10 Appendix A: Stage 1 Survey and Focus Group Transcripts
13 APPENDIX D: WP1 Design Concepts and Recommendations One requirement of evaluation was to consider how, or whether, design concepts and
13.1.2 D1.1 Chapter 2 Uncertainty
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MODULE THREE: DIFFRENTIATIONS, INTEGRATION AND OPTIMIZATION TECHNIQUES
UNIT 1 Differential Calculus and Some Economic Applications
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Differential calculus studies the rates at which quantities change. The entities involved in the study of differential calculus are the derivative of a function, or differentiation, and the associated applications.
3.2 Derivative of a Function: The derivative of a function defines the rate of change of the function at given value. It is a sequence of differentiation. The geometrical explanation of derivative hinges on the slope of the line tangential to the graph of the function at the point of tangent.
Suppose that x and z are real numbers and that z is a function of x, that is, for every value of x, there is a corresponding value of y (which - y or z?). This mathematical link can be written as:
z = f(x)= c+bx
Thus, the slope of the function is obtained by obtaining the derivative of the function w.r.t. z.
This is shown below:
change in z b change in x b z
x
=
=
Where the symbol Δ symbolizes "change in". It follows that Δz = bΔx. The geometry of the derivative of f at the point z = b is the slope of the tangent line to the function f at the point b.
This is shown in the figure below and is denoted using either Langrange or Leibniz's notation as shown below:
Source: Wikipedia f ′(x) (Langrange notation)
x d
dy
dx = (Leibniz's notation)
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3.3. Rules of Differentiation: These are rule for finding the derivative of a function. They include the constant rule, power rule, product rule, quotient rule, exponential rule, logarithmic rule, trigonometry rule etc. Nevertheless, the derivative
dx dy is
x im y
→0
if the limit exists i. Constant Rule of Differentiation: For any constant k, such that
( )
'( ) 0 ( ) 0 f x k
f x d k
dx
=
= =
ii. Power Rule of Differentiation: The derivative of f(x), from the first principles is as follows:
1
4
1 3
5
1 2
6
1 5
( ) , '( )
( )
'( ) 4
( ) 3
'( ) 15
( ) 2
'( ) 12
n n
n
n
n
f x x f x dy nx dx f x x
f x dy nx x
dx f x x
f x dy nx x
dx
f x x
f x dy nx x
dx
−
−
−
−
= =
=
= =
=
= =
=
= =
For example, if f(x) = x6, then the derivative function f ′(x) would be given as:
'( ) y 6 5
f x x
x
= =
6 5
) (
' x
dx x dy
f =
iii. The derivatives of exponential and logarithmic functions obeys the following rules
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( )
( ) ( ) 1 ( )
log 1 , 0, 1
ln
(ln ) 1, 0
(ln ) 1, 0
( ) (1 ln ), 0
[ ( ) ] ( ) ( ) ( ) ln[ ( )] ( ) 0
ax ax
e
x x
h x h x h x
d e ae dx
d x where k k
dx x k
d x where x
dx x
d x where x
dx x
d x x x where x
dx
d df dh
f x h x f x f x f x provided f x
dx dx dx
−
=
=
=
=
= +
= +
iv. The derivatives of trigonometric functions obeys the following rules of engagement:
2 2
2
(sin ) ' cos (cos ) ' sin
(tan ) ' sec 1 tan (sec ) ' sec tan (cot ) ' (1 cot )
x x
x x
x x x
x x x
x x
=
= −
= +
=
= − +
v. The derivatives of hyperbolic functions obeys the following rules of engagement:
2
2
(sinh ) ' cosh
2 (cosh ) ' sinh
2 (tanh ) ' sec
(sec ) ' tanh sec (coth ) ' csc
x x
x x
e e
x x
e e
x x
x h
hx x hx
x h x
−
−
= +
= −
=
= −
= −
vi. The derivative of implicit functions: The derivative of an implicit function is a derivative of both sides of the function w.r.t. one of the variables while holding the other variables relatively constant. Examples of implicit function is: g z y( , )= +z3 y7+2, then the circle is the set of all pairs (x, y) such that g(z, y) = 0. So,
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( , ) 0
z y 0
z y
z y
g z y g dz g dy g dz g dy
g dy
dz g
=
+ =
= −
= −
vii. The product rule: Given that a functiong(uv)where uand vare two separate functions of the same variablez, the derivative of g(uv)is given as
dz vdu dz udv uv
g'( )= +
viii. Given that uand vare two separate functions of the same variablez, the derivative of the function
v z u g( )=
) 2
(
' v
dz udv dz vdu z g
= −
SELF ASSESSMENT EXERCISE Explain the meaning of an implicit function
How would you differentiate between Langrange notation and Leibniz’s notation?
Differentiate between an exponential and a logarithmic functions 3.4Solving Numerical Problems on Differential Calculus Numerical Example 1: Find the derivative of the given function.
( )i g z( )=5z3−4z+2
Solution to Numerical Example 1: Taking the derivative, we have that:
'( ) g 15 2 4
g z z
z
= = −
Numerical Example 2: Find the derivative of the given function.
( )i g z( )=(5z−2)(9z+z2)
Solution to Numerical Example 2: Taking the derivative, we have that:
2 2
2 2
( ) (5 2)(9 )
(5 2), (9 )
, ,
'( ) (5 2) (9 ) (9 ) (5 2)
g z z z z
let u z v z z
Utilizing the product riule we have dv du
uv u v
dz dz
g d d
g z z z z z z z
z dv dv
= − +
= − = +
= +
= = − + + + −
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(5z 2)(9 2 ) (9z z z2)5
= − + + +
2
2 4 45 5
10 18
45z− + z − z+ z+ z
=
18 86 15 2 + −
= z z
Numerical Example 2
5
5
5 5
5 5
5
( ) '( )
,
, ,
( ) ( ) ( ) ( )
(5 ) ( )(1) (5 1)
z
z
z z
z z
z
g z ze
dv du
g z u v
dz dz let u z v e
Utilizing the product riule we have dv du
uv u v
dz dz
g d d
g z z e e z
z dv dv
ze e
e z
=
= +
= =
= +
= = +
= +
= +
Numerical Example 3: Find the derivative of the given functions.
2
3
6 3
3
( ) ( ) 5
6
5 10 2
( ) ( ) ( ) ( ) 2
i g z z z
z z
ii g z
z
z z z
iii g z
z
= +
− +
=
= − + Solution to Numerical Example 3:
2
2
( ) 5
6
5 , 6
, ,
g z z z
let u z v z
Utilizing the quotient riule we have
= +
= = +
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2
2 2
2
2 2
2 2
2 2
2
( 6) (5 ) (5 ) ( 6)
( ) ( 6)
( 6)10 (5 ) ( 6)
10 60 5
( 6)
5 60
( 6) du dv
v u
g dz dz
z v
d d
z z z z
g dv dv
g z z z
z z z
z
z z z
z
z z
z
−
=
+ − +
= = +
+ −
= +
+ −
= +
= + +
3
3
2
3 3
2
5 10 2
( )
(5 10 2),
, ,
(5 10 2) (5 10 2) ( ) ( )
z z
g z z
let u z z v z
Utilizing the quotient riule we have du dv
v u
g dz dz
z v
d d
z z z z z z
g dv dv
g z z z
− +
=
= − + =
= −
− + − − +
= =
2 3
2
3 3
2 3
2 2
(15 10) (5 10 2)
15 10 5 10 2
10 2
10 2
z z z z
z
z z z z
z z
z z z−
− − − +
=
− − + −
=
= −
= −
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6 3
3
6 3 3
2
( ) 2
(2 ),
, ,
z z z
g z z
let u z z z v z
Utilizing the quotient riule we have du dv
v u
g dz dz
z v
= − +
= − + =
= −
3 6 3 6 3 3
3 3
(2 ) (2 ) ( )
( ) ( ).( )
d d
z z z z z z z z
g dv dv
g z z z z
− + − − +
= =
3 5 2 2 6 3
6
(12 3 1) 3 (2 )
z z z z z z z
z
− + − − +
=
6
3 5 8 3 5
8 3 6 3 3
12
z
z z z z z
z − + − + −
=
6 3
8 2
6 z
z z −
=
3
2 2
6 − −
= z z
Alternatively, we can divide the numerator of g z( )by the denominator of each function respectively and apply the power function rule as follows:
1 2
3
2 10 2 5
10
5 −
+
− + =
− z z
z z z
(
5z2 −10+2z−1)
=10z−2z−2dz d
2 3
3 3 6
2 2 2
2 −
+
− + =
− z z
z z z z
(
2z3−2+z−2)
=6z2−2z−3dz d
Numerical Example 4: Consider the following function
2 3
32z −8y +29=0
Solution to Numerical Example 4: Carrying out an implicit differentiation of the function, we have as follows:
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2 3
2 3
2 3
2
2
2
32 8 29 0
(32 8 29) (0)
(32 ) (8 ) (29) 0
64 24 0
64 24 64 24
z y
d d
z y
dz dz
d d d
z y
dz dz dz
z y dy dz z y dy
dz
dy z
dz y
− + =
− + =
− + =
− =
=
= − Alternatively,
2
( , ) 0
64 24
z y
f z y f dy
dz f
dy z
dz y
=
= −
= −
Numerical Example 5: Find the derivative of the following functions:
i. y=exp( )x ex ii. y=x7lnx
iii. y=ln(4x2−5x+6)
Solutions: Given that y=exp( )x ex 2.71828...
' exp( ) x
where e
y dy x e
dx
=
= =
Given thaty=x7lnx
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7
7 6
6 6
6
ln
, :
1 ln .7 7 ln (1 7 ln ) let u x
v x
Applying the product rule we have as follows
dy dv du
u v
dx dx dx
x x x
x
x x x
x x
=
=
= +
= +
= +
= +
Given thaty=ln(4x2−5x+6)
2
2
2
(4 5 6)
ln
, :
1 .(8 5)
4 5 6
8 5
4 5 6
let u x x
y u
Applying the chain rule we have as follows dy dy du
dx du dx x x x
x
x x
= − +
=
=
= −
− +
= −
− + SELF ASSESSMENT EXERCISE
Describe the relationship between an explicit and an implicit derivative. Give examples
3.5. Application of Differentiation to Economic Problems: Recall that the slope of the graph