Mathematics Formulary
Version 0.4
December 11, 2019
Dan Mønster, Marie Dyveke, Thomas Hoffmann
mathematics formulary
Contents
Differentiation
4
Basic rules for differentiation . . . . . . . . . . . . . . . . . .
4
Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4
Increasing and decreasing functions . . . . . . . . . . . . . .
5
Convex and concave functions . . . . . . . . . . . . . . . . .
5
Implicit differentiation . . . . . . . . . . . . . . . . . . . . . .
5
Differentiating the Inverse . . . . . . . . . . . . . . . . . . . .
6
L’Hôpital’s rule . . . . . . . . . . . . . . . . . . . . . . . . . .
6
Functions of several variables and partial derivatives
6
Convex and concave functions of two variables . . . . . . . .
7
Chain rule for functions of several variables . . . . . . . . .
7
Implicit differentiation along a level curve. . . . . . . . . . .
7
Optimization
8
Optimization without constraints . . . . . . . . . . . . . . . .
8
Optimization with constraints . . . . . . . . . . . . . . . . . .
9
Integration
9
General rules . . . . . . . . . . . . . . . . . . . . . . . . . . . .
9
Rules for definite integrals . . . . . . . . . . . . . . . . . . . .
10
Rules for composite functions . . . . . . . . . . . . . . . . . .
10
Linear algebra
Matrix algebra . . . . . . . . . . . . . . . . . . . . . . . . . . .
11
11
2
mathematics formulary
Determinant . . . . . . . . . . . . . . . . . . . . . . . . . . . .
12
Inverse matrix . . . . . . . . . . . . . . . . . . . . . . . . . . .
13
Systems of linear equations . . . . . . . . . . . . . . . . . . .
14
Appendix: prerequisite knowledge
16
Fractions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
16
Identities involving squares . . . . . . . . . . . . . . . . . . .
16
Powers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
16
Quadratic equation . . . . . . . . . . . . . . . . . . . . . . . .
17
Exponential and logarithmic functions . . . . . . . . . . . . .
17
We gratefully acknowledge Peter Le whose compendium we have been
allowed to use as inspiration. Disclaimer: This formulary is work in
progress and may contain errors. If you find one, please contact us at
danm@econ.au.dk.
3
mathematics formulary
Differentiation
The table below gives rules and examples for differentiating some
basic functions.
Rule
Example
f (x)
f 0 (x)
f (x)
f 0 (x)
xa
ex
ln( x )
e ax
ax
ax a−1
ex
1/x
ae ax
a x ln( a)
x6
ex
ln( x )
e3x
5x
6x5
ex
1/x
3e3x
5x ln(5)
Basic rules for differentiation
Rule
Example
F(x)
F0 (x)
F(x)
F0 (x)
A
0
13
0
A + f (x)
f 0 (x)
4 + x2
2x
A · f (x)
A · f 0 (x)
3x2
6x
f ( x ) ± g( x )
f 0 ( x ) ± g0 ( x )
2x + x2
2 + 2x
f ( x ) g( x )
f 0 ( x ) g( x ) + f ( x ) g0 ( x )
x4 e5x
4x3 · e5x + x4 · 5e5x
f (x)
g( x )
f 0 ( x ) g( x ) − f ( x ) g0 ( x )
2x3 + 2
x+4
6x2 · ( x + 4) − (2x3 + 2) · 1
( x + 4)2
f ( g( x ))
f 0 ( g( x ) · g0 ( x )
(2x2 + 4x )3
3(2x2 + 4x )2 · (4x + 4))
( g( x ))
2
Notation
Leibniz’s notation for the derivative of f ( x ) w.r.t. x
f 0 (x) =
df
dx
In general the n-th order derivative is written
f (n) ( x ) =
dn f
dx n
4
mathematics formulary
The chain rule
( f ( g( x )))0 = f 0 ( g( x )) · g0 ( x )
can be rewritten by introducing y = f (u) and u = g( x ) using Leibniz’s notation
dy du
dy
=
dx
du dx
Increasing and decreasing functions
The derivative of a function indicates whether the function is increasing, decreasing or constant in an interval I if the following inequalities are satisfied for all x ∈ I
f 0 ( x ) ≥ 0 ⇐⇒ f ( x ) is increasing
f 0 ( x ) ≤ 0 ⇐⇒ f ( x ) is decreasing
f 0 ( x ) = 0 ⇐⇒ f ( x ) is constant
Convex and concave functions
The second order derivative of a function can be used to determine
if the function is convex or concave in an interval I if one of the
following inequalities hold for all x ∈ I
f 00 ( x ) ≥ 0 ⇐⇒ f ( x ) is convex
f 00 ( x ) ≤ 0 ⇐⇒ f ( x ) is concave
Implicit differentiation
(From the book, page 223)
To find y’ when an equation relates two variables x and y (total differentiation):
(i) Differentiate each side of the equation w.r.t. x, considering y as a function
of x.
(ii) Solve the resulting equation for y ’
Alternatively, y0 can be found with partial differentiation:
(i) Isolate all terms that contain x or y on side of the equation to get f ( x, y) = c,
∂F
(ii) Partial differentiate the equation w.r.t both y and x to get ∂F
∂x and ∂y
(iii) y0 can be found by applying the formula y0 =
dy
dx
∂F
∂x
= − ∂F
∂y
5
mathematics formulary
Differentiating the Inverse
To find the inverse function of f(x):
(i) Write f(x) as y
(ii) Swap x and y
(iii) Solve the resulting equation for y
To differentiate the inverse function:
(iv) Differentiate the equation from (iii) w.r.t. x
g 0 ( y0 ) =
f0
1
( x0 )
L’Hôpital’s rule
L’Hôpital’s rule can be used to find the limit of a quotient of two
functions when both functions tend towards 0, such as:
lim
x→a
”0”
f (x)
=
g( x )
0
Note that “0/0” is not a valid expression — hence the quotation
marks. In such a case one can apply l’Hôpital’s Rule:
If f ( a) = g( a) = 0, and g0 ( a) 6= 0 then:
lim
x→a
f (x)
f 0 (x)
= 0
g( x )
g (x)
Functions of several variables and partial derivatives
The partial derivatives of f ( x, y) are written:
∂f
= f x0 ( x, y) = f 10 ( x, y)
∂x
∂f
= f y0 ( x, y) = f 20 ( x, y)
∂y
Examples of higher order derivatives:
∂2 f
00
00
( x, y)
= f xx
( x, y) = f 11
∂x2
∂2 f
00
00
= f yy
( x, y) = f 22
( x, y)
∂y2
∂2 f
00
00
= f xy
( x, y) = f 12
( x, y)
∂y∂x
6
mathematics formulary
7
Convex and concave functions of two variables
A function of two variables is said to be convex if
2 2
∂2 f
∂2 f
∂2 f ∂2 f
∂ f
≥ 0,
≥ 0,
· 2−
≥0
2
2
2
∂x∂y
∂x
∂y
∂x ∂y
and concave if:
∂2 f
≤ 0,
∂x2
∂2 f
≤ 0,
∂y2
∂2 f ∂2 f
·
−
∂x2 ∂y2
∂2 f
∂x∂y
2
≥ 0.
Note: This must be true for all points ( x, y) in the domain.
Chain rule for functions of several variables
The chain rule for the composite function F ( x, y), where x = f (t)
and y = g(t) is:
dF
= F10 ( x, y) f 0 (t) + F20 ( x, y) g0 (t)
dt
In Leibniz’s notation:
dF
∂F dx ∂F dy
=
+
dt
∂x dt
∂y dt
If x = f (t, s) and y = g(t, s):
∂F
∂F ∂x ∂F ∂y
=
+
∂t
∂x ∂t
∂y ∂t
∂F
∂F ∂x ∂F ∂y
=
+
∂s
∂x ∂s
∂y ∂s
The general chain rule for a function F ( x1 , x2 , . . . , xn ) where each of
the n variables are functions of m other variables xk = f k (t1 , t2 , . . . , tm ):
∂F ∂x1
∂F ∂x2
∂F ∂xn
∂F
=
+
+···+
∂t j
∂x1 ∂t j
∂x2 ∂t j
∂xn ∂t j
Implicit differentiation along a level curve.
A level curve is given by F ( x, y) = c. Assuming this defines a function y = f ( x ), then the slope of this function can be computed as
dy
∂F
f (x) =
=−
dx
∂x
0
∂F
∂y
Note: There are a total of m formulas
like this (one for each t-variable), where
this is number j out of m.
mathematics formulary
8
Optimization
Optimization without constraints
First-order conditions
∂f
=0
∂x
∂f
=0
∂y
and
A solution ( x0 , y0 ) to the first-order conditions is called stationary
(critical) point.
Global second-order conditions (Theorem 13.2.1)
Maximum: If these conditions hold for all ( x, y) then ( x0 , y0 ) is a
global maximum in S.
∂2 f
≤ 0,
∂x2
∂2 f ∂2 f
·
−
∂x2 ∂y2
Note: To use theorem 13.2.1. it is a requirement that f is a C2 function in a
convex set S in R2
∂2 f
≤0
∂y2
∂2 f
∂x∂y
In words: If f is concave then ( x0 , y0 ) is
a maximum.
2
≥0
Minimum: If these conditions hold for all ( x, y) then ( x0 , y0 ) is a
global minimum in S.
∂2 f
≥ 0,
∂x2
∂2 f ∂2 f
·
−
∂x2 ∂y2
∂2 f
≥0
∂y2
∂2 f
∂x∂y
In words: If f is convex then ( x0 , y0 ) is
a minimum.
2
≥0
Local second-order conditions (Theorem 13.3.1)
A=
∂2 f
( x0 , y0 )
∂x2
C=
∂2 f
( x0 , y0 )
∂y2
B=
∂2 f
( x0 , y0 )
∂x∂y
The type of point is shown in the table.
Point
A
AC − B2
( x0 , y0 )
( x0 , y0 )
( x0 , y0 )
( x0 , y0 )
>0
<0
>0
>0
<0
=0
Type
local minimum
local maximum
saddle point
one of the above
mathematics formulary
Optimization with constraints
Find optimal value for f ( x, y) such that g( x, y) = c.
Lagrange method:
L( x, y) = f ( x, y) − λ( g( x, y) − c)
First-order conditions
Any solution ( x0 , y0 ) to the problem must be a stationary point for
the Lagrangian, and satisfy the constraint, i.e.
∂L
=0
∂x
∂L
=0
∂y
g( x, y) = c
Global SOC (Theorem 14.5.1)
If L is concave then ( x0 , y0 ) is a maximum for f ( x, y).
If L is convex then ( x0 , y0 ) is a minimum for f ( x, y).
Local SOC (Theorem 14.5.2)
00
00
D ( x, y, λ) = f 11
− λg11
g20
2
0 2
0 0
00
00
00
00
− 2 f 12
− λg12
g1
g1 g2 + f 22
− λg22
If D ( x0 , y0 , λ) < 0 then ( x0 , y0 ) is a local solution to the max-problem.
If D ( x0 , y0 , λ) > 0 then ( x0 , y0 ) is a local solution to the min-problem.
Integration
Definition (indefinite integral):
Z
f ( x ) dx = F ( x ) + C,
where: F 0 ( x ) = f ( x ).
Definition (definite integral):
Z b
a
b
F ( x ) = F (b) − F ( a)
f ( x ) dx =
a
General rules
Z
Z
a f ( x ) dx = a
Z
a dx = ax + C
f ( x ) dx
9
mathematics formulary
x a +1
+ C,
a+1
Z
x a dx =
Z
1
dx = ln | x | + C,
x
Z
e x dx = e x + C
a 6 = −1
x 6= 0
1 ax
e + C, a 6= 0
a
Z
1 x
a + C, a > 0, a 6= 1
a x dx =
ln a
Z
e ax dx =
Z
ln x dx = x ln x − x + C,
x>0
Rules for definite integrals
Z b
a
Z a
a
Z b
a
Z c
a
Z b
a
f ( x ) dx = −
Z a
b
f ( x ) dx
f ( x ) dx = 0
α f ( x ) dx = α
f ( x ) dx =
Z b
a
Z b
f ( x ) dx
a
f ( x ) dx +
f ( x ) + g( x ) dx =
Z b(t)
a(t)
Z b
a
Z c
b
f ( x ) dx
f ( x ) dx +
Z b
a
g( x ) dx
f ( x ) dx = F (b(t)) − F ( a(t))
Rules for composite functions
Integration of a sum or difference of two functions
Z
f ( x ) ± g( x ) dx =
Z
f ( x ) dx ±
Z
g( x ) dx
Integration by parts (indefinite integral):
Z
f ( x ) g0 ( x ) dx = f ( x ) g( x ) −
Z
f 0 ( x ) g( x ) dx
Integration by parts (definite integral):
Z b
a
b
f ( x ) g0 ( x ) dx =
f ( x ) g( x ) −
a
Z b
a
f 0 ( x ) g( x ) dx
Integration by substitution (indefinite integral):
Z
f ( g( x )) g0 ( x ) dx =
Z
f (u) du
Integration by substitution (definite integral):
Z b
a
0
f ( g( x )) g ( x ) dx =
Z g(b)
g( a)
f (u) du
10
mathematics formulary
Linear algebra
Matrix algebra
Sum of two matrices with the same dimensions
⇐⇒
C = A+B
cij = aij + bij
Multiplication by a scalar
B = αA
⇐⇒
bij = αaij
Matrix multiplication
n
C = AB
⇐⇒
cij =
∑ aik bkj
k =1
Here, A is a matrix with dimension m × n and B is a matrix with
dimension n × p, then the matrix product C = AB has dimension
m × p.
A, B and C are matrices and α and β are scalars.
(A + B) + C = A + (B + C)
A+B = B+A
A+0 = A
A + (−A) = 0
(α + β)A = αA + βA
α(A + B) = αA + αB
(AB)C = A(BC)
A(B + C) = AB + AC
(A + B)C = AC + BC
(αA)B = A(αB) = αAB
Identity matrix

1 0

0 1
In = 
 .. ..
. .
0 0
···
···
..
.
···

0

0
.. 

.
1
In is a quadratic n × n matrix In with 1 in the diagonal elements, and
0 everywhere else.
For any n × n matrix A:
AIn = In A = A
11
mathematics formulary
Tranpose matrix

a11 a12

 a21 a22
A=
..
 ..
 .
.
am1 am2

a1n

a2n 
.. 

. 
amn
···
···
···

=⇒
a11

 a12
A0 = 
 ..
 .
a1n
a21
a22
..
.
a2n
···
···
···

am1

am2 
.. 

. 
amn
A0 is the transpose of A and has rows and columns swapped relative
to A.
Rules for transposition
(A0 )0 = A
(A + B)0 = A0 + B0
(αA)0 = αA0
(AB)0 = B0 A0
Determinant
Determinant of a 2 × 2 matrix
|A| =
a11
a21
a12
= a11 a22 − a21 a12
a22
Determinant of a 3 × 3 matrix, suing Sarrus’s rule
|A| =
+
a11
+
a12
+
a13
a11
a12
a21
a22
a23
a21
a22
a31
a32
a33
a31
a32
−
−
−
Determinant of a 3 × 3 matrix, using cofactor expansion along the
first row
|A| = a11
a22
a32
a23
a
− a12 21
a33
a31
a23
a
+ a13 21
a33
a31
a22
a32
General determinant using cofactor expansion along column j:
|A| = a1j C1j + a2j C2j + · · · + anj Cnj
where Cij are the cofactors:
Cij = (−1)i+ j Dij
12
mathematics formulary
and Dij is the minor, i.e. the determinant of the matrix obtained by
deleting the i’th row and the j’th column from A.
General determinant using cofactor expansion along row i:
|A| = ai1 Ci1 + ai2 Ci2 + · · · + ain Cin
where Cij are the cofactors.
Example of cofactor expansion by row two:
10
0
2
2
0
1 −1
√
10
1 −1
10 0
1
√
−1
0
2
= (−1)(−1)2+2 2
0
4 + 2(−1)2+4 2 1
0
1
0
4
2 −1
5
2 0 −1
0 −1
5
!
1
1 −1
2+3 10
+ 4(−1)
= − 2(−1)
2 −1
−1
5
!
√
1
0
1+3 2 1
+ 2 10(−1)1 + 1
+ 1(−1)
2 0
0 −1
√
= − (−2 · 4 − 4 · (−12)) + 2 (10 · (−1) + 1 · (−2))
√
= −40 − 12 2
Rules for determinants
|A0 | = |A|
|AB| = |A||B|
|αA| = αn |A|
where A has dimension n × n
Inverse matrix
A quadratic matrix A has an inverse matrix A−1 if |A| 6= 0, and
A−1 A = AA−1 = I
where I is the identity matrix.
Rules for matrix inversion
If A and B are invertible n × n matrices, then
( A −1 ) −1 = A
(AB)−1 = B−1 A−1
( A 0 ) −1 = ( A −1 ) 0
(αA)−1 = α−1 A−1
13
mathematics formulary
Inverse of a 2 × 2 matrix
a11
a21
A=
a12
a22
!
If |A| 6= 0, then
A −1
1
=
|A|
a22
− a21
− a12
a11
!
Systems of linear equations
A system of m linear equations with n unknowns
a11 x1 + a12 x2 + · · · + a1n xn = b1
a21 x1 + a22 x2 + · · · + a2n xn = b2
..
.
am1 x1 + am2 x2 + · · · + amn xn = bm
The system can also be written in matrix form: Ax = b, where
 
 


b1
x1
a11 a12 · · · a1n
 
 


 b2 
 x2 
 a21 a22 · · · a2n 
 
 
A=
..
.. 
 x =  ..  b =  .. 
 ..
 . 



 .
.
.
.
bm
xn
am1 am2 · · · amn
A system of n linear equations with n unknowns has a unique solution if the determinant is non-zero: |A| 6= 0.
Cramer’s rule for two equations with two unknowns:
a11 x1 + a12 x2 = b1
a21 x1 + a22 x2 = b2
If |A| 6= 0 then the solution is
x1 =
b1
b2
a12
a22
|A|
x2 =
a11
a21
b1
b2
|A|
Cramer’s rule for three equations with three unknowns:
a11 x1 + a12 x2 + a13 x3 = b1
a21 x1 + a22 x2 + a23 x3 = b2
a31 x1 + a32 x2 + a33 x3 = b3
If |A| 6= 0 then the solution is
x1 =
b1
b2
b3
a12 a13
a22 a23
a32 a33
|A|
x2 =
a11
a21
a31
b1 a13
b2 a23
b3 a33
|A|
x3 =
a11
a21
a31
a12
a22
a32
|A|
b1
b2
b3
14
mathematics formulary
Solution by using the inverse matrix
x = A −1 b
15
mathematics formulary
Appendix: prerequisite knowledge
Fractions
b
a·b
=
c
c
a
a·c
=
b
b
c
a·
a
b
c
a
b
c
d
=
a
b·c
=
a·d
b·c
a·c
a c
· =
b d
b·d
Identities involving squares
( a + b)2 = a2 + b2 + 2ab
( a − b)2 = a2 + b2 − 2ab
( a + b)( a − b) = a2 − b2
Powers
am · an = am+n
am
= am−n
an
( am )n = am·n
( a · b)m = am · bm
a m
am
= m
b
b
a0 = 1
a−m =
√
m
√
n
1
am
1
a = am
m
am = a n
√
√ √
a·b = a· b
r
√
a
a
= √
b
b
√
1
a = a2
16
mathematics formulary
Quadratic equation
ax2 + bx + c = 0
The discriminant is defined as
D = b2 − 4ac
If D > 0 there are two solutions (roots) given by the formula
√
−b ± b2 − 4ac
x=
2a
If D = 0 there one solution given by
x=
−b
2a
If there are two solutions x1 and x2 the quadratic equation can be
written as
a( x − x1 )( x − x2 ) = 0
This is called factorising the quadratic equation, since it is now written as a product of two factors.
Exponential and logarithmic functions
The natural logarithm
y = ln( x ) ⇔ x = ey
ln(e) = 1
ln( a · b) = ln( a) + ln(b)
a
= ln( a) − ln(b)
ln
b
ln ( ar ) = r · ln( a)
Logarithmic function with base 10
y = log( x ) ⇔ x = 10y
log(10) = 1
log( a · b) = log( a) + log(b)
a
log
= log( a) − log(b)
b
log ( ar ) = r · log( a)
17
mathematics formulary
The natural exponential function
y = exp( x ) = e x
y = e x ⇔ x = ln(y)
e x ey = e x +y
ex
= e x −y
ey
Exponential function with base a
y = ax
ekx = (ek ) x = a x ,
a x ay = a x +y
ax
= a x −y
ay
where a = ek
18