MATH FUNDAMENTALS
1. Derivative of a Function
Basically, the derivative of a function is another function which is derived from the original or primitive
function. The derivative is a limit of the difference quotient. To explain this, first we must understand what a
difference quotient is. The difference quotient measures the rate of change in the dependent variable 𝑦 per
unit change in the dependent variable 𝑥. The primitive function is generally denoted by 𝑦 = 𝑓(𝑥), and the
Δ𝑦
rate of change in 𝑦 per unit change in x is denoted by . When 𝑥 changes from an initial value 𝑥0 to 𝑥0 + ∆𝑥,
Δ𝑥
then the value of the function 𝑦 = 𝑓(𝑥) changes from 𝑓(𝑥0 ) to 𝑓(𝑥0 + ∆𝑥). That is,
∆𝑦 = 𝑓(𝑥0 + ∆𝑥) − 𝑓(𝑥0 )
Thus, the difference quotient is presented as follows:
Δ𝑦 𝑓(𝑥0 + ∆𝑥) − 𝑓(𝑥0 )
=
Δ𝑥
∆𝑥
For example, let
𝑦 = 𝑓(𝑥) = 4𝑥 2 − 5
Then,
𝑓(𝑥0 ) = 4𝑥02 − 5
and
𝑓(𝑥0 + ∆𝑥) = 4(𝑥0 + ∆𝑥)2 − 5
Hence, the difference quotient is
Δ𝑦 4(𝑥0 + ∆𝑥)2 − 5 − (4𝑥02 − 5)
=
Δ𝑥
∆𝑥
Δ𝑦 8𝑥0 ∆𝑥 + 4(∆𝑥)2
=
Δ𝑥
∆𝑥
Δ𝑦
= 8𝑥0 + 4∆𝑥
Δ𝑥
Let 𝑥0 = 6 and ∆𝑥 = 10. Then change in 𝑦 per unit change in 𝑥 (the average rate of change in 𝑦) is
Δ𝑦
= 8(6) + 4(10) = 88
Δ𝑥
When 𝑥 increases from 6 to 6 + 1 = 16, 𝑦 changes from 139 to 1019. Thus, the average rate of change in 𝑦 is
Δ𝑦 1019 − 139 880
=
=
= 88
Δ𝑥
10
10
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y
16, 1019
∆y = 880
∆x = 10
6, 139
x
Now, note what happens to the difference quotient as ∆𝑥 becomes smaller and smaller.
∆𝑥
10
5
2
0.5
0.1
0.01
0.001
0.0001
∆𝑦⁄∆𝑥 = 8𝑥0 + 4∆𝑥
𝑥0 = 6
88
68
56
50
48.4
48.04
48.004
48.0004
As ∆𝑥 gets smaller, the second term (4∆𝑥) on the right hand side of the difference quotient ∆𝑦⁄∆𝑥 = 8𝑥0 +
4∆𝑥 begins to disappear. In other words, as ∆𝑥 approaches zero, 8𝑥0 + 4∆𝑥 will approach the value 8𝑥0 =
8(6) = 48. Thus, for an infinitesimally small ∆𝑥 we may simply take the term 8𝑥0 on the right hand side of
the difference quotient as an approximation of ∆𝑦⁄∆𝑥. In short, as ∆𝑥 approaches zero, ∆𝑦⁄∆𝑥 approaches
8𝑥0 . In symbols, this fact is expressed by
Δ𝑦
𝑓(𝑥0 + ∆𝑥) − 𝑓(𝑥0 )
= lim
∆𝑥→0 Δ𝑥
∆𝑥→0
Δ𝑥
lim
lim
Δ𝑦
∆𝑥→0 Δ𝑥
= lim 8𝑥0 + 4∆𝑥
∆𝑥→0
Δ𝑦
= 8𝑥0
∆𝑥→0 Δ𝑥
lim
This reads as: “The limit of ∆𝑦⁄∆𝑥 as ∆𝑥 approaches zero is 8𝑥0 .” This limit is the derivative of the function
𝑦 = 𝑓(𝑥).
In explaining the derivative of the function 𝑦 = 𝑓(𝑥) we have used the subscripted symbol 𝑥0 to point out the
fact that any change in 𝑥 must start from some specified value of 𝑥. Once the concept is explained, we
dispense with the subscript 0. The derivative, like the primitive function, is itself a function of 𝑥. To indicate
that it is a derivative function, it is usually expressed as 𝑦′ or 𝑓′(𝑥).
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Also, since the derivative is simply a limit of the difference quotient ∆𝑦⁄∆𝑥, which measures a rate of change
in 𝑦, the derivative is also a measure of rate of change. But since the ∆𝑥 is infinitesimally small (∆𝑥 → 0), the
rate of change measured by the derivative is an instantaneous rate of change. Therefore, to distinguish
between the difference quotient and derivative as a measure of instantaneous rate of change we change the
Δ𝑦
𝑑𝑦
symbols from to . Gathering all symbols representing the derivative of the function 𝑦 = 𝑓(𝑥), we have
Δ𝑥
lim
Δ𝑦
∆𝑥→0 Δ𝑥
≡
𝑑𝑥
𝑑𝑦
≡ 𝑦′ ≡ 𝑓′(𝑥)
𝑑𝑥
Now referring back to the function used in defining the derivative,
𝑦 = 𝑓(𝑥) = 4𝑥 2 − 5
𝑑𝑦
= 𝑓 ′ (𝑥) = 8𝑥
𝑑𝑥
Note that 8𝑥 is simply obtained by multiplying the coefficient of 𝑥, 4, by the exponent of 𝑥, 2, and then, raised
𝑥 to the exponent of 2 − 1 = 1.
𝑓 ′ (𝑥) = 2(4)𝑥 2−1 = 8𝑥
1.1. The Derivative and the Slope of the Curve of a Function
Consider the function 𝑦 = 𝑓(𝑥) = 4𝑥 2 − 5. In the diagram below, the points B and C on the curve represent
the values of 𝑦 when 𝑥 is changed from the initial position A (𝑥 = 6) to 16 (∆𝑥 = 10) and then to 13 (∆𝑥 = 7).
Note that as ∆𝑥 gets smaller, the line connecting the point A to points B and C on the curve gets closer to the
curve. When ∆𝑥 is infinitesimally small around point A, this line will become tangent to the curve at that
𝑑𝑦
Δ𝑦
point. We have defined the derivative as the limit of when ∆𝑥 → 0. Thus, the derivative of 𝑦 evaluated at
𝑑𝑥
Δ𝑥
point A is the same as the slope of the line tangent to the curve at point A.
y
B
16, 1019
C
13, 671
13, 475
A
6, 139
x
It was shown above that 𝑓 ′ (𝑥) = 8𝑥. When 𝑥 = 6, then 𝑓 ′ (𝑥) = 48. In the diagram the slope of the line
tangent at point A is
Δ𝑦 475 − 139
=
= 48
Δ𝑥
13 − 6
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1.2. Rules of Differentiation for a Function of One Variable
1.2.1.
Constant Function Rule
The derivative of a constant function 𝑦 = 𝑓(𝑥) = 𝑘 is zero. A constant function means that for all values of 𝑥,
𝑦 can take on a single value (a constant). For example, 𝑦 = 5 is a constant function depicting a horizontal line.
Since a horizontal line has zero slope, then the derivative must be zero.
1.2.2. Power Function Rule
The general form of a power function is 𝑦 = 𝑓(𝑥) = 𝑥 𝑛 . The derivative of this function is:
𝑦′ = 𝑓′(𝑥) = 𝑛𝑥 𝑛−1
Examples
𝑦 = 𝑓(𝑥) = 𝑥 4
𝑓 ′ (𝑥) = 4𝑥 3
𝑦 = 𝑓(𝑥) = 𝑥
𝑓 ′ (𝑥) = (1)𝑥 1−1 = (1)𝑥 0 = 1
𝑦=
1
= 𝑥 −1
𝑥
𝑦 = √𝑥 = 𝑥 1⁄2
𝑑𝑦
1
= (−1)𝑥 −1−1 = −𝑥 −2 = − 2
𝑑𝑥
𝑥
𝑑𝑦 1 −1⁄2
1
= 𝑥
=
𝑑𝑥 2
2√ 𝑥
1.2.3. Power Function Rule Generalized
Include a constant coefficient for 𝑥 in the power function above so that 𝑦 = 𝑓(𝑥) = 𝑐𝑥 𝑛
The derivative of y is:
𝑑𝑦
𝑑𝑥
= 𝑐𝑛𝑥 𝑛−1
Examples
𝑦 = 2𝑥
𝑑𝑦
=2
𝑑𝑥
𝑦 = 4𝑥 3
𝑑𝑦
= 12𝑥 3
𝑑𝑥
𝑦=
3
= 3𝑥 −2
𝑥2
𝑑𝑦
= −6𝑥 −3
𝑑𝑥
1.2.4. Sum-Difference Rule
The derivative of a sum (difference) of two functions is the sum (difference) of the derivative of the two
functions:
𝑑
𝑑
𝑑
[𝑓(𝑥) ± 𝑔(𝑥)] =
𝑓(𝑥) ±
𝑔(𝑥)
𝑑𝑥
𝑑𝑥
𝑑𝑥
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Example
𝑦 = 7𝑥 4 + 2𝑥 3 − 3𝑥 + 5
𝑑
(7𝑥 4 + 2𝑥 3 − 3𝑥 + 5) = 28𝑥 3 + 6𝑥 2 − 3
𝑑𝑥
1.2.5. Product Rule
The product rule applies to the derivative of the product of two functions f(x) and g(x). The rule is:
𝑑
[𝑓(𝑥)𝑔(𝑥)] = 𝑓 ′ (𝑥)𝑔(𝑥) + 𝑓(𝑥)𝑔′(𝑥)
𝑑𝑥
Examples
(a) 𝑦 = (2𝑥 + 3)(3𝑥 2 )
𝑑
[(2𝑥 + 3)(3𝑥 2 )] = 2(3𝑥 2 ) + (2𝑥 + 3)(6𝑥) = 18𝑥 2 + 18𝑥
𝑑𝑥
(b) 𝑦 = (2𝑥 + 3)2
𝑦 = (2𝑥 + 3)(2𝑥 + 3)
𝑑
[(2𝑥 + 3)(2𝑥 + 3)] = 2(2𝑥 + 3) + (2𝑥 + 3)(2) = 4(2𝑥 + 3)
𝑑𝑥
1.2.6.
Quotient Rule
The quotient rule applies to the quotient of two function,
𝑓(𝑥)
𝑔(𝑥)
The rule is,
𝑑 𝑓(𝑥)
𝑓 ′ (𝑥)𝑔(𝑥) − 𝑔(𝑥)𝑔′(𝑥)
[
]=
𝑑𝑥 𝑔(𝑥)
𝑔2 (𝑥)
Examples
𝑦=
2𝑥 − 3
𝑥+1
𝑑𝑦 2(𝑥 + 1) − (2𝑥 − 3)
5
=
=
(𝑥 + 1)2
(𝑥 + 1)2
𝑑𝑥
𝑦=
5𝑥
2
𝑥 +1
𝑑𝑦 5(𝑥 2 + 1) − 5𝑥(2𝑥) 5(1 − 𝑥 2 )
=
= 2
(𝑥 2 + 1)2
(𝑥 + 1)2
𝑑𝑥
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1.2.7. Chain Rule
If we have a function 𝑧 = 𝑓(𝑦), where 𝑦 is in turn a function of another variable 𝑥—𝑦 = 𝑔(𝑥)—then the
following rule of derivatives apply:
𝑑𝑧 𝑑𝑧 𝑑𝑦
=
= 𝑓 ′ (𝑦)𝑔′(𝑥)
𝑑𝑥 𝑑𝑦 𝑑𝑥
Examples
(a) Let 𝑧 = 3𝑦 2 and 𝑦 = 2𝑥 + 5
𝑑𝑧
= 6𝑦 = 6(2𝑥 + 5)
𝑑𝑦
𝑑𝑦
=2
𝑑𝑥
𝑑𝑧
= 12(2𝑥 + 5)
𝑑𝑥
(b) Find the derivative of 𝑦 = (2𝑥 + 3)2
Let 𝑤 = 2𝑥 + 3, then
𝑦 = 𝑤2
𝑑𝑦 𝑑𝑦 𝑑𝑤
=
= 2𝑤(2) = 4(2𝑥 + 3)
𝑑𝑥 𝑑𝑤 𝑑𝑥
1.2.8. Inverse Function Rule
If the function 𝑦 = 𝑓(𝑥) is such that a different value of 𝑥 will always yield a different value of 𝑦, then the
function will have an inverse function 𝑥 = 𝑓 −1 (𝑦). That is, 𝑥 is an inverse function of 𝑦. Consider the
following counter example. The function 𝑦 = 4 + 30𝑥 − 3𝑥 2 does not have an inverse function because there
are more than one value of 𝑥 associated with each value of 𝑦, for all values of 0 ≤ 𝑥 ≤ 10. For example 𝑥 = 3
and 7 yield 𝑦 = 67.
y
67
0
1
2
3
4
5
6
7
8
9
10
11
x
Any function that yields a different value of y for a different value of x is called a monotonic function. The
functions 𝑦 = 50 + 3𝑥 2 and 𝑦 = 300 − 3𝑥 2 , as graphed below, are examples of the monotonic function. The
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first is monotonically increasing, and the latter is monotonically decreasing. Both functions have an inverse
function.
y
y
0
1
2
3
4
5
6
7
8
9
10 11
x
0
1
2
3
4
5
6
7
8
9
10 11
x
For inverse functions, the rule of differentiation is
𝑑𝑥
1
=
𝑑𝑦 𝑑𝑦⁄𝑑𝑥
Example
Find the
𝑑𝑥
𝑑𝑦
for
(a) 𝑦 = 50 + 3𝑥 2
𝑑𝑦
= 6𝑥
𝑑𝑥
𝑑𝑥
1
=
𝑑𝑦 6𝑥
(b) 𝑦 = 300 − 3𝑥 2
𝑑𝑦
= −6𝑥
𝑑𝑥
𝑑𝑥
1
=−
𝑑𝑦
6𝑥
1.3. Inflection Point, Relative Maximum and Relative Minimum of a Function
Example
A typical short run production function in microeconomics can be represented by the following function,
where output Q is a function of the variable input L:
𝑄 = 𝑓(𝐿) = 60𝐿 + 48𝐿2 − 4𝐿3
𝑄 = 60𝐿 + 48𝐿2 − 4𝐿3
The graph of the function is as follows.
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1600
Q
1400
1200
1000
800
600
400
200
0
0
1
2
3
4
5
6
7
8
9
10
11
L
Find the marginal product function and the average product function.
Note that marginal product represents the rate of change in output per unit change in input. For
infinitesimally small changes in L, then marginal product is the derivative of the short-run production
function. Using the sum-difference rule of derivatives, we have:
𝑀𝑃𝐿 ≡
𝑑𝑄
𝑑
(60𝐿 + 48𝐿2 − 4𝐿3 )
=
𝑑𝐿 𝑑𝐿
𝑀𝑃𝐿 = 60 + 96𝐿 − 12𝐿2
Average product is simply output per unit of labor.
𝐴𝑃𝐿 ≡
𝑄 60𝐿 + 48𝐿2 − 4𝐿3
=
𝐿
𝐿
𝑀𝑃𝐿 = 60 + 48𝐿 − 4𝐿2
Graph of 𝑀𝑃𝐿 and 𝐴𝑃𝐿 is as follows:
Q
300
252
250
200
AP
150
100
MP
50
0
0
1
2
3
4
5
6
7
8
9
10
11
L
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The relationship between 𝑄 = 𝑓(𝐿) and 𝑀𝑃𝐿 = 𝑓′(𝐿) is very instructive in showing the relationship between
the primitive function and its derivative. Note that 𝑀𝑃𝐿 rises, reaches a maximum [𝑓′(4) = 252], and then
begins to fall. The maximum point of 𝑀𝑃𝐿 is the point of diminishing marginal productivity of labor (the point
of “diminishing return”). It was pointed out that the derivative of a function evaluated at a given point
represents the slope of (the line tangent to) the primitive function at the point. Thus, when the derivative
function is increasing for a given interval of the independent variable values, then the primitive function is
rising at an increasing rate (the tangent at each point is becoming steeper). When the derivative function is
falling, the primitive function is rising at a decreasing rate. Therefore, the maximum point of the 𝑀𝑃𝐿 , the
derivative function, is associated with the inflection point on the primitive function (the point at which the
algebraic sign of the rate of change of primitive function changes).
We can find the value of the independent variable corresponding to the inflection point by finding the value at
which the slope of the derivative function is zero—the point at which 𝑀𝑃𝐿 reaches the maximum. To do this
we take the derivative of derivative function—the second derivative of the primitive function—and set it
equal to zero and solve for the independent variable value.
𝑓 ′′ (𝐿) ≡
𝑑2𝑄
𝑑 𝑑𝑄
≡
( )
𝑑𝐿2 𝑑𝐿 𝑑𝐿
𝑑2𝑄
𝑑
(60 + 96𝐿 − 12𝐿2 ) = 96 − 24𝐿
=
𝑑𝐿2 𝑑𝐿
96 − 24𝐿 = 0
𝐿=4
Now find the maximum of the production (primitive) function. Again, find the L value at which the slope of
the primitive function is zero. To find this, set MPL = f ′(L) = 0.
𝑀𝑃𝐿 = 60 + 96𝐿 − 12𝐿2 = 0
The latter is a quadratic equation. You can use the quadratic formula or, using the Excel Tools, Goal Seek, you
can find L = 8.6.
Now find the L value at which 𝐴𝑃𝐿 is the maximum. This can be done in two ways. Note that, mathematically,
when the marginal value exceeds the average, the latter rises, and when the marginal value is below average,
the average falls. Thus marginal and average values are equal when the average is at its maximum.
Therefore, find the 𝐿 value such that 𝑀𝑃𝐿 = 𝐴𝑃𝐿 :
60 + 96𝐿 − 12𝐿2 = 60 + 48𝐿 − 4𝐿2
8𝐿2 − 48𝐿 = 0
𝐿=6
The alternative approach is to set
𝑑
(𝐴𝑃𝐿 ) = 48 − 8𝐿 = 0
𝑑𝐿
𝐿=6
which gives us the same answer.
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1522
1600
Q
1400
1200
1000
800
752
600
400
200
0
4
Q
8.6
L
300
252
250
204
200
AP
150
100
MP
50
0
4
6
8.6
L
Practice Problems
1. Find the derivative of each of the following functions.
(a) 𝑦 = 𝑥 19
(b) 𝑦 = 4𝑥 6
(c) 𝑦 = −4𝑥 3
(d) 𝑦 = 3𝑥 −1
2.
Find the following:
(a)
𝑑
(−𝑥 −4 )
𝑑𝑥
(b)
𝑑
2𝑥 1⁄3
𝑑𝑥
(c)
𝑑
𝑐𝑥 2
𝑑𝑥
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3.
4.
Find f (1) and f (2) from the following functions:
(a)
𝑦 = 𝑓(𝑥) = 8𝑥
(b)
𝑓(𝑥) = −3𝑥 −2
(c)
𝑓(𝑥) =
1 1⁄2
𝑥
4
(d)
𝑓(𝑥) =
2 3⁄2
𝑥
3
Use the following cost function
𝐶 = 𝑓(𝑄) = 𝑄3 − 𝑄2 + 60𝑄
(a) Determine the marginal cost function and the average cost function.
(b) Find the inflection point by finding the values for Q and C. How is the inflection point related to the
marginal cost?
(c) Find the minimum average cost.
5.
Differentiate the following:
(a)
𝑦 = (9𝑥 2 − 2)(3𝑥 + 1)
(b)
𝑦 = (3𝑥 + 11)(6𝑥 2 − 𝑥)
(c)
𝑦 = 𝑥 2 (4𝑥 + 6)
(d)
𝑦 = (𝑥 2 + 3)𝑥 −1
(e)
𝑦=
𝑥+7
𝑥
(f)
𝑦=
4𝑥
𝑥+5
6.
Given 𝑦 = 𝑢3 + 1, where 𝑢 = 5 − 𝑥 2 , find 𝑑𝑦⁄𝑑𝑥 by the chain rule.
7.
Given 𝑤 = 𝑎𝑦 2 and 𝑦 = 𝑏𝑥 2 + 𝑐𝑥, find 𝑑𝑤 ⁄𝑑𝑥 by the chain rule.
8.
Use the chain rule to find 𝑑𝑦⁄𝑑𝑥 for the following:
(a) 𝑦 = (3𝑥 2 − 7)3
(b) 𝑦 = (8𝑥 3 − 7)3
(c) 𝑦 = (𝑎𝑥 + 𝑏)4
9.
Given 𝑦 = (16𝑥 + 3)−2 , use the chain rule to find 𝑑𝑦⁄𝑑𝑥 . Then rewrite the function 𝑦 = 1⁄(16𝑥 + 3)2
and find 𝑑𝑦⁄𝑑𝑥 by the quotient rule. Are the answers identical?
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2.
Partial Derivatives
Partial derivatives (differentiation) apply when the dependent variable is a function of 2 or more
independent variables:
𝑦 = 𝑓(𝑥1 , 𝑥2 , ⋯ , 𝑥𝑛 )
The variables 𝑥𝑖 are all independent of one another such that each can vary by itself without affecting the
others. If the variable 𝑥1 varies while all the other independent variables remain fixed, then 𝑦 would vary
solely due to the change in 𝑥. Starting with the difference quotient,
Δ𝑦
𝑓(𝑥1 + ∆𝑥1 , 𝑥2 , ⋯ , 𝑥𝑛 ) − 𝑓(𝑥1 , 𝑥2 , ⋯ , 𝑥𝑛 )
=
Δ𝑥1
∆𝑥1
Partial derivative of y with respect to x1 is then defined as
𝜕𝑦
Δ𝑦
≡ lim
𝜕𝑥1 Δ𝑥1 →0 Δ𝑥1
Example
Let 𝑦 = 𝑓(𝑥1 , 𝑥2 ) = 3𝑥12 + 3𝑥1 𝑥2 + 4𝑥22
𝜕𝑦
= 6𝑥1 + 3𝑥2
𝜕𝑥1
𝜕𝑦
= 3𝑥1 + 8𝑥2
𝜕𝑥2
Example
Let 𝑦 = 𝑓(𝑥1 , 𝑥2 ) = (𝑥1 + 4)(3𝑥1 + 2𝑥2 )
𝑓1 ≡
𝜕𝑦
= (𝑥1 + 4)(3𝑥1 + 2𝑥2 ) [using the product rule]
𝜕𝑥1
𝑓2 ≡
𝜕𝑦
= 2(𝑥1 + 4)
𝜕𝑥2
Example
Let 𝑦 = 𝑓((𝑥1 , 𝑥2 ) =
3𝑥1 − 2𝑥2
𝑥12 + 3𝑥23
𝑓1 =
3(𝑥12 + 3𝑥2 ) − 2𝑥1 (3𝑥1 − 2𝑥2 ) −3𝑥12 + 4𝑥1 𝑥2 + 9𝑥2
=
(𝑥12 + 3𝑥2 )2
(𝑥12 + 3𝑥2 )2
𝑓1 =
−2(𝑥12 + 3𝑥2 ) − 3(3𝑥1 − 2𝑥2 ) −𝑥1 (2𝑥1 + 9)
=
(𝑥12 + 3𝑥2 )2
(𝑥12 + 3𝑥2 )2
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3.
Differentials
𝑑𝑦
As explained, the derivative of the function 𝑦 = 𝑓(𝑥) is denoted by = 𝑓′(𝑥). We can denote the change in 𝑦
𝑑𝑥
as the product of rate of change in 𝑦 times the change in 𝑥:
𝑑𝑦
𝑑𝑦 = 𝑓 ′ (𝑥)𝑑𝑥 = ( ) 𝑑𝑥
𝑑𝑥
The symbols 𝑑𝑦 and 𝑑𝑥 are called the differentials of 𝑦 and 𝑥, respectively.
Example
𝑦 = 3𝑥 2 + 7𝑥 − 5
𝑑𝑦
= 6𝑥 + 7
𝑑𝑥
𝑑𝑦 = (6𝑥 + 7)𝑑𝑥
The process of finding the differential 𝑑𝑦 is called differentiation.
3.1. Point Elasticity of a Function
The point elasticity of a function is defined as the proportionate (percentage) change in 𝑦 in response to a
proportionate or percentage change in 𝑥.
𝜀=
𝑑𝑦⁄𝑦
𝑑𝑥 ⁄𝑥
In microeconomics the concept of elasticity is applied particularly within the context of the price elasticity of
demand: the percentage change is quantity demanded relative to a percentage change in price of a product.
The difference between “pure” elasticity and point elasticity is that in the point elasticity the change in 𝑦 is
measured in response to an infinitesimal change in 𝑥. Thus,
𝜀=
𝑑𝑦⁄𝑦
∆𝑦 ⁄𝑦
≡ lim
𝑑𝑥 ⁄𝑥 ∆𝑥→0 ∆𝑥 ⁄𝑥
We can write the point elasticity of 𝑦 as
𝜀=
𝑑𝑦⁄𝑑𝑥 𝑑𝑦 𝑥
=
𝑦 ⁄𝑥
𝑑𝑥 𝑦
Note that in the middle part of the above expression,
𝑑𝑦 ⁄𝑑𝑥
𝑦⁄𝑥
, the numerator measures the change in 𝑦 per unit
change in 𝑥 and the denominator measures 𝑦 per unit of 𝑥. Thus, point elasticity is simply the ratio of the
marginal to the average function. For example, consider the production function 𝑄 = 𝑓(𝐿), where output is a
function of the variable input. We can define the point elasticity of output with respect to the variable input
as
𝜀=
𝑑𝑄 ⁄𝑑𝐿 marginal product
=
𝑄 ⁄𝐿
average product
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Note that when marginal product is greater than the average product, 𝜀 > 1, then the output is elastic and
output per unit of input will rise.1
Practice Problems
10. Find 𝜕𝑧⁄𝜕𝑥 and 𝜕𝑧⁄𝜕𝑦 for each of the following functions:
(a) 𝑧 = 2𝑥 3 − 11𝑥 2 𝑦 + 3𝑦 2
(b) 𝑧 = 6𝑥 − 16𝑥𝑦 2 + 9𝑦 3
(c) 𝑧 = (2𝑥 + 3)(𝑦 − 2)
(d) 𝑧 = (2𝑥 + 3)⁄(𝑦 − 2)
11. Find 𝑓𝑥 and 𝑓𝑦 from the following
(a)
𝑓(𝑥, 𝑦) = 𝑥 2 + 5𝑥𝑦 − 𝑦 3
(b)
𝑓(𝑥, 𝑦) = (𝑥 2 − 7𝑦)(𝑥 − 2)
(c)
𝑓(𝑥, 𝑦) =
2𝑥 − 3𝑦
𝑥+𝑦
(d)
𝑓(𝑥, 𝑦) =
𝑥2 − 1
𝑥𝑦
12. From the answers to the preceding questions, find 𝑓𝑥 (1,2).
13. If the utility function of an individual takes the form
𝑈 = 𝑈(𝑥, 𝑦) = (𝑥 + 2)2 (𝑦 + 3)3
where 𝑈 is total utility, and 𝑥 and 𝑦 are the quantities of two commodities consumed:
(a) Find the marginal utility function of each of the two commodities.
(b) Find the value of the marginal utility of the first commodity when 3 units of each commodity are
consumed.
14. Find the differential dy, given:
(a)
𝑦 = −𝑥(𝑥 2 + 3)
(b)
𝑦 = (𝑥 − 8)(𝑥 + 5)
(c)
𝑦=
𝑥2
𝑥
+1
15. Given the consumption function 𝐶 = 𝑎 + 𝑏𝑌,
(a) Find the marginal and the average functions.
1
When margin exceeds the average, the average will rise.
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(b) Find the income elasticity of consumption 𝜀𝐶𝑌
(c) Show that this consumption function is inelastic at all positive income levels.
16. Given,
𝑄=
𝑘
𝑃𝑛
where 𝑄 is the quantity demanded, 𝑃 is the price of the product, and 𝑘 and 𝑛 are constants, find the point
elasticity of demand. Is the value of elasticity dependent on the price 𝑃?
4. Exponential and Logarithmic Functions
The term exponent means an indicator of the power to which a variable is to be raised. A function whose
independent variable appears as an exponent is called an exponential function.
𝑦 = 𝑓(𝑥) = 𝑏 𝑥
𝑏>1
For example, let 𝑦 = 2𝑥 . Note that there are no restrictions on 𝑥 (−∞ < 𝑥 < ∞). The following is the graph of
this function.
8
7
6
5
4
3
2
1
-4
-3
-2
-1
0
1
2
3
4
Note the function is monotonically increasing.2 Thus the function must have an inverse function, which is a
logarithmic function, to be explained later. Also because the function is monotonic, there is a unique value of
𝑥 for a given value of 𝑦. This implies that any positive number is a unique power of a base 𝑏 > 1. For
example, if 𝑏 = 10, then 𝑦 = 1,000 is 10 raised to the third power. Thus, 1,000 is unique power of base 10.
Or, in the graph above, 8 is a unique power of base 2, the power being 𝑥 = 3. If the base were, say 𝑏 = 10,
then 𝑦 = 8 is 10 raised to the power 𝑥 = 0.9031—100.9031 . 3 The following table shows that 𝑦 = 8 can be
expressed as a unique power of 𝑏 = 2, 3, ⋯ , 10.
The restriction b > 1 provides that the function is monotonically increasing. When 0 < b < 1, the function would be
monotonically decreasing.
3 This, of course, is: log10 8 = 0.9031
2
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𝑏
2
3
4
5
6
7
8
9
10
𝑥
3.000
1.893
1.500
1.292
1.161
1.069
1.000
0.946
0.903
𝑦 = 𝑏𝑥
8
8
8
8
8
8
8
8
8
Note that one can select any number 𝑏 > 1 as the base and find the value 8 as the power of that base. Now,
consider the following outlandish base and the corresponding power that results in 𝑦 = 8:
𝑦 = 𝑏 𝑥 = 𝟐. 𝟕𝟏𝟖𝟐𝟖2.07944 = 8
The base shown here is not just any base. In fact, this happens to be the most common base used in
mathematics and all its applications. Since it is so common, we usually do not pay attention to the actual
numeric value, rather we know it as 𝑒, the base for natural logarithm,
log 𝑒 8 ≡ ln 8 = 2.07944
The term ln 8 = 2.07944 implies that raising the base 𝑒 to the power 2.07944 will yield 8—e2.07944 = 8.
𝑒 2.07944 = 8. We will come back to the discussion of ln in more detail later.
Where does 𝑒 come from? To explain the origin of 𝑒, consider the function
1 𝑥
𝑓(𝑥) = (1 + )
𝑥
Then 𝑒 is defined as,
1 𝑥
𝑒 ≡ lim 𝑓(𝑥) = lim (1 + )
𝑥→∞
𝑥→∞
𝑥
To provide a more intuitive explanation of 𝑒, consider the following “investment” example. Suppose you have
an investment capital of 𝐾 = $1 invested at an interest rate of 𝑟 = 100% = 1 per annum. At the end of the
year your $1 will grow into:
𝐾 + 𝐾𝑟 = 𝐾(1 + 𝑟) = $1(1 + 1) = $2
If, however, the interest was compounded semiannually, then at end of the first six months you would have
𝑟
𝑟
1
𝐾 + 𝑘 ( ) = 𝐾 (1 + ) = (1 + )
2
2
2
and at the end of the second six months your investment would grow into
𝑟
𝑟 𝑟
𝑟
𝑟
𝑟 2
1 2
𝐾 (1 + ) + 𝐾 (1 + ) = 𝐾 (1 + ) (1 + ) = 𝐾 (1 + ) = (1 + ) = 2.25
2
2 2
2
2
2
2
Let’s do one more compounding example. Assume now interest is being compounded quarterly—
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At the end of 𝑄1 :
𝑟
𝐾 (1 + )
4
At the end of 𝑄2 :
𝑟
𝑟 𝑟
𝑟 2
𝐾 (1 + ) + 𝐾 (1 + ) = 𝐾 (1 + )
4
4 4
4
At the end of 𝑄3 :
𝑟 2
𝑟 2𝑟
𝑟 3
𝐾 (1 + ) + 𝐾 (1 + ) = 𝐾 (1 + )
4
4 4
4
At the end of 𝑄4
𝑟 3
𝑟 3𝑟
𝑟 4
1 4
𝐾 (1 + ) + 𝐾 (1 + ) = 𝐾 (1 + ) = (1 + ) = 2.44141
4
4 4
4
4
Following the same methodology, you can increase the number of compounding periods to as high as you like
(nanoseconds?). The following table shows the results obtained by methodically increasing the compounding
periods to every second of the year. The last value, 2.71828, is practically the value obtained through
continuous compounding.
𝑥
Annual
Semiannual
Quarterly
Monthly
Daily
Hourly
Every minute
Every second
1
2
4
12
365
8,760
525,600
31,536,000
1 𝑥
𝑓(𝑥) = (1 + )
𝑥
2
2.25
2.44141
2.61304
2.71457
2.71813
2.71828
2.71828
Now that we understand the origin of 𝑒, let’s change the value of K from $1 to $100 and r from 100% to a
more down-to-earth number, say, 𝑟 = 10% or 0.1. The interest being compounded every second, at the end
of one year 𝐾 will grow to
𝑦 = 100 (1 +
0.1 𝑥
) = 110.52
𝑥
𝑥 = 365(24)(60)(60) = 31,536,000
The same figure is obtained by
𝑦 = 𝐾𝑒 𝑟 = 100𝑒 0.1 = 110.52
4
If interest were compounded every second for two years (𝑥 = 31,536,000, 𝑛 = 2), then at the end of the twoyear investment period, 𝐾 would grow to:
𝑦 = 𝐾 (1 +
𝑟𝑛 𝑥
0.1(2) 𝑥
) = 100 (1 +
) = 122.14
𝑥
𝑥
Again, the same figure is obtained by
𝑦 = 𝐾𝑒 𝑟𝑛 = 100𝑒 0.1(2) = 122.14
4
In Excel, use =exp(0.1) for 𝑒 0.1 .
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Practice Problems
17. Write an exponential expression, 𝑦 = 𝐾𝑒 𝑟𝑛 , for each of the following and compute the future value:
(a) $10, compounded continuously at the APR of 5% for 3 years.
(b) $700, compounded continuously at the APR of 4% for 2 years.
4.1.
Logarithms
To understand what a logarithm is, let’s use a very simple example. When you raise a number, say, 10, to the
third power, the result is 1,000—103 = 1,000. Logarithm here is defined with respect to or applied to the
figure 1,000. Log of 1,000 is simply the power to which we raise the base 10 to get the value 1,000, which is 3.
This is expressed as:
log10 1000 = 3
Therefore, generally, denoting the base by 𝑏 and the power to which 𝑏 is raised by 𝑥, then the logarithm of 𝑦
is defined as
log 𝑏 𝑦 = 𝑥
In log applications two numbers are commonly used as the base—the number 10 and 𝑒. When 𝑏 = 10, then
the logarithm is known as common log. When 𝑏 = 𝑒, it is known as natural log. By convention, log 𝑦 is
understood as the common log of 𝑦 (the base 10 is not shown). The notation for natural log is 𝐥𝐧 𝒚.
log10 𝑦 ≡ log 𝑦
log 𝑒 𝑦 ≡ ln 𝑦
Note that according to the exponential function 𝑦 = 𝑒 𝑥 , 𝑥 is the power to which the base 𝑒 is raised to obtain
𝑦. This means that, by definition, the natural log of 𝑦 is 𝑥. That is,
𝑦 = 𝑒𝑥 ⇔
ln 𝑦 = ln 𝑒 𝑥 = 𝑥
Also note ln 𝑒 = log 𝑒 𝑒 = 1
4.2. Rules of Logarithms
4.2.1.
RULE I (Log of a Product)
𝐥𝐧 𝒖𝒗 = 𝐥𝐧 𝒖 + 𝐥𝐧 𝒗
Let 𝑢 = 𝑒 𝑚 and 𝑣 = 𝑒 𝑛
ln 𝑢𝑣 = ln(𝑒 𝑚 𝑒 𝑛 )
ln 𝑢𝑣 = ln 𝑒 𝑚+𝑛
ln 𝑢𝑣 = 𝑚 + 𝑛
ln 𝑢𝑣 = ln 𝑢 + ln 𝑣
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4.2.2.
𝐥𝐧
RULE II (Log of a Quotient)
𝒖
= 𝐥𝐧 𝒖 − 𝐥𝐧 𝒗
𝒗
Let 𝑢 = 𝑒 𝑚 and 𝑣 = 𝑒 𝑛
ln(𝑢⁄𝑣 ) = ln(𝑒 𝑚 𝑒 −𝑛 )
ln(𝑢⁄𝑣 ) = ln 𝑒 𝑚−𝑛
ln(𝑢⁄𝑣 ) = 𝑚 − 𝑛
ln(𝑢⁄𝑣 ) = ln 𝑢 − ln 𝑣
4.2.3.
RULE III (Log of a Power)
𝐥𝐧 𝒖𝒂 = 𝒂 𝐥𝐧 𝒖
Let 𝑢 = 𝑒 𝑚
ln 𝑢𝑎 = ln(𝑒 𝑚 )𝑎
ln 𝑢𝑎 = ln 𝑒 𝑎𝑚
ln 𝑢𝑎 = 𝑎𝑚
ln 𝑢𝑎 = 𝑎 ln 𝑢
Example
If a nation’s GDP grows at 3% a year, how many years would it take for the GDP to double?
1(1 + 0.03)𝑥 = 1.03𝑥 = 2
ln 1.03𝑥 = ln 2
𝑥 ln 1.03 = ln 2
𝑥=
ln 2
= 23.5
ln 1.03
4.2.4.
RULE IV (Conversion of Log Base)
𝐥𝐨𝐠 𝒃 𝒖 = (𝐥𝐨𝐠 𝒃 𝒆)(𝐥𝐨𝐠 𝒆 𝒖) = (𝐥𝐨𝐠 𝒃 𝒆) 𝐥𝐧 𝒖
Let 𝑢 = 𝑒 𝑚
log 𝑏 𝑢 = log 𝑏 𝑒 𝑚 = 𝑚 log 𝑏 𝑒 = (ln 𝑢)(log 𝑏 𝑒)
log 𝑏 𝑢 = log 𝑏 𝑒 𝑚
log 𝑏 𝑢 = 𝑚 log 𝑏 𝑒
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log 𝑏 𝑢 = (ln 𝑢)(log 𝑏 𝑒)
4.2.5.
𝐥𝐨𝐠 𝒃 𝒆 =
RULE V (Inversion of Log Base)
𝟏
𝟏
=
𝐥𝐨𝐠 𝒆 𝒃 𝐥𝐧 𝒃
From Rule IV we have
log 𝑏 𝑢 = (log 𝑏 𝑒)(log 𝑒 𝑢)
Let u = b. Then
log 𝑏 𝑏 = (log 𝑏 𝑒)(ln 𝑏)
Since log 𝑏 𝑏 = 1, then
log 𝑏 𝑒 =
1
ln 𝑏
Practice Problems
18. What are values of the following logarithms?
(a) log10 10,000
(b) log10 1,000,000
(e) log 2 8
(f) log 5 3125
(c) log 3 81
(d) log10 0.0001
19. What are the values of the following logarithms?
(a) ln 𝑒 −4
(b) ln 𝑒 5
(c) ln
1
𝑒
(d) ln
1
𝑒3
20. Evaluate the following by application of the rules of logarithms:
1
(a) log 10014
(b) log
(e) ln(𝐴𝐵𝑒 −4 )
(f) (log 4 𝑒)(log 4 64)
100
(c) ln
3
𝐵
(g) 1⁄log16 2
(d) ln(𝐴𝑒 2 )
(h) log 32 2
21. Which of the following are valid?
𝑢
𝑒2
(a)
ln(𝑢) − 2 = ln
(b)
3 + ln 𝑣 = ln
(c)
ln 𝑢 + ln 𝑣 − ln 𝑤 =
(d)
ln 3 + ln 8 = ln 8
𝑒3
𝑣
Derivatives and Other Math Concepts
ln 𝑢𝑣
𝑤
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5.
Log Functions and Exponential Functions
When 𝑦 is expressed as a function of the logarithm of another variable, then it is a logarithmic function. When
the independent variable appears as an exponent, then 𝑦 is an exponential function of 𝑥.
As noted above, the exponential function 𝑦 = 𝑒 𝑥 is monotonically increasing. Therefore, it has an inverse
function. The inverse function of 𝑦 = 𝑒 𝑥 is 𝑥 = ln 𝑦.
5.1. Derivative of the Logarithmic Function
Define the logarithmic function as
𝑦 = 𝑓(𝑥) = ln 𝑥
then
𝑓 ′ (𝑥) =
𝑑𝑦
𝑑
1
=
ln 𝑥 =
𝑑𝑥 𝑑𝑥
𝑥
(See APPENDIX A for the proof)
5.2. Derivative of the Exponential Function
The derivative of the exponential function 𝑦 = 𝑒 𝑥 is
𝑓 ′ (𝑥) =
𝑑𝑦
𝑑 𝑥
=
𝑒 = 𝑒𝑥
𝑑𝑥 𝑑𝑥
Since 𝑦 = 𝑒 𝑥 is a monotonic function, then it has an inverse which is 𝑥 = ln 𝑦. The derivative of this inverse
function is:
𝑑𝑥
𝑑
1
=
ln 𝑦 =
𝑑𝑦 𝑑𝑦
𝑦
Now using the inverse function rule for the exponential function, we have
𝑑𝑦
1
=
𝑑𝑥 𝑑𝑥 ⁄𝑑𝑦
𝑑𝑦
1
=
= 𝑦 = 𝑒𝑥
𝑑𝑥 1⁄𝑦
5.3. Chain Rule and Derivatives of Logarithmic and Exponential Functions
You apply the chain rule when the logarithmic function is in the form of 𝑦 = ln 𝑓(𝑥), or the exponential
function is 𝑦 = 𝑒 𝑓(𝑥) .
For logarithmic functions,
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𝑑𝑦
𝑑
=
ln 𝑓(𝑥)
𝑑𝑥 𝑑𝑥
𝑑𝑦
1 𝑑
=
𝑓(𝑥)
𝑑𝑥 𝑓(𝑥) 𝑑𝑥
𝑑𝑦 𝑓′(𝑥)
=
𝑑𝑥 𝑓(𝑥)
For exponential functions,
𝑑𝑦
𝑑 𝑓(𝑥)
=
𝑒
𝑑𝑥 𝑑𝑥
𝑑𝑦
𝑑
= 𝑒 𝑓(𝑥)
𝑓(𝑥)
𝑑𝑥
𝑑𝑥
𝑑𝑦
= 𝑓′(𝑥)𝑒 𝑓(𝑥)
𝑑𝑥
Examples
𝑦 = ln(𝑥 2 + 2𝑥)
𝑑𝑦
2𝑥 + 2
= 2
𝑑𝑥 𝑥 + 2𝑥
𝑦 = 𝑒𝑥
2 +2𝑥
𝑑𝑦
2
= (2𝑥 + 2)𝑒 𝑥 +2𝑥
𝑑𝑥
5.4.
The Case of Base 𝒃
(a) 𝑦 = 𝑏 𝑥
𝑑𝑦
= (ln 𝑏)𝑏 𝑥
𝑑𝑥
Note that,
𝑏 = 𝑒 ln 𝑏
Then,
𝑏 𝑥 = 𝑒 𝑥 ln 𝑏
𝑑 𝑥
𝑏 = (ln 𝑏)𝑒 𝑥 ln 𝑏 = (ln 𝑏)𝑏 𝑥
𝑑𝑥
(b) 𝑦 = log 𝑏 𝑥
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𝑑𝑦
𝑑
1
=
log 𝑏 𝑥 =
𝑑𝑥 𝑑𝑥
𝑥 ln 𝑏
Note that,
log 𝑏 𝑥 = (log 𝑏 𝑒)(log 𝑒 𝑥)
log 𝑏 𝑥 =
1
ln 𝑥
ln 𝑏
Therefore,
𝑑
𝑑 1
log 𝑏 𝑥 =
(
ln 𝑥)
𝑑𝑥
𝑑𝑥 ln 𝑏
𝑑
1 1
log 𝑏 𝑥 =
( )
𝑑𝑥
ln 𝑏 𝑥
Examples
Find the derivative of 𝑦 = 121−𝑥
𝑑𝑦
= (ln 12)𝑏1−𝑥 (−1)
𝑑𝑥
Find the derivative of
𝑦 = 𝑥 log 5 [𝑥 2 ⁄(1 + 𝑥)]
Let
𝑢 = log 5 [𝑥 2 ⁄(1 + 𝑥)]
𝑢 = 2 log 5 𝑥 − log 5 (1 + 𝑥)
Thus,
𝑦 = 𝑥𝑢
𝑦 = 𝑥[2 log 5 𝑥 − log 5 (1 + 𝑥)]
𝑑𝑦
𝑑𝑢
=𝑥
+𝑢
𝑑𝑥
𝑑𝑥
𝑑𝑢
2
1
2+𝑥
=
−
=
𝑑𝑥 𝑥 ln 5 (1 + 𝑥) ln 5 𝑥(1 + 𝑥) ln 5
𝑑𝑦
2+𝑥
=
+ log 5 [𝑥 2 ⁄(1 + 𝑥)]
𝑑𝑥 (1 + 𝑥) ln 5
Practice Problems
22. Find the inverse function of 𝑦 = 𝑎𝑏 𝑐𝑥 .
Derivatives and Other Math Concepts
23 of 28
23. Transform the following functions to their natural exponential forms:
(a) 𝑦 = 83𝑥
(b) 𝑦 = 2(7)2𝑥
(c) 𝑦 = 5(5) 𝑥
(d) 𝑦 = 2(15)4𝑥
24. Transform the following functions to their natural logarithmic forms:
(a) 𝑥 = log 7 𝑦
(b) 𝑥 = 3 log15 9𝑦
(c) 𝑥 = log 8 3𝑦
(d) 𝑥 = 2 log10 𝑦
25. Find the derivatives of:
(a) 𝑦 = 𝑒 2𝑥+3
(d) 𝑦 = 3𝑒 2−𝑥
(b) 𝑦 = 𝑒 1−5𝑥
2
(e) 𝑦 = 𝑒 𝑎𝑥
(g) 𝑦 = 𝑥 2 𝑒 2𝑥
(c) 𝑦 = 𝑒 𝑥
2 +𝑏𝑥+𝑐
2 +1
(f) 𝑦 = 𝑥𝑒 𝑥
(f) 𝑦 = 𝑎𝑥𝑒 𝑏𝑥+𝑐
26. Find the derivatives of :
(a) 𝑦 = ln(3𝑥 5 )
(b) 𝑦 = ln(𝑎𝑥 𝑐 )
(c)
𝑦 = ln(𝑥 + 2)
(d) 𝑦 = 5 ln(𝑥 + 1)2
(e)
(f)
𝑦 = ln[𝑥(1 − 𝑥)3 ]
(g) 𝑦 = ln (
3𝑥
1+𝑥
)
𝑦 = ln 𝑥 − ln(1 + 𝑥)
(h) 𝑦 = 3𝑥 2 ln 𝑥 2
APPENDIX A
𝑑
1
ln 𝑥 =
𝑑𝑥
𝑥
Proof
By definition the derivative of y = f(x) has the following value at x = n:
 x
ln  
f ( x )  f ( n)
ln( x)  ln( n)
n
f ′(n) = lim
 lim
 lim  
xn
x

n
x

n
xn
xn
xn
n
Now, let m =
. Dividing both sides by n, we can write the denominator of the limit on the right as
xn
1
m

xn n
Also, we can rewrite
x
of the ln term in the numerator as,
n
x xnn
xn
1

 1
 1
n
n
n
m
Derivatives and Other Math Concepts
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Therefore,
 x
ln  
m
 n   1 ln  x   m ln 1  1   1 ln 1  1 
 




xn
xn n n  m n  m
Now we can show that,
 x
ln  
m
1 
1
1
1
n
 lim ln 1    ln(e) 
f ′(n) = lim
x n x  n
m  n
m
n
n


n
Here note two points: (1) As x → n, m =
→ ∞. (2) Since n can be any number for which a logarithm is
x n
defined, we can generalize this result and write
f ′(x) =
d
1
ln( x) 
dx
x
APPENDIX B
Solutions Practice Problems
dy
dy
1. (a)
(b)
 19x18
 24x 5
dx
dx
dy
dy 1 2 / 3
2. (a)
(b)
 x
 4 x 5
dx 3
dx
3. (a) f ( x)  8 ;
f (1)  f (2)  8
4.
(b)
f ( x)  6 x 3
(c)
f ( x) 
(c)
f (1)  6(1) 3  6
1
f (1)  (1) 1/ 2  1 / 8
8
dy
 12x 2
dx
(d)
dy
 3x 2
dx
dy
 2cx
dx
f (2)  6(2) 3 
6 3

8 4
1
1
f (2)  (2) 1/ 2 
8
8 2
f (2)  (2)1 / 2  2
(d) f ( x)  x1/ 2
f (1)  (1)1/ 2  1
C = f(Q) = Q3 – 12Q2 + 60Q
dC
C
(a) MC =
AC =
 3Q 2  24Q  60
 Q 2  12Q  60
dQ
Q
(b) The inflection point is where MC reaches the minimum. Take the derivative of MC and set it equal to
zero.
(c)
5.
1 1/ 2
x
8
(c)
(a)
(b)
(c)
(d)
d 2C
d
= f″(Q) = 6Q – 24 = 0
Q=4
(MC ) =
dQ
dQ 2
d
d
( AC ) =
(Q 2  12Q) = 2Q – 12 = 0 Q = 6
dQ
dQ
dy
 18 x(3x  1)  3(9 x 2  2)  3(27 x 2  6 x  2)
dx
dy
 3(6 x 2  x)  (12 x  1)(3x  11)  54 x 2  126 x  11
dx
dy
 2 x(4 x  6)  4 x 2  12 x( x  1)
dx
dy
x2  3
  x  2 ( x 2  3)  2 x( x) 1 
dx
x2
Derivatives and Other Math Concepts
y = (3x + 11)(6x2 – x)
25 of 28
dy 1 x  7
7
  2  2
dx x
x
x
dy
4
4x
20
(f)



2
dx x  5 ( x  5)
( x  5) 2
dy du
6.
 3u 2 (2 x)  3(5  x 2 ) 2 (2 x)  6 x(5  x 2 ) 2
du dx
dw dy
7.
 2ay(2bx  c)  2a(bx 2  cx)( 2bx  c)  2ax(2b 2 x 2  3bcx  c 2 )
dy dx
dy
8. (a)
 3(3x 2  7) 2 (6 x)  18 x(3x 2  7) 2
dx
dy
(b)
 6(8x 3  5) 5 (24 x 2 )  144 x 2 (8x 3  5) 5
dx
dy
(c)
 4(ax  b) 3 (a)  4a(ax  b) 3
dx
dy
9.
 2(16 x  3) 3 (16)  32(16 x  3) 3 . Both give the same answer.
dx
z
z
10. (a)
 6 x 2  22 xy
 11x 2  6 y
x
dy
z
z
 6  16 y 2
(b)
 32 xy  27 y 2
x
dy
z
z
 2x  3
(c)
 2( y  2)
dy
x
z
2
z
2x  3

(d)

x y  2
dy
( y  2) 2
(e)
(b)

f y  5x  3 y 2
f ( x, y)  2 x  5 y
x
f x  2x( x  2)  x 2  7 y  3x 2  4x  7 y
(c)
fx 
11. (a) f x 
2
2x  3y
5y


2
x  y ( x  y)
( x  y) 2
2 x ( x 2  1) y x 2  1

 2
xy
x2 y2
x y
12. (a) 2 + 10 = 12
(b) 3 – 4 – 14 = –15
U
U
 3( x  2) 2 ( y  3) 2
 2( x  2)( y  3) 3
13. (a)
y
x
(d)
fx 
f y  7( x  2)
fy 
3
2x  3y
5 x


2
x  y ( x  y)
( x  y) 2
fy  
x2 1
(c)
xy 2
10/9
(b)
dy  (2 x  3)dx
(d)
1
(b) U x (3,3)  2(3  2)(3  3) 3  2160
14. (a) dy  ( x 2  3  2x 2 )dx  3( x 2  1)dx
 1
2x 2 
1 x2
 2
dx
(c) dy   2
=
dx

2
( x 2  1) 2
 x  1 ( x  1) 
dC
C a  bY
b

15. (a)
dY
Y
Y
dC dY
bY

(b)  CY 
CY
a  bY
(c) Since 𝑏𝑌 < a + 𝑏𝑌, 𝜀𝐶𝑌 < 1. The consumption function is inelastic at all positive income levels.
Derivatives and Other Math Concepts
26 of 28
16.  QP 
 QP 
17.
18.
19.
20.
dQ
 nkP( n1)
dP
dQ dP
Q P
Q
k
 n1
P P
 nkP ( n 1)
 n
kP ( n 1)
No. the value of elasticity is not dependent on the price. Elasticity is constant at all price ranges. Also,
when n = 1, the demand function is Q = k ∕ P, which plots a rectangular hyperbola (a unitarily elastic
demand).
(a) 10e0.05(3) = 10e0.15
(b) 700e0.08
4
(a) 10 = 10,000
log1010,000 = 4
(b) 106 = 1,000,000 log101,000,000 = 6
(c) 34 = 81
log381 = 4
(d) log100.0001 = log10(1/10,000) = log10(10,000)−1 = −1 log1010,000 = −4
(e) 23 = 8
log28 = 3
(f) 55 = 3125
log53125 = 5
(a) ln(e-4) = −4ln(e) = −4
(b) ln(e5) = 5
1
(c) ln   = ln(e-1) = −ln(e) = −1
e
1
(d) ln 3 = ln (e-3) = −3ln(e) = −3
e
(a) 14 log10100 = 14(2) = 28
1
(b) log 10 100 = −log10 100 = −2
(c) ln (3/B) = ln (3) – ln (B)
(d) ln (Ae2) = ln (A) + 2
(e) ln (ABe−4) = ln (A) + ln (B) – 4
(f) (log4 e)(loge 64) = log4 64 = 3
(g) 1/(log16 2) = log2 16 = 4
(h) log32 2 = 1/(log2 32) = 1/5
 u 
21. (a) ln (u) – 2 = ln (u) – ln (e2) = ln  2 
e 
(valid)
 e3 
(b) 3 + ln (v) = ln (e3) + ln (v) = ln (e3v) ≠ ln   (not valid)
 v 
 
 uv 
(c) ln (u) + ln (v) – ln (w) = ln (𝑢𝑣) – ln(w) = ln   (valid)
w
(d) ln (3) + ln (5) = ln (3 × 5) = ln (15) ≠ ln (8)
(not valid)
22. logb y = logb a + 𝑐𝑥logb b = logb a + cx
log b y  log b a
x
c
23. Transform the following functions to their natural exponential forms:
3
(a) y = 83x = (83)x = e ln(8
)x
= e 3 ln(8) x
(b) y = 2(7)2x = 2(72)x = 2e 2 ln(7) x
(c) y = 5(5)x = 5e ln(5) x
(d) y = 2(15)4x = 2e 4 ln(15) x
24. Transform the following functions to their natural logarithmic forms:
1
ln( y )
(a) x = log7 y = (log7 e)(loge y) =
ln( 7)
Derivatives and Other Math Concepts
27 of 28
(b) x = 3log15 9y = 3(log15 e)(loge 9y) =
(c) x = log8 3y= (log8 e)(loge 3y) =
(d) x = 2log10 y=
3
ln(9 y )
ln(15 )
1
ln(3 y )
ln(8)
2
ln( y )
ln(10 )
Derivatives and Other Math Concepts
28 of 28