Microeconomic Theory
Class 1
Functions
A function is a mapping at which each member of domain has exactly one match in codomain.
A. Polynomials
A polynomial is a function of a general form:
𝑓(𝑥) = 𝑎0 + 𝑎1 𝑥 + 𝑎2 𝑥 2 + ⋯ + 𝑎𝑛 𝑥 𝑛
Domain of all the polynomials is the entire set of real numbers.
0. Polynomial of the 0 degree: 𝑓(𝑥) = 𝑎0
1. Polynomial of the 1st degree: 𝑓(𝑥) = 𝑎0 + 𝑎1 𝑥 (graph is called LINE)
Sometimes it is necessary to deduct a line equation that passes through two points
𝐴(𝑥𝐴 , 𝑦𝐴 ), 𝐵(𝑥𝐵 , 𝑦𝐵 ). This equation is obtained using the following formula:
𝑦𝐵 − 𝑦𝐴
𝑦 − 𝑦𝐴 =
(𝑥 − 𝑥𝐴 )
𝑥𝐵 − 𝑥𝐴
Problem 1: Draw a line defined by the following points, and find a line equation:
a) T1(20,800); T2(30,600).
b) T1(800,20); T2(600,30).
c) T1(4,60); T2(6,90).
d) T1(4,60); T2(6,60).
2. Polynomial of the 2nd degree: 𝑓(𝑥) = 𝑎0 + 𝑎1 𝑥 + 𝑎2 𝑥 2 (graph is called parabola)
1
3. Polynomial of the 3rd degree: 𝑓(𝑥) = 𝑎0 + 𝑎1 𝑥 + 𝑎2 𝑥 2 + 𝑎3 𝑥 3 (graph is called cubic
parabola)
Problem 2:
a) Draw a graph of the following function in a 2 dimensional space:
y=100+0,5x.
b) Let us presume y is the quantity, and p is the price in the demand function. What is the
economic meaning of the parameters?
c) Draw the following functions: y=100+x;
y=100+2x;
y=50+0,5x.
B. Rational functions
𝑟(𝑥)
Rational functions are functions of the form 𝑓(𝑥) = 𝑞(𝑥) , 𝑤ℎ𝑒𝑟𝑒 𝑞(𝑥) ≠ 0
They have hyperbolic shape.
C. Irrational functions
2
1
Irrational functions (roots) are functions of the general form 𝑦 = 𝑥 𝑛 , 𝑛 ∈ ℕ+ , for n being even
number 𝐷(𝑓) = ℝ+
0 , for n being an odd number 𝐷(𝑓) = ℝ.
Irrational functions are inverse functions of polynomials. An inverse function of a function f(x) is a
function 𝑥 = 𝑓 −1 (𝑥) = ℎ(𝑥). Graphically, it is a mirror image of the basic function with respect to
the diagonal of the 1st quadrant.
Example 1:
Example 2:
3
2 (𝑝𝑜𝑙𝑦𝑛𝑜𝑚𝑖𝑎𝑙)
𝑥
𝑎𝑛𝑑 𝑖𝑡𝑠 𝑖𝑛𝑣𝑒𝑟𝑠𝑒 √𝑥 (𝑖𝑟𝑟𝑎𝑡𝑖𝑜𝑛𝑎𝑙 𝑓. ) 𝑥 3 (𝑝𝑜𝑙𝑦𝑛𝑜𝑚𝑖𝑎𝑙) 𝑎𝑛𝑑 𝑖𝑡𝑠 𝑖𝑛𝑣𝑒𝑟𝑠𝑒 √𝑥 (𝑖𝑟𝑟𝑎𝑡𝑖𝑜𝑛𝑎𝑙 𝑓. )
D. Exponential functions
Exponential functions have the following form: 𝑓(𝑥) = 𝑎 𝑥 , 𝑎 > 0
The most common is a function where a = e = 2,71… (Euclid constant)
Domain is the entire set of real numbers. If a<1 then function decreases and if a>1 then function
increases:
E. Logarithmic functions
Logarithmic function is an inverse function of an exponential function: 𝑓(𝑥) = log 𝑎 𝑥
If a = e then it is called a natural logarithm: 𝑓(𝑥) = ln 𝑥
Rules:
3
ln 𝐴𝐵 = ln 𝐴 + ln 𝐵
ln
𝐴
= ln 𝐴 − ln 𝐵
𝐵
𝑒 ln 𝑥 = 𝑥
Domain of the function is ⟨0, +∞⟩
F. Trigonometric functions
Trigonometric functions are sin 𝑥 , cos 𝑥 , tan 𝑥 , cot 𝑥.
Some rules:
sin 𝑥
𝜋
tan 𝑥 = cos 𝑥, (Domain: ℝ \ { 2 + 𝑘𝜋, 𝑘 ∈ ℤ})
cot 𝑥 =
cos 𝑥
,
sin 𝑥
(Domain: ℝ \{𝑘𝜋, 𝑘 ∈ ℤ})
sin2 𝑥 + cos 2 𝑥 = 1,
−1 ≤ sin 𝑥 ≤ 1,
−1 ≤ cos 𝑥 ≤ 1
sin 𝑥 and cos 𝑥 have domain equal to ℝ.
Elementary transformations of functions
𝑓(𝑥) + 𝑐
−𝑓(𝑥)
𝑓(𝑥 − 𝑐)
4
𝑎𝑓(𝑥), 𝑎 > 0
𝑓(𝑎𝑥), 𝑎 > 0
𝑓(−𝑥)
Multivariable functions and their graphs
A function f(x,y) can be drawn in 3 dimensional space (2 for x and y and 1 for f), but function with
more than two variables cannot be depicted graphically.
Function that reach local maximum and minimum look as follows:
5
The function that has minimum in 1 variable and maximum in the other looks as follows (saddle
point):
Layer curves are the curves obtained by cutting the function with a horizontal plane. All the points
on a layer curve satisfy the condition 𝑓(𝑥, 𝑦) = 𝑐, 𝑐 = 𝑐𝑜𝑛𝑠𝑡.
Problem 3: For the function 𝑧 = 𝑓(𝑥, 𝑦) = 𝑥𝑦 find layer curves when z = 1, 2 and 3.
This function is a Cobb-Douglas function which looks like this:
6
1
2
3
Its layer curves are 1=xy, 2=xy, 3=xy → 𝑦 = 𝑥 , 𝑦 = 𝑥 , 𝑦 = 𝑥
Slope of the function
A slope of the function between two points T1 and T2 (the following picture) is the slope of the line
that passes through these two points (a secant line):
The value of the slope is:
y f x2   f  x1  f x  x   f x 


x
x2  x1
x
(the change of y caused by the change of x over the change of x).
Problem 4. If x = 1 and x = 1 find slopes of the following functions:
a) y  f x  2x  3
y  1  1

 2
x
1
b) y  f x   16 x  x 2
y 28  15

 13
x
1
7
c) y  f  x   4 x 
1
x2
y 7.75  3

 4.75
x
1
Problem 5: find slopes for the following functions when x increases from 1 to 2:
1. 𝑦 = 𝑥 2
2. 𝑦 = 𝑒 𝑥
3. 𝑦 = 2𝑥 − 3
4. 𝑦 = ln 𝑥
Slope function –derivative
When T1 and T2 merge into one point we obtain the slope at the single point and the line that
passes through that point is a tangent line. Derivative at the point is the slope of that tangent line.
Since ∆𝑥 → 0 it is no longer called a difference, but a differential and it is denoted with dx. A change
caused by dx (a tiny change of x) is now no longer ∆𝑦 but dy. Now the slope of the line is no longer
∆𝑦
∆𝑥
𝑑𝑦
but 𝑑𝑥. It is called a derivative of the function, or the slope function.
𝑦′ =
𝑑𝑦
∆𝑦
= lim
𝑑𝑥 ∆𝑥→0 ∆𝑥
Basic derivatives table
8
Derivative
Function
yI =
y = f(x)
Example
dY
dX
y=C
yI = 0
y = 5; yI = 0
y=x
yI = 1
-
y = kx
yI = k
y =5x; yI = 5
y = kx2
yI =2kx
y= 5 x2; yI =10x
y = kxn
yI = nkxn-1
yI= 2x10; yI = 20x9
y = ln x
yI =
y = f(x)+ g(x)
yI =
1
x
df
dg
+
dX
y = 10x-5 x2;
dX
yI = 10 – 10x
y = f(x)g(x)
yI =f(x)
dg
+ g(x)
dX
df
y = lnx x3);
dX
yI = x2 +3x2 lnx
product rule
y = f(x)/g(x)
df
dg
g x   f x 
dx
dx
yI =
2
g x 
y=
quotient rule
yI =
dg
dz
×
ln x
x
1
 x  ln x  1
1  ln x
x

yI =
2
x
x2
dz
dX
9
y = g ( z ( x ))
chain rule
y = 10z, z = 3 – 5x;
yI = -50
Problem 6: If x = 1 and x → 0 find slopes of the following functions:
a) y  f x  2x  3
dy
 2x  x   3  2 x  3  2x
 lim

 2 . The slope is identical on the entire domain.
dx x0
x
x
y
f(x)=-2x+3
5
4
3
2
1
x
-1.5
-1
-0.5
0.5
1
1.5
2
2.5
3
3.5
4
4.5
5
5.5
6
6.5
7
7.5
8
-1
-2
-3
-4
b) y  f x   16 x  x 2
dy
16x  x   x  x   16 x  x 2
 lim

dx x 0
x
16 x  16x  x 2  2 xx  x 2  16 x  x 2
lim
.
x  0
x
lim 16  2 x  x   16  2 x
2
x  0
The slope of this function changes. At x = 1 it is 14, and on the interval Δx = 1 it is 13.
y
f(x)=16x-x^2
60
50
40
30
20
10
x
-15
-10
-5
5
10
15
20
25
30
35
40
45
50
55
60
-10
10
c) y  f  x   4 x 
1
x2
4 x  x  
1
1
 4x  2
2
dy
x
x  x 
 lim
dx x  0
x
1
1
4 x  4x  2
 4x  2
2
x  2 xx  x
x 
lim
x  0
x
1
1

4
x

x 2  2 xx  x 2
x2 
lim
x  0
x

1
1 
lim  4 
 2  
2
2
x  0
x x  2 xx  x
x x 

4 x  4x 


 2 xx  x 2
2
  2x  0 
 4 4
 4 3

2
2
2
x 0 x  x x  2 xx  x
x
 x  0  0
4  lim


y
f(x)=4x-x^(-2)
15
10
5
x
-10
-8
-6
-4
-2
2
4
6
8
10
12
14
16
18
20
22
24
26
28
-5
-10
-15
-20
Problem 7: Differentiate the following functions, with respect to x:
a) y=3x3-4x2+250x-1500;
b) y=x98-x55;
c) y=x5 × 6x3;
d) y=125+2z3-x2+10x;
e) y=10-4z; z=7x4+15;
Then find the slope of the functions a), b) and c) when x = 1.
11
Problem 8: Differentiate the following functions:
a)
y = 3x + 1
y’ = 3
b)
y
2
x
y'  
2
x2
c)
y  x2  4
y'  2 x
d) f  x  
1  ln x
x
12
Class 2
The tangent and its equation
Y
f(x)
g(x)
T1
y*
x*
0
X
The tangent is the line that has the slope identical to the slope of the function at the tangency point
and its equation is: y  y * 


dy
x  x* .
dx
Find tangents for the functions in Problem 6:
a) y  g x  2x  2  2  3  2x  3
Note that the tangent is identical to the basic function.
b) y  g x  14x 14  15  14 x  1
c) y  g x   6 x  6 +3 = 6x - 3
Multivariable derivatives
Problem 9: Find the first gradient (vector of the first derivatives) of the following functions:
a) f  x, y   24 y  2 y 2  2 y x  2 x
y


2


f  x, y  
x


 2 x  4 y  24
13
b) Total revenue function is TRQ1 , Q2   2Q12  Q22  28Q1  24Q2 , and cost function is
TC Q1 , Q2   5Q12  2Q22  15 . Find the 1st and the 2nd gradient of the profit function.
Q1 , Q2   7Q12  3Q22  28Q1  24Q2  15
 14Q1  28
Q1 , Q2   

  6Q2  24 
Complex derivatives of the single variable functions
Problem 10: Find derivatives of the following functions:
a)

ln x x 2  2 x
y
x
x
y' 
2


2
2
 2x
 2 ln x x 2  2 x 2 x  2
x
x2
b) f x   e 20.5 x


2
Complex derivatives of the multi variable functions
Problem 11: Find the 2nd derivative of the following functions (the 1st is made in the Problem
9)
a) f  x, y   24 y  2 y 2  2 y x  2 x
y


2


f  x, y  
x


 2 x  4 y  24
1 
 y


x 
H   2 f  x, y    2 x x
 1
4 


x
14
b) Profit function is Q1 , Q2   7Q12  3Q22  28Q1  24Q2  15 . Find the Hessian matrix.
 14 0 
H   2 Q1 , Q2   

 0  6
Chain rule for multivariable functions
If f(x,y) has to be differentiated with respect to a and b if x(a,b) and y(a,b) then it has to be done in
the following way:
Scheme:
f
x
a
y
b
a
b
f f x f y
   
a x a y a
f f x f y
   
b x b y b
One has to identify all the paths of influence of the independent variable on the dependant variable.
Problem 12: Find derivatives of the following functions:


a) Find dx/dpx for the function x p x , p y , U if U ( p x , p y , I ) .
x
x
f U



p x p x U p x
b) Find df/dk for the function f(x,y) = 2xy if x =k+2l, and y = 3k -l.
15
f
 2 y 1  2x  3  2 y  6x
k
Note the difference in differentiation in the case of a single variable and a multivariable function!
Differential and total differential
1st differential dy of a single variable function is a the approximate change of a function f(x)
caused by a small change in the variable x:
y' 
dy
dx
dy  y ' dx
Total differential od the 1st order for function is: df 
f
f
f
dx1 
dx 2  ... 
dx n
x1
x 2
x n
Tangent plane equation at the point (x*, y*, z*) for function z(x,y) is:
z  z* 



dz
dz
x  x* 
y  y*
dx
dy

Problem 13: Solve the following exercises:
a) For f(x) = x2 at the point x = 2 find the approximate change of a function when dx = 0.01.
dy  2 x  dx  4  0.01  0.04
y  dy  4  0.04  4.04
15
10
5
1
2
3
4
16
b) For f(x) = x3 at the point x = 2 find the approximate change of a function when dx = 0.01.
dy  3x 2  dx  12  0.01  0.12
y  dy  8  0.12  8.12
60
40
20
1
2
3
4
Problem 14. Utility function u(x,y) = x0.5 y0.5.
a) Find the approximate utility for the basket (1.99, 2.02).




du  u x x* , y*  dx  u y x* , y*  dy
du  0.5
y
x
 dx  0.5
 dy  0.01e
x
y
dz  0.5   0.01  0.5  0.02  0.005
z  dz  2.005
b) Find the tangent plane at T(2,2,2).
u  2  0.5x  2  0.5 y  2
u  0.5 x  1  0.5 y  1  2
u  0.5 x  0.5 y
Problem 15. Solve the following exercises:
a) Find the approximate value of a function z(1.01, 0.98) if z(x,y) =exy using a total
differential. Find the tangent plane equation at point A(1,1,e).
17
z x  e xy y
z y  e xy z
z x 1,1  e
z y 1,1  e




dz  z x x* , y*  dx  z y x* , y*  dy
dz  e  0.01  e   0.02  0.01e
z  dz  0.99 z



z  z *  z x x  x*  z y y  y*
z  ex  1  e y  1

z  ex  ey  2e
b) Find total differential of a function f  x, y  
function f(3.05,3.95) without a calculator.
zx 
zy
x
x

2
2

 y 2  xx  y 
x
2
x
2
 y2


1 .5
 y 2  yx  y 
 y2

1.5
z x 3,4  
4
125
-3
z y 3,4  
125
18
x y
x2  y2
. Find the approximate value of a




dz  z x x * , y *  dx  z y x * , y *  dy
dz 
4
3
 0.05 
  0.05  0.0028
125
125



z  z *  z x x  x*  z y y  y*

4
x  3  3  y  4
125
125
4
3
z
x
y
125
125
z
Integration
Basic undefined integrals
 dx
xC
n
 x dx
x n 1
C
n 1
e
ex  C
x
dx
1
 x dx
f ' x 
ln x  ln C
 f x  dx
ln f ( x)  ln C
 af x dx
a  f  x dx
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  f x   g x dx
 f x dx   g x dx
Partial integration:
 u x   dvx 
u x  vx    v x du x 
Newton-Leibniz formula:
b

f x dx F x  a  F b   F a 
b
F x  a  F b   F a 
b
a
Undefined integral 𝑭(𝒙) = ∫ 𝒇(𝒙)𝒅𝒙 is a function for which 𝑭′(𝒙) = 𝒇(𝒙) holds.
𝒃
Defined integral ∫𝒂 𝒇(𝒙) 𝒅𝒙 is an area bellow the curve. One sums up all the rectangles bellow
the curve between a and b, by cutting that zone in many pieces, each being Δx wide:
The smaller Δx the more correct the area. If the difference between xi and xi+1, Δx, becomes
very small, the area of rectangles covers the entire area below the curve:
𝒃
Finally, the solution is ∫𝒂 𝒇(𝒙) 𝒅𝒙 = 𝑭(𝒃) − 𝑭(𝒂)
Differential equation is an equation which contains the function y, the variable x and the
derivative of y, y’. One has to rescribe y’ as dy/dx and then separate all y’s and x’s and then
apply integration.
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Problem 16. Solve the exercises:

a) Find ln xdx
u  x   ln x
du  x  
dx
x
dv x   dx v( x)   dx  x
 ln xdx  x ln x   x dx  x ln x  x  C  xln x  1  C
x
b) Find all functions y(x) which have the elasticity equal to 4 and we know that f(1) = 2. The
formula the elasticity is je E y , x 
x dy
.
y dx
It is a differential equation. The variables have to be separated:
x dy
4
y dx
dy
dx
4
y
x
dy
dx
 y 4 x
ln y  4 ln x  ln C
ln y  ln Cx 4
y  Cx4 (general solution)
y 1  C14  2
C2
yx   2 x 4
(particular solution)
c) Total surplus in an economy is the area between supply and demand curve. Find total surplus if
supply is p = eq, and demand p = e-q+10, if the market is ints competitive equilibrium (it means, on the
intercept of supply and demand).
e q  e  q 10
ln e q  ln e q 10
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q   q  10
q5
Hence one has to find the grey shaded area:
p
S
D
0
8
8
8
0
0
q
P   e q dq   e q 10  e q  e q 10  e 8  e 2  2
8
8
0
0
Marginal and average value
𝑑𝑦
Marginal value of a function y(x) is a derivative of a function, 𝑀𝑉 = 𝑑𝑥.
Average value of a function y(x) is a function divided with its independent variable: 𝐴𝑉 =
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𝑦(𝑥)
𝑥