left of zero is zero as any number to the left of zero is actually –ve number? This is
mainly due to any number out of the absolute value is positive. And this case is as x
approaches to zero so that it is almost zero

iii) f(0) = f(x) = 0

Therefore, f(x) = /x
/x/ is continuous.
Step 2: To check its differentiability, take the limit of the difference quotient.
 If the limit exists, it is differentiable at x = 0.
 If the limit of the difference quotient does not exist, f(x) is not differentiable at x = 0

Now check both the right and left side limits.

 Since the right and left side limits are not the same we say f is not differentiable at x = 0.
 This shows that continuity does not guarantee differentiability.
Let us make clearer the example 1 given above:
above:
Absolute value function is piecewise function
So we can write it like:

So,

and

Therefore,

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 But i.e = =1 = = -1

So the limit from the right is 1, while the limit from the left is −1.
This means the two sided limit does not exist.

That is, the derivative does not exist at .

Take example:

Therefore as the limit doesn’t exist

So the

See using derivative of it:

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 , two different value of f derivative

Therefore,

Example 2:
 Prove that the function f is given by

Solution

 So the

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 Now

= = =1

= = -1
Since the left hand limit is not equal to the right hand limit f(x) is not differentiable at x = 1
Example 3:
 Show that the function f (x
(x) = /x
/x – 2/ + 1 is continuous at x = 2 but not differentiable at
x=2
Solution
The function is continuous at x = 2 because
i) f(x) is defined

ii) f(x) exists

iii) f(2) = f(2) =1

 Next show that the limit of the difference quotient does not exist and f is not
differentiable at x =2

is the derivative of the function f (x

(x) = /x
/x-2/ + 1

 The next step is to check the left and right side limits.

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 Since the right and left side limits are not equal the limit of the difference quotient does
not exist.
exist.
 Hence it is not differentiable at x = 2

Example 4

 Show that the following function is differentiable or not at x = 0

Solution

 H is not continuous at x = 0
 So it is not differentiable at 0

Example 5:

 Check the given equation is correct

Class Work
Which of the following is true about the function given below at x =2?

A. Continuous but not differentiable

B. Differentiable but not continuous

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 C. Both continuous and differentiable
D. Neither continuous nor differentiable
Solution
a). First checking for continuity at x=2
 For the function to be continuous at x=2, we need the two-sided limit ) to exist

and be equal to f(2)

 This is the same as requiring that the two one-sided limits and

exst and are equal to f(2)

 Then =

 And =

 The two limits exits, but they are not equal.

 Therefore, the function is not continuous at x = 2
b). Second checking for differentiability at x=2
 Since the function isn't continuous at x =2, it cannot be differentiable at that point.
 In conclusion, the function is neither continuous nor differentiable at x=2
Simple Quiz:
 Check if the given below function is differentiable

 Hint check the continuous and limit of the function

Theorem: A differentiable function is continuous.

 If f(x) is differentiable at x =a, then f(x) is also continuous at x = a
 If f not continuous at x = a, then f is not differentiable at x = a

Check Your Progress Exercise – 2

Show that the following functions are continuous but not differentiable at the given value of
x
a) f(x) = /x
/x-1/ : x = 1
b) f(x) = x1/3 : x = 0

3.5 RULES OF DIFFERENTIATION

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 The d/t rules of differentiation simplifies the task of finding the derivative of a function

 The process of finding the derivative of a function is called differentiation.

 Up to now we found the derivative of a function by using the steps discussed in page

 But it is a tedious process and time -consuming.

 Hence the following rules will help to simplify the process.

1. Constant Function Rule

The derivative of a constant function y = f(x) = k is zero.

Example 1: Find the derivative of f (x

(x) = 3
Solution: f1(x) = 0
Solution:
2. The power Rule
For any number n, the derivative of a power function f(x) = xn is nxn-1

i.e.

Example 2. Find the derivative of the following functions

a) f (x ) = x 2

Solution: f1(x) =

b) f (x ) = x 3

Solution: f1(x) =
Solution:

c)

Solution:
Solution:

d) f(x) = x25

Solution:
Solution:

3. The Sum – Difference Rule

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The derivative of a sum (difference) of two functions is the sum (difference) of the individual
derivatives.

If both f1(x) and g1(x) exist then

Example 3: Find the derivative dy/

dy/dx for each of the following functions.

a) f(x) = x2 + 2x
2x

Solution:
Solution:

But

Therefore

b) (x) = x3 + x2 + 1
f (x

Solution:
Solution:

But

c) f(x) = x5 + x3 + x + 10
Solution:

d) f(x) = x2 – 5x

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Solution:
Solution:

Check Your Progress Exercise - 3

Find the derivatives of the following functions by using the above rules of differentiation.

a) f(x) = 1/5 b)

c) d) f(x) = x

4. The Product Rule

 The derivative of the product of two (differentiable) functions is equal to the first function
times the derivative of the second function plus the second function times the derivative of
the first function.

Example 4 Find the derivative of each of the following functions

a) y =(x (x2 + 5)
=(x + 1) (x

Solution: Let f (x (x) = x2 + 5

(x) = x + 1 and g (x
Then f1(x) = 1 and g1(x) = 2x
2x

, Substituting the above values we get

b) 2x2 (3x
y = 2x (3x4 – 2)

Solution:
Solution: Let f (x 2x2 and g(x) = 3x
(x) = 2x 3x4 – 2

(x2 – 5) (1-2x
c) y = (x (1-2x)

Solution: Let f(x) = x2 – 5 and g (x

(x) = 1 – 2x
2x

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Note that the product rule can be extended to more than two functions. For instance for the
case of three functions,
y = f(x) . g(x) . h(x)

Example 5: Find the derivative of the function

(3x2 + 2) (3x
y = (3x (3x) (6x + 7)

Solution: Let f (x 3x2 + 2, g(x) = 3x

(x) = 3x 3x and h(x) = 6x
6x + 7

5. The Quotient Rule

The derivative of the quotient of two functions, is

Example 5 y =

Solution: Let f (x) = x + 1 and g(x) = x – 2. Then

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Example 6 y=

Solution:

Example 7

Solution:

Correction: The denominator (x-3)3 is

to mean that (x-3)2

Check Your Progress Exercise – 4

1. Differentiate the following functions by using the product rule

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 a) f(x) = (x
(x – 3) (2x
(2x – 1) (5x2) (x
b) f(x) = (5x (x3 +2)
c) f( (x2 +1) (2x
f(x) = (x (2x – 3) d) f(x) = (2x (x2 – 1) (3x
(2x + 1) (x (3x2)

2. Differentiate the following by using the quotient rule

6. The Chain Rule:

Rule:
 Will develop a rule for differentiating function of one variable, which in turn is a
function of a second variable.
 Suppose y is a differentiable function of u and u is a differentiable function of x.
 Then y can be regarded as a function of x and the derivative of y with respect to x is the
derivative of y with respect to u times the derivative of u with respect to x.
 That is,

……………………… Leibniz notation

Or

 If g is differentiable at x and f is differentiable at g(x), then the composite function F =

f ∘ g defined by F(x) = f(g(x)) is differentiable at x and F’ is given by the product
F’(x) = f’(g(x)) · g’(x)
 So the chain rule the
t Chain Rule can be written either in the prime notation
 (f ∘ g)’(x) = f’(g(x))·g’(x)
Or if
 y = f(u) and u = g(x), in Leibniz notation:

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Example 1 Find the if y = u2 + 1 and u = 3x
3x –2

Solution: First find

(by power rule)

Then find

Therefore,

 Since you are thinking of y as function of x, it is more natural to express as a function

of x.
 Substituting the value of u, we get.

Example 2: Find if y = u – 5, where u = x3

Solution:

Example 3: Given the function Y = , find ?

Solution:

Example 4: Find (2x – 5)10

if y = (2x

Solution:

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 The chain rule is very useful to differentiate such kind of functions.
 Note that the dy/
dy/dx can be found by multiplying 2x
2x – 5 ten times.
 But that will be a tedious process.
 The first step is to take the function inside the bracket as one function, say u and apply the
chain rule.

Let 2x – 5 = u, then y = u10

 Substituting the value of u, 2x – 5, we get

Example 5: Find the derivative of the function

Solution;
Combining the Power Rule, Chain Rule, and Quotient Rule, we get

so that

= * =
Example 6:
6: Differentiate the function given as;
y=

In this example we must use the Product Rule before using the Chain Rule:

= +

= (2x+1)

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 = (2)
=

7: Differentiate y =
Example 7:

Solution

y=

Apply chain rule to derivate . This can be differentiate using the

chain rule, which states that [f(g(x)] is f’(g(x))g’(x), Where f(x)=sec(x)

and g(x)= x3

let Hence, find

 The derivative of sec(u) with respect to u is sec(u)tan(u).

 And the derivative of .

Therefore

Now apply all the given derivatives;

= 3

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 =

7. Inverse Function Rule

 Will learn the rules of finding the derivative of inverse function by applying inverse
function rule.
 This rule is applicable if the function has an inverse.
 The function f will have an inverse function, written as x = f-1(y), if there is a one -to –one
mapping (for monotonic functions).
 A function f(x) is called one-to-one if every element of the range corresponds
to exactly one element of the domain.

 Similar to the Vertical Line Test (VLT) for functions, we have the Horizontal Line Test
(HLT) for the one-to-one property.
 The Horizontal Line Test Theorem. A function is one-to-one if and only if there is no
horizontal line that intersects its graph more than once.
 Given a function f(x), if x1 > xf2 is
andone-to-one;
f(x1) > f(x2) then thenot
g is function f is called an increasing
or monotonically increasing function.
 If x1 > x2 and f(x1) < f(x2), then the function is said to be a decreasing (or monotonically
decreasing) function.
 In both cases an inverse function exists.
 However for instance “Parabola is Not One-to-one”
 The parabola f(x) = x2 is not one-to-one because it does not satisfy the Horizontal Line
Test.
 For example, the horizontal line y=1and y=1 intersects the parabola at two points, when x
= −1 and x=1.

f(x) = x2

 
X
-1 1

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The graph of the function f(x) = x2
 This test states that a function f has an inverse function if and only if every horizontal line
intersects the graph of f at most once
 QUESTION:
QUESTION: What is the relationship b/n derivatives of a function and its inverse????
 The function f (x
(x) = x + 3 from A = {1, 2, 3, 4} to B = {4, 5, 6, 7} can be written as

 By interchanging the first and second coordinates of each ordered pair, you can form the
inverse function of f.
 This function is denoted by f –1.
 It is a function from B to A, and can be written as

 Note that
 the domain of f is equal to the range of f –1, and vice versa, as shown in Figure.
 the functions f and f –1 have the effect of “undoing” each other.
 that is, when you form the composition of f with f –1 or the composition of f –1 with f,
you obtain the identity function.

(f –1(x)) = x
f (f and f –1(f (x
(x)) = x

 Here are some important observations about inverse functions.

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 1. If g is the inverse function of f, then f is the inverse function of g.
g.
2. The domain of f –1 is equal to the range of f, and the range of f –1 is equal to the domain
of f.
3. A function need not have an inverse function, but if it does, the inverse function is
unique.
 The rule of differentiation is: the

 This means that the derivative of the inverse function is the reciprocal of the derivative
of the original function.
 Or use the following way to find the derivative of inverse functions

When applying chain rule for this we get:

 Hence we can generalize this as follows:

 And we can derivate the invers function using implicit differentiation approach
Example 1 Given y = 3x
3x + 5, find dx/
dx/dy
Solution:
 Since the function is a monotonic function (its slope is positive)
positive) an inverse function
exists.

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 Hence the derivative of the inverse function is

Example 2 Given: y = x3 + x; find dx/

dx/dy
Solution:
Solution:

 First check whether an inverse function exists

 This value is greater than zero for any value of x.

 That is, it is a monotonically increasing function and an inverse function exists.

Example 3: If a function f (x
(x) = 2x
2x + cos x, find dx/
dx/dy
 First check whether an inverse function exists

Y=

Example 4: The function given y = x2, find dx/

dx/dy
 First check whether an inverse function exists :
 This function is not monotonic.
 Because it is not a one-to- one mapping.
 It does not have an inverse and the inverse function rule cannot be applied.
 Note that the inverse function rule is strictly speaking applicable only when the function
involved is a one-to-one mapping.
 Note that the inverse function rule is strictly speaking applicable only when the function
involved is a one-to-one mapping.
mapping.
Example 5: show that the functions are inverse functions of each other

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 and
Solution:
Þ Because the domains and ranges of both f and g consist of all real numbers, you can
conclude that both composite functions exist for all x.
Þ The composition of f with g is given by:

Þ The composition of g with f is given by

Þ Because f (g(x)) = x and g(f (x)) = x, you can conclude that f and g are inverse
functions of each other
Þ See the below graph

Þ In the above figure, the graphs of f and g = f –1 appear to be mirror images of each
other with respect to the line y = x.
Þ The graph of f –1 is a reflection of the graph of f in the line y = x.
Þ This idea is generalized in the next theorem

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 The following guidelines suggest a procedure for finding an inverse function

 Example 6: Find the inverse function of:

Solution:
 From the graph of f the given function (depicted below), it appears that f is increasing over
its entire domain, (3/2, ∞)

 To verify this, note that is positive on the domain of f.

 So, f is strictly monotonic and it must have an inverse function.

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 To find an equation for the inverse function, let y = f (x) and solve for x in terms of y

 The domain of f –1 is the range of f which is .

 You can verify this result as shown.

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

Example 7: Let

a) What is the value of f –1(x) when x = 3?

b) What is the value of (f –1)'(x) when x = 3?
Solution:
Notice that f is one-to-one and therefore has an inverse function.
a. Because f (x) = 3 when x = 2, you know that f –1(3) = 2
b. Because the function f is differentiable and has an inverse function, you can apply the
rule of it to write

 Note that at the point

(2, 3) the slope of the
–1
graph of f is 4 and at the point (3, 2) the slope of the graph of f is as shown in the next
figure

 In general, if y = g(x) = f –1(x), then f (y) = x and f'(y) =

 The rule says that

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  This reciprocal relationship is sometimes

 Example 8:

df df  1
Given: f 3 5 3 6 Find: 5
dx dx

 Example 9:

. , find

df  1 1

5 
dx 6

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Example 10:

 If f (x) = 2x5 + x3 + 1, find (a) f (1) and f '(1) and (b) (f -1 )(4) and (f -1)'(4).
Solution:
Þ Y = 2x5 + x3 + 1
Þ y’ = 10x4 + 3x2 is positive everywhere y is strictly increasing, thus f (x) has an
inverse
Þ Since f(1) = 4 implies the point (1, 4) is on the curve f(x)=2x5 +x3 +1
Þ Hence, the point (4, 1) (which is the reflection of (1, 4) on y =x) is on the curve (f -1)
(x)
Þ Thus, (f -1)(4)=1.
Example 11:
 If f(x) = 5x3 + x +8, find (f - 1
)'(8).
Solution:
Þ Since y is strictly increasing near x =8, f(x) has an inverse near x =8.
Þ Note that f(0)=5(0)3 +0+8=8 which implies the point (0, 8) is on the curve of f(x).
Þ Thus, the point (8, 0) is on the curve of (f -1)(x).
Þ f '(x)=15x2 +1
Þ f '(0)=1
Þ Therefore:

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Check Your Progress Exercise – 5

1. Use the chain rule to compute the derivative dy/

dy/dx

2. Check whether the following functions are monotonic or not and find the inverse
functions if it exists.
a) y = ½x
½x + 3 b) y = x2

3.6 THE RULE

 L‘ Hôpital Rule is a method of Computing the Limit of

indeterminate forms/functions.

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  It is pronounced as Lo-ppi-talls rule.

x3  8
lim 4
x  2 x  2 x  20
In order to use L’Hôpital’s Rule L’Hôpital’s Rule applies
3
direct substitution must result
in 0/0 or ∞/∞.
2 8
24  22  20
 0
0
since this is
indeterminate form.
an

x  8  3x 2
Differentiate the numerator d 3
& the denominator. dx
Since the result is finite
d
dx x  2 x  20 4 x3  2
4
or infinite, the result is
valid.
1. Analytically evaluate the following limit:
Example 1.

Find the limit of the quotient of

3 x2 322

the derivatives.
lim 38
3
423  2
1234 176
x  2 4 x 2
Example 2:
2: Find the limit of as x→1

Solution:

Evaluate this limit by factoring, but let’s see the application of the rule.

Since f (x) = -2x and g

g(x) = -1

Example3:
Example3: Find the limit of as

Solution: f (x) and g (x) becomes infinite as ,( )

= = =2

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 Solution:

and
Since both numerator and the denominator are indeterminate we can apply L‘Hôpital’s
L‘Hôpital’s Rule:

Note:
Note:

Solution:
Solution:
The given limit is indeterminate because, as x  0+, the first factor (x
(x) approaches 0 while the
second factor (ln x) approaches -∞
Writing x = 1/(1/x 1/x  ∞ as x  0+, so l’Hospital’s Rule gives as:
1/(1/x), we have 1/x

Limitations of L’Hôpital’s Rule

• L’Hôpital’s Rule cannot be applied to the finite limit or to the limit which
Repetition:
• On applying L’Hôpital’s Rule one time
can be solved simply.
• If we get another indeterminate form then L’Hôpital’s Rule can be applied more
times•tillApplying
we get a L’Hôpital’s
determinateRule
formto such limits results in wrong answers.
• And then we apply limit on that expression and we get our answer

Example:
• That is, if the right-side expression ( ) again falls into or format,

same as the left side expression, we may reapply the L’Hôpital’s rule (

40
 Example 6: Analytically evaluate the following limit:

In order to use L’Hôpital’s Rule direct 0

1 cos 0

L’Hôpital’s Rule applies s
substitution must result in 0/0 or ∞/∞. is an indeterminate form.
02 0

Differentiate the numerator and the

d
dx L’Hôpital’s
1  cos x  sin x
d
dx x  2
2x
In order to use L’Hôpital’s Rule direct 0
1 cos 0

denominator. Ruleapplies since this
substitution must result in 0/0 or ∞/∞. is an indeterminate form.
2
Find the limit of0
the quotient of the derivatives.
0
 lim 
1 lim
sin x sin 0 0
Differentiate the numerator and the
denominator.
d
dx cos
x 0 x 
2x  sin x d
2
x  0 dx

0 x 2
 0 2x
This is still an indeterminate
L’Hôpital’s Rule again to th

Differentiate
Find the limit of the quotient ofthe
denominator.
thenew numerator
derivatives. and the d
dx sin x  cos x
d
dx 2 x  2

lim sin x
lim sin 0
 0 lim cos x
 cos
the second derivatives.
2 2
0
 12
Find the limit of the quotient of Since the result
infinite, the
This is still an indeterminate form, apply result
2x 20 0 x 0
x 0 x 0 L’Hôpital’s Rule again to the new limit.

Differentiate the new numerator and the

denominator.
d
dx sin x  cos x
d
dx 2 x  2

Find the limit of the quotient of

the second derivatives. lim cos2 x  cos2 0  12 Since the result is finite or
infinite, the result is valid.
x 0

41
Check Your Progress Exercise - 6

Use the rule to evaluate the following limits.

a) b)

c) d)

3.7 Higher Order Derivatives

 In our previous discussion we have considered only the first derivative of a function

 Next we will see the concept of second derivative, and derivatives of even high orders.
orders.
 The derivative of a function is the first derivative.
 The derivative of the first derivative is called the second derivative.
derivative.

 The second derivative of the function f denoted by or indicates that has

been differentiated with respect to x twice.

 If exists for all x values in the domain, the function is said to be twice
differentiable;
 If, in addition is continuous, the function is said to be twice continuously
differentiable.

 We use the notation f for the derivative  f 

d
f ( x )  f ( x )
 That is, dx

 We call f   the second derivative of f. For

 For y  f ( x )  x 5  3 x 4  x, the second derivative is given by

y  f ( x )  20 x 3  36 x 2 .
Continuing in this manner, we have

42
 f ( x )  60 x 2  72 x, the third derivative of f
f ( x )  120 x  72, the fourth derivative of f
f ( x )  120, the fifth derivative of f .
Example1: Find the first and second derivative of the function f (x 3x2-5x
(x) = 3x -5x+15
Solution:
Solution:
The first derivative of the function is f’(x) = 6x
6x-5
And the second derivative is =6
Since is a measure of the rate of change of the function f, is also the measure of
the rate of change of the first derivative,
The second derivative can be differentiated with respect to x to produce a third derivative and
so on.

Example 2: Find the first through the fifth derivative of the following function.

Solution
f ' ( x) 24 x 3  3
f ' ' ( x) 72 x 2
f ' ' ' ( x) 144 x
f 4 ( x) 144
f 5 ( x ) 0

Example 3. Find the first and second derivatives of the rational function f (x
(x) =

Solution:
First find the first derivative by applying the quotient rule.

The second derivative can also be found by using quotient rule

43
Example 4: For
y ( x 2  10 x) 20 , find and

Solution

By the Extended Chain Rule: y 20( x
2
 10 x )19 (2 x  10).
Using the Product Rule and Extended Chain Rule,

y 20( x 2  10 x )19 2  20(2 x  10) 19( x 2  10 x)18 (2 x  10)


40( x 2  10 x )18 ( x 2  10 x )  19( x  5)(2 x  10) 
40( x 2  10 x )18  x  10 x  19(2 x  20 x  50) 
2 2

40( x 2  10 x )18  x  10 x  38 x  380 x  950) 

2 2

y 40( x 2  10 x )18 39 x  390 x  950 .

2

Example 5:
d4  1
4 
Find dx  x


Solution:

d4  1
dx 4  x 
d  1
   x  2
dx  x 
d2  1 d
2      x  2  2 x  3
dx  x  dx

d3  1 d
3     2 x  3   6 x  4
dx  x  dx
d4  1 d 4 5 24
   6 x  24 x 
dx  x 
4 
dx   x5

Check Your Progress Exercise -7

1). Find the second derivative of the function
a).

44
b).
2). Find the first four derivatives of the following function.
a).

b).

3.8 Derivatives of Exponential and Logarithmic Functions

 The derivative formulas for logarithmic and exponential functions
 First will discuss on the simple derivatives of y = x and y =
 And later we will extend it to any base b
 Before that let us see some of the properties of exponential and logarithmic functions

45
46
 3.8.1. Log Function Rule

 The derivative of the natural logarithmic function y = x is

 Let’s proof this formula.

Proof 1: Using the definition of derivative

f(x+x) f(x)

= f(x)

x x x+x
=

Since , then

Now let n =  = = .

47
 As n = or
since ,
Now make substitution on the above limit equation: That is:-

= highlighted in yellow color is “e”

= , but

=
Derivative of lnx
Therefore,

Proof 2: The Second and Simplest way

y ln x
y
e x dy 1
 y
d y d dx e
e   x  u
b
S

dx dx
Implicit Differentiation dy 1

dyy
dx x
e 1 48

dx
Proof 3:
3:

Thus the expression to the right of the limit sign above can be converted to the form

From , as x approaches to N, then k will tend to infinity.

Therefore taking the limit of as x approaches N is the same as finding the limit of

Generalizing the above result, we get

If x is replaced by some function of x, say f(x), then the generalized rule for finding the
derivative of ln f(x) with respect to x is given as

49
For bases other than e (i.e. logbx), the rule is as follows

If b = e then

Proof:

logbx =

Generalization the above result

Correction: If b = e not 1

the formula would be like that

d 1
 log a x  
Or

dx x ln a
How?

d 1 du
 log a u   
dx u ln a dx

d 50 1 1
Ex   log 2 5 x   5 
dx 5 x ln 2 x ln 2
Example 1 Find the derivatives of the following logarithmic functions

a) f(x) = ln 2x
2x

Solution:

f’(x) =

b) f (x ) =
Solution:

c)

Solution:

= = = =

= .Therefore, = =

d. Let + ) find its second derivative

Solution:
Let us first do its first derivative

51
= = =

= =

Then find the second derivative of it

= = =

= =

d) f(x) = log2x

Solution: by formula 3.10.1 (c) or formula 3.10.1 (d)

e) f(x) = log25x

d 1 1
Ex   log 2 5 x   5 x ln 2 5  x ln 2
dx
f) (x+ x2)
f(x) = log3 (x

Solution: by formula (3.10.1. d)

Check Your Progress Exercise – 8

1. Differentiate the following functions by using logarithmic rule.

3.8.2. Exponential function Rule

Now we learn to differentiate exponential functions of the form .
The derivative of the exponential function y = ex is ex.

52
We can proof this by using the inverse function rule.
Let y = ex
 The inverse of the function will be x= log ey = lny

Then, by inverse function rule:-

d
Ex   e 2 x   e 2 x 2 2e 2 x
dx

(x) is a differentiable function of x, then the derivative of eh(x)

Generally if h (x h(x)
can be found by
using the following formula.

That is the derivative of eh(x) is multiplying eh(x) by the derivative of the exponent h(x).

Note that in y = ex , h (x
(x) = x and h1(x) = 1 hence

Generalizing the above result to any base b, that is, in the form of bh(x), then the derivative can
be found by using the following formula.

- If b = e then the formula will reduce to the above formula. (Since ln b = lne
lne = 1).
- If h(x) = x, then the derivative of bh(x) = bx is equal to bx ln b.
- This can be proved as follows.
Let b = elnb
lnb

53
 bx = exlnb
lnb

Differentiating both sides with respect to x,

d
 a x  a x ln a, w/ a  0, 1
dx
d du
 a u  a u ln a 
dx dx
d
Ex   2sin x   2sin x ln 2 cos x 
dx
Examples 2 Find the derivatives of the following functions by using exponential rules.

a) (x) = ex + 1
f (x
Solution: (x), then f(x) = eh(x)
Solution: Let x + 1 = h (x
f1(x) = h1(x)e
(x)eh(x) = ex+1 --------by rule 3.8.2.b
3.8.2.b

b) f(x) = e4x
Solution: 4x, then f1(x) = h1(x)eh(x)
Solution: Let h(x) = 4x
f1(x) = 4 e4x …… by rule 3.8.2.b

c) f (x ) =
2x2 + 3x
Solution: Let h(x) = 2x 3x + 1 , h1(x) = 4x
4x + 3
f(x) = eh(x) and f1(x) = h1(x) eh(x) = (4x
(4x + 3) -------- by rule 3.8. 2b

54
d) f(x) = 5x-2
Solution: Let x – 2 = h (x (x) = 5h(x)
(x), then f (x
f1(x) = h1(x) bh(x) lnb
lnb = (1) 5x-2 ln5 = 5x-2 ln5

e) f (x ) =
Solution:
Let x2 – 3x
3x = h(x) h1(x) = 2x
2x – 3.
Therefore, f1(x) = (2x
(2x – 3) ln12.

Check Your Progress Exercise – 9

1. Differentiate the following functions by using exponential rule.

(Hint: use the product rule for question 1g).

55
 How do we differentiate a function when both the base and exponent contain the
variable???
 Example: , find the derivative of y with respect to x

56
 For a function like this we have to use Logarithmic Differentiation
 Procedures to use Logarithmic Differentiation are:
are:
a. Take the natural logarithm of both sides of the equation
b. Use the properties of logarithms to simplify the equation
c. Differentiate (sometimes implicitly!) the simplified equation
Solution:
Solution: y x x

ln y ln x x =x
dy
ln y  x ln x  y ln x  1
dx
d d dy
ln y    x ln x   x x ln x  1
dx dx dx
Some Additional Rules including Trigonometric Functions
The derivative of ln(y) is 1/y:

Or if we return to the use of the more common letter x, we have that

Example 1
Let's consider the function

Here we'll need to apply the chain rule . What is the derivative of the outside
function? It is one over the argument:

57
And the derivative of cos(x) is -sin(x):

Example 2
Let's calculate the derivative of

Here a and b are constants. We'll apply the chain rule:

And the derivative of the inside function is a:

Example 3
Let's calculate the derivative of

58
We can actually solve this problem using the chain rule. However, if we go that
route, this will become a long and hairy calculation.

Instead, we'll use an old trigonometric trick that will make our life easier. Inside
the square root, let's multiply both the numerator and denominator by 1+sin(x):

The expression in the denominator is interesting. It actually equals cos 2(x).

That simplified the expression. Now we can simplify it a little more, applying a
property of logarithms

Now we are finally ready to calculate the derivative. We apply the chain rule

59
We can leave at that and accept that expression as the answer. In this case,
though, we can simplify it

So the derivative equals

3.9 The sign of the Derivative

 You are now well acquainted of what the derivatives of higher orders mean
 Form which the second derivative has a special importance.
 It is possible to decide whether the given function is an increasing or decreasing by
observing the sign of the derivative of the given function.
Conditions:-
a) First order derivative
If f(x) is a differentiable function at any point and/ or interval, then

1. If f’(x)(= then the value of the function tends to increase(or f(x) is an

increasing function at a given point and/or interval or the rate of change of f(x) with
respect to x is positive.)

2. If f’(x)(= then the value of the function tends to decrease (or f(x) is

decreasing function at a given point and/or interval) ; (or the rate of change of
change of f(x) with respect to x is negative)

60
b) Second order derivative

1. If f”(x) (= , then the slope of the curve tends to increase (positive).

2. If f”(x) (= , then the slope of the curve tends to decrease (negative).

c) The possible combination of 1st and 2nd order derivatives and their interpretations.
1. If f’(x)>0 and f” (x) >0
The value of the function is increasing at increasing rate/the slope of the curve is
positive and increasing as the value of x increases/
2. If f’(x)>0 but f”(x) <0
The value of the function is increasing at decreasing rate/ the slope of the curve is
negative but decreasing as the value of x increases/
3. If f’(x) <0 and f” (x)<0
The value of the function is decreasing at decreasing rate/the slope of the curve is
negative and decreasing as x increases/
4. If f’(x) <0 but f”(x)>0
The value of the function is decreasing at an increasing rate/the slope of the curve is
negative but increasing as x increases/
5. What will be the conclusion, if f”(x) =0?
a. Concave and convex curve.
We can distinguish the concavity and convexity of the given function/curve/ simply by
looking at the sign of the second order derivative.
Hence:-
a. A function is said to be concave up ward (convex down ward) curve, if it increases at
an increasing rate or f” (x)>0.
b. A function is said to be concave down ward/convex up ward) curve, if it increases at a
decreasing rate or f” (x) <0
c. A function can be neither concave nor convex curve at a given point and it is called
point of inflection it is true when f”(x) = 0
Example. We can illustrate in diagram form about concavity, convexity & point of
inflection.

61
y

F -
B-
,
D

G
E1

A1
-C

0 x

 We can see different up ward sloping functions/ curves.

 Curve EGF is concave up ward (convex downward), that shows, the functions is
increasing at and increasing rate because f”(x) > 0.
 Curve CGD is concave downward (convex up ward), that shows, the function is
increasing at a decreasing rate because f” (x) <0.
 Curve CGF and EGD are neither concave nor convex at point G.
 At point G, f”(x) = 0, G is point of inflection or inflection tangent for CGF and EGD.
Example:
1. Identify the following function whether they are increasing or decreasing at a given
points.

a. f(x) = ,at x=2 and x=4

Þ f’(x)= 2x-6
Þ When x=2 f’ (2) =2(2)-6=-2<0, the function is decreasing.
Þ When = 4 f’ (4) -6 = 2>0, the function is increasing.
b. f(x) = x2+4x+5, at x =-3 and x = -1, and x =-2
Þ f’(x) = 2x+4
Þ When x = -3 f’(x) =2(-3) +4 = -2<0, the function is decreasing.
Þ When x =-1 f’(x) =2(-1) +4 = 2>0, the function is increasing.

62
 Þ When x = -2 f’(x) = 2(-2) + 4=0, the slope of the curve is horizontal and
stationary.
2. Identify the following function whether they are concave, convex curve or has point of
inflection.
a) f(x)=2x2-4x
f’(x) = 4x-4
f” (x) = 4>0 , then the curve is convex downward.
b). f (x) = x3-3 x2+4

f”(x) = 3x2-6x
f”’(x) = 6x – 6, point of inflection is where f”(x) = 0, hence it will be at x=1.

3.10 EXAMPLES ON ECONOMIC APPLICATIONS

 The aim of this section is to illustrate the application of differential calculus in solving
economic problems.

 Finding the marginal revenue and marginal cost from their total cost function are
discussed.

 In addition how to find the point elasticity demand or supply for a good is presented.

 Let us see the importance of derivatives in solving such kind of problems by taking
different examples.

I. Demand, Supply and Market equilibrium

a) Point Elasticity of demand

 The relative change in demand in response to a relative change in price.

 If the changes are small and the demand function Q=f(p) is continuous of price P, we
express elasticity of demand as:

ed = =-

63
 In fact, the ratio gives us the average elasticity of demand over the price range of

p. According to the law of demand, if price increases demand falls (the change is
negative). Thus,

ed = (-)  /ed/ = .

 Observe that is the derivative of the given demand function with respect to price.

 For each price, the expression will give the elasticity of demand.
Example 1: Suppose the demand function for a commodity is given by Q = 75 – p 2 where Q
is the quantity of the commodity and p is the unit price. Then find the point price elasticity of
demand.

 The meaning of this number (i.e. = 1) is as price of a good increase by one percent, then
quantity demanded decreases by the same percent.
Example 2:
 1) Find ed if the demand function is Q=7-2p
i) P=1 ii) p=2

ed = .  -2 × =

i) For p=1, ed = = - , hence demand is inelastic at p = 1.

ii) For p=2, ed = = , demand is elastic at p = 2.

b) Point elasticity of supply

 It is the relative change in supply in response to a relative change in price.

64
 If Q stands for supply and the supply function is written Q = g (p), then elasticity of

supply retain the same as that of the ed. i.e, es= , where Q is supply function.

 In general using the law of demand and supply:

 The slop of the demand curve is negative and hence ed is negative.
 But conventionally we use the absolute value.
 Similarly, the slope of supply curve is generally positive and hence es is positive.
Example: 1) Find es, for supply function Q = 5+2p2
i) For p =2 ii) For p = 3

es = . 

i) For p = 2, es = = , supply is elastic

ii) For p =3, es = = , supply is elastic.

Note: /ed/ = . , is a formula for Point price elasticity of demand.

a) If /ed/ >1, demand is said to be elastic

b) If 0< /ed/< 1, demand is said to be inelastic
c) If /ed/ = 1, demand is unitary elastic
c) The relationship between average revenue, marginal revenue and price elasticity.
Given total revenue function. (TR)
R = P.Q = Price × quantity sold, and Q = f(p)

Then, marginal revenue (MR) = =

;AR MR = P (1- ) = AR (1- ) average revenue

A R = =
Or AR= MR ( )

65
 = 1-

Then, it follows that when:

ed = 1, TR remains constant with a full in price, MR = 0
:. ed =
(or TR constant, P , MR = 0)
ed > 1, MR > 0, TR rises, then Q  or P
ed < 1, MR< 0, TR falls when Q  or P

II. Price and income elasticities

 Give the angle curve equation Q = f(y), where y-is income.

 Then the point income elasticity can express as e y = From the above expression,

we note that:
a) If ey > 1, then it is for luxury good.
b) If ey < 0, then it is for inferior good.
c) If 0 < ey < 1, then it is for necessity good.
Example: 1) Find the income elasticity of demand, if the angle curve equation is 40Q = 20+5y
and if y= 8000, recall the nature of the good.

 The given function can be re-expressed as Q = .

ey =

= , but y= 8000 = =

 0 < ey < 1, hence it is necessity good.

III. Cost Functions

 For producing a given level of output, there is a corresponding outlay of money
expenditure, called costs of production.
 Let us now generalize the total cost function:
 TC = f (Q) + K, (K is fixed cost); (f (Q) is variable cost)

66
 Then average total cost = ATC = ,

 Average variable cost = AVC =

 Average fixed cost = AFC = .

 Therefore, ATC = AVC + AFC =

 Marginal Cost = MC =

Relationship between AC and MC

MC = = = Product rule.

MC= Q. +AC

The rate of change of AC with respect to Q (the slope of AC) is:

Characteristics of the Slope of AC Curve:

i. When AC curve slopes downward [ i.e., if the slope of AC is negative.],then

= < 0.

 MC –AC< 0.
 MC < AC; (MC lies below the AC curve.)
ii. When Ac curve is rising [i.e., if the slope of AC is positive.], then

> 0.

 MC –AC > 0
 MC > AC (MC curve is above AC curve.)
iii. At the minimum point of AC curve, the slope is zero.

67
 =0

 MC = AC (MC crosses the AC curve at the minimum point of AC curve.)

The MC and AC relationship can also be shown as follows which almost the same
method as above is
Given a total cost function, C = c(Q), the average cost (AC) function can be found by dividing
the total cost by output, Q.

Find the rate of change of average cost (AC) with respect to output, Q.

Solution: The rate of change of AC with respect to Q is

Now use the Quotient rule

(Since

From the above result we can derive important relationship between average cost and
marginal cost.

The economic meaning of this is that the slope of the average cost curve will be positive,
zero, or negative if and only if the marginal cost curve lies a bove, intersects, or lies below
the average cost (Ac) curve.
Since 𝑄>0, it follows that the slope of the AC curve at each value of 𝑄 has the same sign
as MC−AC, which is what we wanted to prove.

68
 TC = C(Q)
TC = C(Q) C
B
A

Q1 Q0 Q11 Q

AC, The MC curve always intersects the ATC

curve at its minimum
MC MC

C ATC

A B

Q1 Q0 Q11 Q

Example: - 1) If TC (Q) = 0.5Q3- 0.7Q2 -30Q + 300, find

i) AC and its slope
ii) MC and its slope
iii) The values for which MC = AVC

i) AC =

= 0.5Q2- 0.7Q -30+

= Q – 0.7 -

ii) MC = = 1.5Q2 – 1.4Q – 30

69
 = 3Q – 1.4

iii) MC = AVC
1.5Q2 – 1.4Q – 30 = 0.5Q2 – 0.7Q – 30
Q2 - 0.7Q = 0
Q (Q - 0.7) = 0
Q = 0 or Q = 0.7, hence if the production process is running, then Q = 0.7 units.

IV. The profit equation

 It is known that profit (π) = Total revenue – Total cost.

Π (Q) = TR (Q) – TC (Q)

 If the TR of a firm equals its TC; then we can say that the firm breaks even.
I.e. TR = TC  TR – TC = 0.
 Marginal profit: It’s the incremental point obtained when the unit of output is produced
and sold.

 Hence Mπ = =

 Example: 1) A firm assumed a cost function C (Q) = , where Q is the

monthly output. And its revenue function is given by R (Q) = 1500Q -1.5Q 2 .If the firm
decides to produce with a marginal cost of 330, find the level of output per month of the
cost to the firm?

 Solution: - Given TC = ; TR = 1500Q – 1.5 Q2; MC = 330

Required to find:
Q =?
TC =?

MC = ,but MC = 330

 thus therefore the level of output is 20units per month.

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And TC = , but Q=20  Total cost =

ii. Find the level of output that maximizes the firm’s profit.
Mπ = 0
But π = R(Q) – TC(Q)

= (1500Q -1.5Q2) – ( = 1290Q – 1.5 Q2 -

 Mπ = = 1290 -3Q - Q2

 - 3Q + 1290 = 0

3Q2 + 30Q – 12900 = 0

Q2 + 10Q – 4300 = 0, using general quadratic formula
Q = -70.76 or Q = 60.76473
 Q ≈ 60.76473 units
iii. Find the MR and MC at this level of output (60.76) and comment on the result.
At Q = 60.76

MC =

MC (60.76) =

= 1107.7057 + 210 = 1317.7057≈ 1317.71

MR (60.76) = 1500 – 3(60.76473)
= 1500 – 182.29419
= 1317.7059
≈ 1317.71
, because R (Q) = 1500Q – 1.5 Q2 and MR = 1500 – 3Q.
 Therefore, MR = MC at the level of output that makes Mπ = 0
Check Your Progress Exercise – 10
1. The total cost function of producing a certain commodity is given as follows:
C(Q 2Q2 + 5Q
C(Q) = 2Q 5Q + 15

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 a) Find the average cost function
b) Find the marginal cost (MC) function

2. The total revenue from the sale of x units of a commodity is

R(x 0.05x2
R(x) = 50 x – 0.05x
a) Find the marginal revenue function
b) What is the value of the marginal revenue at x = 400

3. If the demand function is given as follows

a) Find

b) Compute when the price is 90.

c) Compute the price elasticity of demand at P = 90

3.11 SUMMARY

 This section tries to summarize the important concepts, terms, symbols and formulas of
the unit.

- Derivative:

- Geometric Interpretation of derivative: It is the slope of the tangent line.

- Limits, continuity and differentiability: continuity is a necessary but not a sufficient
condition for differentiability.
- Rules of differentiation

- power rule:

- Constant rule:

- The sum difference rule:

- The product rule:

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 - The Quotient rule:

- The chain Rule: if y = f(u) and u = f(x)

- The inverse function Rule: if it has an inverse

- Logarithmic function:

If

Exponential function and the rules of differentiation

If , then

If then

If then

If , then

Marginal revenue:

Marginal cost (MC) = where Q is output

Point elasticity, where P is price and Q is quantity of a good

3.12 ANSWERS TO CHECK YOUR PROGRESS EXERCISES

Check Your Progress Exercise – 1

a) f1(x) = 3 b) f1(x) = 2x
2x

c) f1(x) = d) f1(x)=8x
)=8x-5

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Check Your Progress Exercise - 2
First check whether the conditions for continuity are satisfied and then show that the limit of
the difference quotient does not exist.
Check Your Progress Exercise – 3

b)

c) d) f1(x) = 1

Check Your Progress Exercise - 4

1. 2.

Check Your Progress Exercise - 5

1.

2) a) Since for all x, it is a monotonically increasing function. Its inverse function

is x = 2 (y
(y – 3)
b) This function is not monotonic. Because it is not a one-to- one mapping. It does not
have an inverse and the inverse function rule cannot be applied.

Check Your Progress Exercise - 6

74
a) ∞ b) 1 c) 0 d) -1/6
Check Your Progress Exercise - 7
1 a) b)
2.a). b).

Check Your Progress Exercise - 8

Check Your Progress Exercise – 9

Check Your Progress Exercise – 10

1. a) AC = 2Q + 15/Q + 5 b) MC = 4Q + 5

2. a) MR = 50 – 0.1x b) MR = 10

3. c) = -0.9

3.13 MODEL EXAMINATION QUESTIONS

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1. Find the first derivatives of the following functions

a) b)

c) d)

e) f)

g) h)

i) j)

k) l)

m) n)

o)
2) Compute the derivative dy/
dy/dx for the given value of x.

a)

b)
c)

3) Let f (x)=/x
)=/x/. Show that continuity does not imply differentiability at x = 0.
4) Are the following functions monotonic?

a) b)

c) Find dx/
dx/dy for each monotonic functions (apply the inverse function rule)
5) Suppose y is a differentiable function of x. Express the derivative of the given function
with respect to x interms of x, y, and dy/
dy/dx

a) y3 b) 2x3y4
6) Find the derivatives of the following exponential and logarithmic functions.

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 a) b)

c) d)

e) f)

g) h)

i) j)

k) l) ( for x > 0)

7) Given the average cost function AC= q2-4q+214, find the marginal cost (MC) function?

8) The gross national product (GNP) of a certain country was N(t )=t2+5t
N(t)=t +5t+106 billion birr
t years after 1980

a) At what rate was the GNP changing with respect to time in 1988?
b) At what percentage rate was the GNP changing with respect to time in 1988?

9) Suppose a revenue function R is given by R(x 40x-x2 , for 0<x

R(x)= 40x 0<x<40
Find the marginal revenue at x =10.

10) At a certain factory, the daily output is Q=3,000K 1/2 L1/3 units, where K denotes the
firms capital investment measured in units of birr 1,000 and L denotes the size of the labor
force measured in worker-hours. Suppose that the current capital investment is birr
400,000 and that 1,331 worker hours of labor are used each day. Use marginal analysis to
estimate the effect that an additional capital investment of birr 1,000 will have on the daily
output if the size of the labor force is not changed.

11) Given the total cost function as and the price is given by

, find the marginal cost and the marginal revenue.

12) Suppose the demand q and price p for a certain commodity are related by the equation
p = 60-2q
60-2q ( for )

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 a) Express the elasticity of demand as a function of q.
b) Calculate the elasticity of demand when q =10. Interpret your answer

3.14 REFERENCES

1. Barnett, Naymod A., and Ziegier, Vichael R.; Applied calculus for Business, Economics,
sciences, 6th ed..
Life sciences, and social sciences,
2. Economics, 3rd ed.
Chiang, Alpha C. : Fundamental methods of Mathematical Economics,
3. Applications, 5th ed.
Hoffmann, Laurence D., and Bradcey, Gerald L.: Brief calculus with Applications,
4. Hunt, Richard A.: Calculus, 2nd ed.
5. Stancl, Donald L., and Stancl, Mildred L.: Calculus for management and the life and
sciences, 2nded.
social sciences,

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