Math Review for ECON 222 - Queen's Economics …qed.econ.queensu.ca/walras/custom/200/222/winter15/math_review... · Math Review for ECON 222 Winter Term, 2015 Wenbo Zhu Queen’s

Math Review for ECON 222Winter Term, 2015

Wenbo Zhu

Queen’s Economic Department

January 13, 2015

Wenbo Zhu (QED) Math Review for ECON 222 January 13, 2015 1 / 37

Outline

1 Exponents

2 Logarithms (Natural logarithms)

3 Growth rates

4 Differentiation

5 Elasticities

6 Examples


ExponentsSome graphs

The function: f (x) = ax , a is the “base” and a > 0.

−3 −2 −1 0 1 2 3

05

1015

20

The exponential functions

x

y

●

base = 5base = ebase = 2base = 0.2


ExponentsSome basic information about the graph

f (x) = ax has the following properties, regardless the value of a:

the domain of the function is (−∞,∞);

the range of the function is (0,∞);

the function goes through the point (0, 1).


ExponentsGeneral Properties of Exponents

an · am = an+m

22 · 23 = 25 = 32(1)

an

am= an−m

23

22= 23−2 = 2

(2)

(an)m = an·m

(22)3 = 22·3 = 64(3)


ExponentsQuestion

How about 24 · 42??I first, notice that 2 6= 4, so we cannot apply the equation (1)

above directly ;I however, 4 = 22, so by equation (3), we have

42 = (22)2 = 22·2 = 24;I then by equation (1), 24 · 42 = 24 · 24 = 24+4 = 256.


LogarithmsSome graphs

The function: f (x) = loga x , again a is the “base” and a > 0, a 6= 1.

0 2 4 6 8

−4

−3

−2

−1

01

2

The logarithm functions

x

y

●

base = 10base = ebase = 2base = 0.5

*note: in this course, most likely we deal with natural logs, whichmeans the base is “e”.


LogarithmsSome basic information about the graph

f (x) = loga x , again regardless the value of a:

the domain of the function is (0,∞);

the range of the function is (−∞,∞);

the function goes through the point (1, 0)


LogarithmsThe relation with Exponents: exp ln(a) = a or ln(exp a) = a,∀a > 0

−4 −2 0 2 4

−4

−2

02

4

x

y ●

●

y = e^xy = ln(x)y = x

The exponential and logarithm functions are the inverse of oneanother (i.e. they appear to cancel each other out).


LogarithmsUseful Rules for logs

Here are some useful rules for calculation involving logs:

ln(xy) = ln x + ln y (4)

lnx

y= ln x − ln y (5)

ln xp = p ln x (6)

ln 1 = 0 (7)


LogarithmsAn example

Please simplify the following expression:

exp[ln(x2)− 2 ln y

]Here is the solution:

exp[ln(x2)− 2 ln y

]= e[ln(x2)−2 ln y]

= e [ln(x2)−ln(y2)]

=e ln(x2)

e ln(y2)

=

(x

y

)2


Growth ratesFormulas

Let X and Z be any two variables, not necessarily related by afunction, that are changing over time.

Let ∆XX

and ∆ZZ

represent the growth rates (percentage changes)

I you know that the two-period growth rate of X is Xt+1−Xt

Xt= ∆X

Xt

Rule 1. The growth rate of the product of X and Z equals thegrowth rate of X plus the growth rate of Z .

∆(XZ )

XZ=

∆X

X+

∆Z

Z(8)


Growth ratesProof of Rule 1

Proof. Suppose that X increases by ∆X and Z increases by∆Z . Then the absolute increase in the product of X and Z is(X + ∆X )(Z + ∆Z )− XZ , and the growth rate of the productof X and Z is:

∆(XY )

XY=

(Xnew )(Znew )− XZ

XZ

=(X + ∆X )(Z + ∆Z )− XZ

XZ

=XZ − (∆Z )Z + (∆X )X + ∆X∆Z − XZ

XZ

=∆X

X+

∆Z

Z+

∆X∆Z

XZ

The last term can be ignored if both term deviations are small(which they usually are).Wenbo Zhu (QED) Math Review for ECON 222 January 13, 2015 13 / 37

Growth ratesProof of Rule 1: Graphical


Growth ratesRule 2

Rule 2. The growth rate of the ratio of X to Z is the growthrate of X minus the growth rate of Z .

Proof. Let W be the ratio of X to Z , so W = X/Z . ThenX = ZW . By Rule 1, as X equals the product of Z and W , thegrowth rate of X equals the growth rate Z plus the growth rateof W:

∆X

X=

∆Z

Z+

∆W

W∆W

W=

∆X

X− ∆Z

Z


Growth rates and logarithms

For small values of x , the following approximation holds :

ln(1 + x) ≈ x

From that property let us show that the growth rate is alsoequal to the difference in logarithms:

Xt+1 − Xt

Xt≈ lnXt+1 − lnXt


Growth rates and logarithmsProof

lnXt+1 − lnXt = ln

(Xt+1

Xt

)= ln

(Xt+1 − Xt + Xt

Xt

)= ln

(1 +

Xt+1 − Xt

Xt

)≈ Xt+1 − Xt

Xt


DifferentiationSingle-variable Differentiation

Definition:

f ′(x) = limh→0f (x+h)−f (x)

h

There are several ways to denote derivatives

f ′(x) denotes the derivative of f (x) with respect to (wrt) x

f ′x (x , y) denotes the partial derivative of f (x , y) wrt x

you can also denote derivatives with df (x)dx

... and partial derivatives with ∂f (x ,y)∂x


DifferentiationRules (with a single function)

Two important derivatives for the course:

f (x) = xa ⇒ f ′(x) = axa−1

f (x) = ln x ⇒ f ′(x) =1

x


DifferentiationMore Rules: (with two functions)

Take the derivative wrt each term separately

F (x) = f (x) + g(x) ⇒ F ′(x) = f ′(x) + g ′(x)

F (x) = f (x)− g(x) ⇒ F ′(x) = f ′(x)− g ′(x)

Example:

y = A ⇒ y ′ = 0

y = A + f (x) ⇒ y ′ = f ′(x)


DifferentiationMore Rules: (with two functions) Cont’d

Product Rule:

F (x) = f (x)g(x) ⇒ F ′(x) = f ′(x)g(x) + f (x)g ′(x)

Example: y = A · f (x)⇒ y ′ = A · f ′(x)

Quotient Rule:

F (x) =f (x)

g(x)⇒ F ′(x) =

f ′(x) · g(x)− f (x) · g ′(x)

[g(x)]2


DifferentiationMore Rules: (with two functions) Cont’d

The Chain Rule: If y is a function of u, and u is a function of x , then

dy

dx=

dy

du

du

dxExample:

y = (ln x)2

dy

dx=

dy

du

du

dx,where y = u2, and u = ln x

= 2 · ln x︸︷︷︸dy

du

· 1

x︸︷︷︸du

dx


DifferentiationMarginal Products with Calculus

Finding the MPK of a Cobb-Douglas production:Y = AKαN1−α

MPK is the change in output for a one unit change in capital(∆Y∆K

when ∆K = 1)

keeping everything else equal.

The MPK is the slope of the tangent line to the productionfunction and is equal to the derivative of the function.

Example: Y = 4K 0.5 Find the MPK when K = 100

MPK =∂Y

∂K= 2K−0.5

=2√100

=2

10=

1

5

Apply the same method to find the marginal product of labour


DifferentiationMaximization and Minimization

π(q) = 1000q − cq2, where 1000 > c > 0

to maximize the above objective function, find the partialderivative w.r.t. q and set it equal to zero

π′(q) = 1000− 2cq = 0

q∗ =1000

2c

the procedure for maximization and minimization is the same

a function has been maximized if π′′(q) < 0

why? At the max the slope of the objective function is zero. Anincrease or decrease in q decreases π.

a function has been minimized if π′′(q) > 0


DifferentiationSimple Constrained Maximization

π(p, q) = pq − cq where c > 0

subject to p(q) = 1000− 4q

substitute the constraint (p(q)) directly into the π(p, q)objective function and set the partial derivative equal to zero

π(q) = (1000− 4q)q − cq

= 1000q − 4q2 − cq

π′(q) = 1000− c − 8q = 0

q∗ =1000− c

8

the procedure for maximization and minimization is the same

a function has been maximized if π′′(q) < 0

a function has been minimized if π′′(q) > 0


ElasticitiesDefinitions

The elasticity of Y with respect to N is defined to be thepercentage change in Y , ∆Y /Y , divided by the percentagechange in N , ∆N/N :

ηY ,N = ∆Y /Y∆N/N

Or if we use derivatives:

ηY ,N = NY∂Y∂N

= ∂ lnY∂ lnN

How so? (Hint: exploit the chain rule):

N

Y

∂Y

∂N=

∂ lnY∂Y

∂Y∂ lnN∂N

∂N=∂ lnY

∂ lnN


Example 1

Find the first and second order derivative

f (x) = x2

f ′(x) = 2x

f ′′(x) = 2


Example 2


f (x) = 3x

f ′(x) = 3

f ′′(x) = 0


Example 3

Find the first and second order partial derivative

f (x , y) = yx3

f ′x (x , y) = 3yx2

f ′′x (x , y) = 6yx

*note: you can think of f (x , y) is a production function and y beingthe labour input and x being the capital input.


Example 4


f (x) = ln(x)

f ′(x) =1

x

f ′′(x) =−1

x2


Example 5

f (x) = ex

f ′(x) = ex

why?

y = ex

ln(y) = ln ex

ln(y) = x

1

y

dy

dx= 1

dy

dx= y

dy

dx= ex


Example 6

Find the first order derivative

f (x) =ex

x2

f ′(x) =exx2 − ex2x

(x2)2

f ′(x) =ex(x2 − 2x)

x4


Example 7

Find the first order derivative

f (x) = ln(x5 − 5x3 − 100

)f ′(x) =

5x4 − 15x2

x5 − 5x3 − 100


Example 8

f (x , y) = [αxω + βyω]1ω

∂f (x , y)

∂x=

1

ω· [αxω + βyω]

1ω−1 · αωxω−1


Back to growth rate rules (Optional)

Rule 3. Suppose that Y is a variable that is a function of twoother variables X and Z . Then ∆Y

Y= ηY ,X

∆XX

+ ηY ,Z∆ZZ

Proof.: informal. The overall effect on Y is approximately equalto the sum of the individual effects on Y of the change in X andthe change in Z .

Rule 4. The growth rate of X raised to the power a, or X a, is atimes the growth rate of X , growth rate of (X a) = a∆X

X

Proof.: Let Y = X a. Using Rule 3, we find the elasticity of Ywith respect to X equals a.


Elasticity of substitution (Optional)

elasticity of substitution is the elasticity of the ratio of twoinputs to a production (or utility) function with respect to theratio of their marginal products (or utilities): σyx = η( y

x ),MRSy,x

We know that the marginal rate of substitution between y and x

is: MRSy ,x =F ′y (x ,y)

F ′x (x ,y)

So if, for example, we want to find σK ,N for the followingCobb-Douglas production function: Y = AKαN1−α

MRSK ,N =MPK

MPN=

AαKα−1N1−α

A(1− α)KαN−α

=α

1− αK

N


Elasticity of substitution (Optional) Cont’d

Rearrange the above equation, we get

K

N=

(1− αα

)MRSK ,N .

Then we can apply the formula of σyx and get

σK ,N =MRSK ,N

1−αα

MRSK ,N

· 1− αα

= 1.


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Math Review for ECON 222 - Queen's Economics …qed.econ.queensu.ca/walras/custom/200/222/winter15/math_review... · Math Review for ECON 222 Winter Term, 2015 Wenbo Zhu Queen’s