1
MATH134 - Quantitative Methods: Calculus
Mapundi K. Banda, Dr. rer. nat.(Darmstadt)
E-mail: [email protected]
School of Mathematical Sciences
University of KwaZulu-Natal (Pietermaritzburg)
03. Apr, 2006
Mathematical Sciences - 2006
MATH134 - Quantitative methods: Calculus.
1 Differentiation
1.1 Functions - Some functions in Economics
➠ A function is a rule that assigns an object (x) in a collection of objects (A) to one and only one
object (y) in a corresponding collection of objects (B). We write y = f (x): x belongs to A Domain of the function and y belongs to B - the Range.
➠ Demand is a function of price or demand depends on price, written:
QD = f (p)
where QD is the quantity demanded per time period, p is the unit price and f is the function
relating p to QD .
➠ Also one may write QD instead of f : QD = QD (p)
➠ A monopolist is a sole supplier of a product who may be able to express a price as a function
of quantity demanded i.e.
p = p(QD ) i.e. p is a function or rule that assigns QD to p.
Banda–UKZN
2
Mathematical Sciences - 2006
MATH134 - Quantitative methods: Calculus.
➠ Example (Example 1A in Book)
Let p3 + 4QD = 150. Find (a) QD (p) (b) p(QD ).
Solution:
(a) Solve for QD in the equation (Make QD subject of the formula):
p3 + 4QD = 150;
4QD = 150 − p3 ;
QD
150 − p3
=
.
4
(b) Solve for p in the equation (Make p subject of the formula):
p3 + 4QD = 150;
p3 = 150 − 4QD ;
(p3 )1/3 = (150 − 4QD )1/3 ;
p = (150 − 4QD )1/3 .
Banda–UKZN
3
Mathematical Sciences - 2006
MATH134 - Quantitative methods: Calculus.
Total Revenue function of a Monopolist
➠ Total Revenue (R) = Price of a commodity (p) × Quantity (Q)
i.e. R = p × Q.
➠ Recall a monopolist determines price as a function of quantity: i.e. p = p(Q).
➠ Hence R = R(Q) = p × Q = p(Q) × Q.
1
➠ Example (Example 2A in Book) Let QD = 170 − p. Find R of a monopolist.
2
Solution:
Find p(Q): i.e.
1
QD = 170 − p;
2
1
p = 170 − QD ;
2
p = 340 − 2QD .
Hence R
Banda–UKZN
= R(Q) = (340 − 2Q)Q = 340Q − 2Q2 .
4
Mathematical Sciences - 2006
MATH134 - Quantitative methods: Calculus.
➠ Example (Example 3B in Book): Let pQD = 1000. Find R.
Solution:
To free up p from QD take log10 of the equation:
pQD = 103 ;
log10 (pQD ) = log10 103 = 3;
QD log10 p = 3;
log10 p =
10
log10 p
3
;
QD
= 10
3
QD
;
p = 10
Hence R
Banda–UKZN
= 10
3
QD
3
QD
;
× QD .
5
Mathematical Sciences - 2006
MATH134 - Quantitative methods: Calculus.
Total Cost, Average Cost
➠ Total Cost per time period (C): total expenditure incurred in producing a given quantity of
commodity (Q) over a time period i.e.
C = C(Q)
➠ Total Cost curves tend to approximate a cubic function.
➠ Average Cost (AC ): cost per unit of commodity produced.
AC = C/Q = C(Q)/Q
➠ If total cost is approximated by a cubic equation without a constant term, then average cost is a
quadratic function.
➠ Example (Example 5A in Book): Given C(Q) = 2Q3 − 3Q2 + 4Q, find AC .
Solution: AC = (2Q3 − 3Q2 + 4Q)/Q = 2Q2 − 3Q + 4.
Banda–UKZN
6
MATH134 - Quantitative methods: Calculus.
Mathematical Sciences - 2006
1.2 The Algebra of Functions - Sum, Difference, Product and Quotients of
Functions
➠ Example - Budget Deficit or Surplus:
Let C(t) and R(t) denote, respectively, the SA government’s Costs (Spending) and Revenue at
time t, measured in billions of rands.
Can derive another function D(t) = R(t) − C(t), the difference between Revenue and
Costs.
➠ If D(t) = R(t) − C(t) > 0, we have a Surplus.
➠ If D(t) = R(t) − C(t) < 0, we have a Deficit.
➠ i.e. We have defined a function D, the difference of the functions C , R written D = R − C .
The domain of D is the same as the domain of R and C .
Banda–UKZN
7
Mathematical Sciences - 2006
MATH134 - Quantitative methods: Calculus.
The Sum, Difference, Product and Quotients of Functions
Let f and g be functions which depend on x:
(f + g)(x) = f (x) + g(x)
Sum
(f − g)(x) = f (x) − g(x)
Difference
(f g)(x) = f (x)g(x)
³f ´
f (x)
(x) =
g
g(x)
Product
Quotient, provided g(x)
6= 0.
If the domain of f is A and domain of g is B then the new functions have domain A
Banda–UKZN
T
B.
8
MATH134 - Quantitative methods: Calculus.
Mathematical Sciences - 2006
Example - not in Book
Let f (x)
= x + 1 and g(x) = 2x + 1. Find the sum s, the difference d, the product p, and the
quotient q of the functions f and g .
s(x) = (f + g)(x) = f (x) + g(x) = (x + 1) + (2x + 1) = 3x + 2;
d(x) = (f − g)(x) = f (x) − g(x) = (x + 1) − (2x + 1) = −x;
p(x) = (f g)(x) = f (x)g(x) = (x + 1)(2x + 1) = 2x2 + x + 2x + 1
= 2x2 + 3x + 1;
³f ´
f (x)
= (x + 1)/(2x + 1); Quotient, provided x 6= − 12 .
q(x) =
(x) =
g
g(x)
Banda–UKZN
9
MATH134 - Quantitative methods: Calculus.
Mathematical Sciences - 2006
Example - not in Book
Suppose P, a manufacturer of water filters, has a monthly fixed cost of R10, 000 and a variable
cost of
−0.0001Q2 + 10Q; (0 ≤ Q ≤ 40, 000)
rands, where Q denotes the number of filters manufactured per month. Find a function C that
gives the total cost incurred by P in the manufacture of Q filters.
Solution:
P’s monthly fixed cost (regardless of level of production)- constant function:
P’s variable cost:
P’s Total Cost
F (Q) = 10, 000.
V (Q) = −0.0001Q2 + 10Q
:= P’s Variable Cost + P’s Fixed Cost
Hence:
C(Q) = V (Q) + F (Q)
= −0.0001Q2 + 10Q + 10, 000; (0 ≤ Q ≤ 40, 000)
Banda–UKZN
10
Mathematical Sciences - 2006
MATH134 - Quantitative methods: Calculus.
The Algebra of Functions - Composition
➠ Functions can be built up from other functions through composition of two functions.
➠ Let f and g be two functions. Then the composition of f and g is the function g ◦ f defined by
(g ◦ f )(x) = g(f (x))
➠ The function g ◦ f (read ’g circle f ’) is also called a composition.
√
➠ Example - not in Book Let f (x) = x2 − 1 and g(x) = x + 1. Compute:
(a) Composition function: g ◦ f .
(b) Composition function: f ◦ g .
Solution:
2
√
◦ f )(x) = g(f (x)) = g(x − 1) = x2 − 1 + 1.
√
√
(b) (f ◦ g)(x) = f (g(x)) = f ( x + 1) = ( x + 1)2 − 1.
(a) (g
Banda–UKZN
11
Mathematical Sciences - 2006
MATH134 - Quantitative methods: Calculus.
Let’s Log in I (Revision)
➠ Consider the exponential equation of the form:
by = x (b > 0, b 6= 1).
Then y is called the logarithm of x to base b written logb x.
= logb x implies x = by (i.e. x = blogb x ); (ii) x = by implies
y = logb x (i.e. y = logb (by )) (x > 0)
Hence (i)y
➠ Notation:
log x = log10 x Common logarithm
ln x = loge x Natural logarithm and
e = 2.71828....
➠ Examples
a) log 100 = 2; b) log5 125 = 3; c) log3 1/27 = −3.
d) Solve for x: log3 x = 4.
Banda–UKZN
12
MATH134 - Quantitative methods: Calculus.
Mathematical Sciences - 2006
Let’s Log in II (Revision)
➠ Laws of Logarithms:
If M , N , n > 0 and for a, b > 0:
loga M N = loga M + loga N
M
loga
= loga M − loga N
N
loga M n = n loga M
logb M
loga M =
(change of base from a to b)
logb a
1
loga M =
logM a
loga 1 = 0
loga a = 1.
Banda–UKZN
13
Mathematical Sciences - 2006
MATH134 - Quantitative methods: Calculus.
∆y
1.3 Difference Quotients ∆x
➠ We have y + ∆y = f (x + ∆x)
y = f (x)
Hence ∆y = y + ∆y − y = f (x + ∆x) − f (x)
Difference Quotient:
f (x + ∆x) − f (x)
∆y
=
∆x
∆x
i.e. slope of a straight line joining 2 points on graph y
The points are (x, y) and (x + ∆x, y
Banda–UKZN
= f (x).
+ ∆y).
14
MATH134 - Quantitative methods: Calculus.
Mathematical Sciences - 2006
Example (Example 7A in Book)
= x3 .
∆y
a) Compute
.
∆x
Consider y
Solution:
∆y = f (x + ∆x) − f (x);
= (x + ∆x)3 − x3 ;
= x3 + 3x2 ∆x + 3x(∆x)2 + (∆x)3 − x3 ;
= 3x2 ∆x + 3x(∆x)2 + (∆x)3 .
Then
¢
∆y
1 ¡ 2
2
3
=
3x ∆x + 3x(∆x) + (∆x) ;
∆x
∆x
= 3x2 + 3x∆x + (∆x)2 .
(b) Difference Quotient: x = 1.5 and ∆x = 0.1.
Solution:
Put x
= 1.5 and ∆x = 0.1 in
∆y
∆x
∆y
= 3 × (1.5)2 + 3 × 1.5 × 0.1 + (0.1)2 = 7.21.
∆x
Banda–UKZN
15
MATH134 - Quantitative methods: Calculus.
Mathematical Sciences - 2006
Example (Example 9B in Book)
Consider y
=
1
∆y
. Find
.
x+1
∆x
∆y = f (x + ∆x) − f (x)
1
1
x + 1 − x − ∆x − 1
=
−
=
(x + ∆x) + 1 x + 1
((x + ∆x) + 1)(x + 1)
−∆x
=
((x + ∆x) + 1)(x + 1)
³ 1 ´³
´
−∆x
−1
∆y
=
=
.
∆x
∆x ((x + ∆x) + 1)(x + 1)
((x + ∆x) + 1)(x + 1)
Banda–UKZN
16
MATH134 - Quantitative methods: Calculus.
Mathematical Sciences - 2006
1.4 Limits and Derivatives
➠ Recall: Difference Quotients :=
∆y
f (x + ∆x) − f (x)
=
∆x
∆x
➠ Difference Quotients approximate an average slope of a function at point x.
➠ What is the true value of the slope at point ’x’?
➠ Idea: Make ∆x small, very small or ’let it get close to zero’, ’let it tend to zero’ (it need not
be equal zero).
➠ Technically we write: lim and say ’the limit as ∆x tends to zero’.
∆x→0
Banda–UKZN
17
Mathematical Sciences - 2006
MATH134 - Quantitative methods: Calculus.
Example (Example 11A in the Book)
∆y
Consider y = f (x) = x − 4x, find
, evaluate it when x = 3, and when ∆x = 0.5, 0.1,
∆x
0.005, and 0.001.
2
∆y = f (x + ∆x) − f (x) =
=
(x + ∆x)2 − 4(x + ∆x) − f (x)
x2 + 2x∆x + (∆x)2 − 4(x + ∆x) − x2 + 4x
= 2x∆x + (∆x)2 − 4∆x.
Hence
1
∆y
=
(2x∆x + (∆x)2 − 4∆x) = 2x + ∆x − 4
∆x
∆x
Thus when x
Banda–UKZN
= 3 and ∆x = 0.5,
∆y
= 2 × 3 + 0.5 − 4 = 2.5
∆x
18
Mathematical Sciences - 2006
MATH134 - Quantitative methods: Calculus.
Similarly, compute for other values of ∆x:
∆x
∆y
∆x
0.5
0.1
0.05
0.01
0.005
0.001
2.5
2.1
2.05
2.01
2.005
2.001
∆x → 0.0
∆y
→ 2.0
∆x
∆y
Guess: lim
= 2.0.
∆x→0.0 ∆x
We say the limit as ∆x
Banda–UKZN
→ 0 of the difference quotient is 2.0.
19
Mathematical Sciences - 2006
MATH134 - Quantitative methods: Calculus.
Rules for computing Limits and Examples:
➠ Building Blocks (these are obvious):
(1)
lim ∆x = 0
∆x→0
(2) If E does not contain ∆x:
lim E = E
∆x→0
e.g.
lim 4 = 4;
∆x→0
lim x = x;
∆x→0
lim (x + 4) = (x + 4)
∆x→0
➠ Properties: If f = f (∆x) and g = g(∆x) and lim f = L and lim g = M then:
∆x→0
(3)
(4)
(5)
(6)
∆x→0
lim (f + g) = lim f + lim g = L + M ;
∆x→0
∆x→0
∆x→0
lim (f − g) = lim f − lim g = L − M ;
∆x→0
∆x→0
∆x→0
lim (f × g) = lim f × lim g = LM ;
∆x→0
∆x→0
∆x→0
lim∆x→0 f
f
= L/M if M 6= 0.
lim ( ) =
∆x→0 g
lim∆x→0 g
Banda–UKZN
20
Mathematical Sciences - 2006
MATH134 - Quantitative methods: Calculus.
Example (Example 12 A in Book): Find
Solution:
lim (x + ∆x) = lim x + lim ∆x = x + 0 = x
∆x→0
∆x→0
Example (Example 13 A in Book):
Solution:
Hence
lim (x + ∆x).
∆x→0
∆x→0
lim (∆x)2 = 0.
∆x→0
lim (∆x)2 = lim ∆x lim ∆x = 0
∆x→0
n
∆x→0
∆x→0
lim (∆x) = 0
∆x→0
¡ ∆x − x∆x ¢
Example (Example 15 B in Book): Find lim
.
∆x→0 (∆x)3 + 2∆x
Solution:
¡ ∆x − x∆x ¢
¡ ∆x(1 − x) ¢
¡ 1−x ¢ 1−x
lim
= lim
= lim
=
.
∆x→0 (∆x)3 + 2∆x
∆x→0 ∆x((∆x)2 + 2)
∆x→0 (∆x)2 + 2
2
Banda–UKZN
21
Mathematical Sciences - 2006
MATH134 - Quantitative methods: Calculus.
1.5 A special limit - The Derivative
In general, for a point x:
f 0 (x)
y
0
dy
dx
=
f (x + ∆x) − f (x)
∆y
= lim
∆x→0
∆x→0 ∆x
∆x
lim
is the derivative of ’y ’ with respect to ’x’
Geometrically, f 0 (x) is the slope of the straight line which is tangent to (i.e. which touches)
the curve y
= f (x) at the point (x, y).
The difference quotient gives average rate of change of y with respect to x in the interval
(x, x + ∆x).
The derivative gives the rate of change of y with respect to x precisely at the point x.
Banda–UKZN
22
MATH134 - Quantitative methods: Calculus.
Mathematical Sciences - 2006
Applications in Economics: Marginal Cost and Marginal Revenue
➠ Recall the Total Cost Function: C = C(Q) and Average Cost function: AC = C(Q)/Q.
∆C
dC
=
.
∆Q→0 ∆Q
dQ
➠ Marginal Cost(MC) = lim
➠ i.e. the rate of change of cost with respect to quantity produced (how expensive it will be to
increase production).
➠ Recall R(Q) = Q × p(Q), the total revenue function.
∆R
dR
=
.
∆Q→0 ∆Q
dQ
➠ Marginal Revenue(MR) = lim
➠ i.e. the rate of change of revenue with respect to quantity sold.
Banda–UKZN
23
MATH134 - Quantitative methods: Calculus.
Mathematical Sciences - 2006
More Examples from the Book:
➠ Example 17A: Derivative of y = x2
Solution:
∆y
∆y
∆x
dy
dx
Banda–UKZN
=
f (x + ∆x) − f (x) = (x + ∆x)2 − x2
=
x2 + 2x∆x + (∆x)2 − x2 = 2x∆x + (∆x)2 .
=
2x∆x + (∆x)2
= 2x + ∆x.
∆x
=
∆y
lim
= lim (2x + ∆x) = 2x.
∆x→0 ∆x
∆x→0
24
MATH134 - Quantitative methods: Calculus.
Mathematical Sciences - 2006
More Examples from the Book:
➠ Example 18A: Derivative of y = 1/x
Solution:
∆y
=
Banda–UKZN
1
1
−
(x + ∆x) x
x − x − ∆x
−∆x
=
.
x(x + ∆x)
x(x + ∆x)
= f (x + ∆x) − f (x) =
∆y
∆x
=
dy
dx
=
1 ³ −∆x ´
−1
=
.
∆x x(x + ∆x)
x(x + ∆x)
∆y
−1
−1
= lim
=
∆x→0 ∆x
∆x→0 x(x + ∆x)
lim∆x→0 (x + ∆x)x
1
= − 2.
x
lim
25
MATH134 - Quantitative methods: Calculus.
Mathematical Sciences - 2006
1.6 Rules for Differentiation
Rule 1: The derivative of a power (The Power Rule):
dy
➠ If y = x , then
= nxn−1 .
dx
➠ Rule 1 is true for all values of n (excluding n = 0).
n
➠ To prove this rule, use the definition of derivatives:
f (x + ∆x) − f (x)
∆y
= lim
f (x) = lim
∆x→0
∆x→0 ∆x
∆x
0
where ∆y
= (x + ∆x)n − xn .
➠ See derivation on page 94 in Book.
Banda–UKZN
26
MATH134 - Quantitative methods: Calculus.
Mathematical Sciences - 2006
➠ Example (Example 22B in Book): Find the derivatives of y = x3 and y = x.
√
➠ Example (Example 23C(d) in Book): Find the derivative of y = x (assume rule 1 applies
for all values of n).
➠ Example (Example 24C (a), (b), (d) in Book):
➛ If u = w5 find du/dw.
➛ If s = t2 find ds/dt.
➛ If A = H x where x is now an integer, find dA/dH .
Banda–UKZN
27
MATH134 - Quantitative methods: Calculus.
Mathematical Sciences - 2006
Rule 2: The derivative of functions multiplied by a constant:
dy
= cf 0 (x).
dx
➠ If y = cf (x), then
➠ To prove this rule, use the definition of derivatives:
➠ Firstly consider the new function defined by multiplying f (x) by c: i.e. g(x) = cf (x)
∆y
g(x + ∆x) − g(x)
y = g (x) = lim
= lim
∆x→0 ∆x
∆x→0
∆x
0
where ∆y
0
= cf (x + ∆x) − cf (x) = c(f (x + ∆x) − f (x)).
➠ Hence
c(f (x + ∆x) − f (x))
f (x + ∆x) − f (x)
∆y
= lim
= c lim
= cf 0 (x).
lim
∆x→0
∆x→0
∆x→0 ∆x
∆x
∆x
Banda–UKZN
28
MATH134 - Quantitative methods: Calculus.
Mathematical Sciences - 2006
➠ Example (Example 26B in Book): Find the derivatives of y = 6x2 and y = 17x3 .
➠ Example (Example 27C(c) in Book): Find the derivative of y = axb .
➠ Example (Example 28C (a)in Book):
If s = 1/2gt2 , where g is a constant, find ds/dt.
Banda–UKZN
29
MATH134 - Quantitative methods: Calculus.
Mathematical Sciences - 2006
Rule 3: The derivative of a sum of two functions:
➠ If y = f (x) + g(x), then
dy
= f 0 (x) − g 0 (x).
dx
dy
= f 0 (x) + g 0 (x) and if y = f (x) − g(x), then
dx
➠ To prove this rule, use the definition of derivatives:
➠ Firstly consider the new function - sum of two functions: y = f (x) + g(x).
∆y
f (x + ∆x) + g(x + ∆x) − f (x) − g(x)
y = lim
= lim
.
∆x→0 ∆x
∆x→0
∆x
0
➠ Rearrange terms and take limits of a sum of two functions:
∆y
f (x + ∆x) − f (x)
g(x + ∆x) − g(x)
lim
= lim
+ lim
= f 0 (x) + g 0 (x).
∆x→0 ∆x
∆x→0
∆x→0
∆x
∆x
Banda–UKZN
30
MATH134 - Quantitative methods: Calculus.
Mathematical Sciences - 2006
➠ Example (Example 31B in Book): Find the derivatives of y = 4x4 + 5x and
y = x4 + 5x3 − 3x2 + 10x.
➠ Example (Example 32C in Book): Find the derivative of C = 2Q3 − 3Q2 + 4Q.
Banda–UKZN
31
MATH134 - Quantitative methods: Calculus.
Mathematical Sciences - 2006
Rule 4: The derivative of a constant function:
➠ If y = f (x) = c, then
dy
= f 0 (x) = 0.
dx
➠ To prove this rule, use the definition of derivatives:
➠ Firstly consider the function: y = f (x) = c.
∆y
f (x + ∆x) − f (x)
c−c
= lim
= lim
= 0.
∆x→0 ∆x
∆x→0
∆x→0 ∆x
∆x
➛ Example (Example 35B and 36B in Book): Find the derivatives of y = 1 and
y = 3x2 − 2x + 5.
y 0 = lim
➛ Example (Example 37C in Book): Find the derivative dA/dr of A = π 2 + 2rc if c is a
constant, and dA/dc if r is a constant.
Banda–UKZN
32
MATH134 - Quantitative methods: Calculus.
Mathematical Sciences - 2006
Rule 5: The derivative of a product of two functions (Product Rule):
➠ If y = f (x)g(x), then
dy
= f 0 (x)g(x) + f (x)g 0 (x).
dx
➠ To prove this rule, use the definition of derivatives:
➠ Firstly consider the new function - product of two functions: y = f (x)g(x).
∆y
f (x + ∆x)g(x + ∆x) − f (x)g(x)
= lim
.
∆x→0 ∆x
∆x→0
∆x
y 0 = lim
➠ Use a trick (add zero to the numerator): −f (x)g(x + ∆x) + f (x)g(x + ∆x).
➠ Hence
f (x + ∆x)g(x + ∆x) − f (x)g(x + ∆x) + f (x)g(x + ∆x) − f (x)g(x)
.
y = lim
∆x→0
∆x
0
➠ Rearrange terms
£
¤
£ and take limits: ¤
g(x + ∆x) − g(x)
f (x + ∆x) − f (x)
0
y = lim
g(x + ∆x) + f (x) lim
.
∆x→0
∆x→0
∆x
∆x
Banda–UKZN
33
MATH134 - Quantitative methods: Calculus.
Mathematical Sciences - 2006
➠ Example (Example 39B, 40C in Book): Find the derivatives of y = (x4 + 1)(3x2 + 1) and
y = (x3 + 6x2 )(9x2 − 3x + 4).
➠ Indeed, the rule extends to the product of three functions: y = f (x)g(x)h(x).
Banda–UKZN
34
Mathematical Sciences - 2006
MATH134 - Quantitative methods: Calculus.
Rule 6: The derivative of a reciprocal of a function:
➠ If y =
1
g(x) , then
dy
g 0 (x)
.
=−
2
dx
(g(x))
➠ To prove this rule, use the definition of derivatives:
➠ Firstly consider the new function: y =
∆y
y 0 = lim
= lim
∆x→0 ∆x
∆x→0
➠ Hence
y 0 = lim
∆x→0
³
1
g(x) .
1
g(x+∆x)
−
∆x
1
g(x)
g(x) − g(x + ∆x)
.
∆x→0 ∆xg(x)g(x + ∆x)
= lim
−1
g(x + ∆x) − g(x) ´
.
g(x)g(x + ∆x)
∆x
➠ Rearrange terms and take limits:
³
´
³ g(x + ∆x) − g(x) ´
−1
g 0 (x)
0
y = lim
lim
=−
.
2
∆x→0 g(x)g(x + ∆x) ∆x→0
∆x
(g(x))
Banda–UKZN
35
MATH134 - Quantitative methods: Calculus.
Mathematical Sciences - 2006
➠ Example (Example 43B, 44C in Book): Find the derivatives of y = 1/(x2 + 1) and
y = (x3 + 6)(3x2 + 4)−1 .
Banda–UKZN
36
MATH134 - Quantitative methods: Calculus.
Mathematical Sciences - 2006
Rule 7: The derivative of a quotient of two functions (Quotient Rule):
dy
(f 0 (x)g(x) − f (x)g 0 (x))
➠ If y = f (x)/g(x), then
=
.
2
dx
[g(x)]
➠ To prove this rule, use the definition of derivatives or the product rule:
➠ Firstly consider the new function - Quotient of two functions: y = f (x)/g(x).
³
g 0 (x)
1 ´0
1
0
0
−
y =
f (x) ×
= f (x)
× f (x)
2
g(x)
g(x) (g(x))
g(x)f 0 (x) − f (x)g 0 (x)
=
(g(x))2
Banda–UKZN
37
MATH134 - Quantitative methods: Calculus.
Mathematical Sciences - 2006
➠ Example (Example 46B, 47C in Book): Find the derivatives of y = 3x/(x4 + 2x) with
respect to x and t = (v + 1)(v 3 − 1)/(v 2 + v − 1) with respect to v .
Banda–UKZN
38
MATH134 - Quantitative methods: Calculus.
Mathematical Sciences - 2006
Rule 8: The derivative of a function of a function (Chain Rule):
dy
➠ If y = f (u) and u = g(x), then
= dy/du × du/dx.
dx
➠ To prove this rule, use the definition of derivatives:
➠ Firstly consider the new function - Composition of two functions: y = f (u); u = g(x)
i.e. y = f (g(x)).
∆y
f (g(x + ∆x)) − f (g(x))
= lim
∆x→0 ∆x
∆x→0
∆x
y 0 = lim
➠ Put u = g(x) so that u + ∆u = g(x + ∆x):
f (u + ∆u) − f (u)
y = lim
∆x→0
∆x
0
➠ Note ∆u = g(x + ∆x) − g(x), and ∆x → 0 so will ∆u → 0.
➠ Multiply the difference quotient by 1 i.e. multiply by ∆u/∆u:
Banda–UKZN
39
MATH134 - Quantitative methods: Calculus.
Mathematical Sciences - 2006
Rule 8: The derivative of a function of a function (Chain Rule) - (contd.):
➠ Multiply the difference quotient by 1 (trick!!!) i.e. multiply by ∆u/∆u:
y
0
=
=
=
f (u + ∆u) − f (u) ∆u
lim
×
∆x→0
∆x
∆u
f (u + ∆u) − f (u) ∆u
lim
×
∆x→0
∆u
∆x
f (u + ∆u) − f (u) g(x + ∆x) − g(x)
lim
×
.
∆x→0
∆u
∆x
➠ Note as ∆x → 0 so does ∆u → 0.
y0
=
=
since y
Banda–UKZN
g(x + ∆x) − g(x)
f (u + ∆u) − f (u)
× lim
∆x→0
∆u→0
∆u
∆x
dy
du
×
.
du dx
lim
= f (u) and u = g(x).
40
MATH134 - Quantitative methods: Calculus.
Mathematical Sciences - 2006
➠ Example (Example 49B in Book): Find the derivatives of y = (x4 − 2x2 + 7)3 with respect
³ 1 + 4x ´4
with respect to x.
to x and y =
x2
Banda–UKZN
41
MATH134 - Quantitative methods: Calculus.
Mathematical Sciences - 2006
Economic applications of the derivative:
➠ Example (Example 51A in Book):
Total Cost function: C = Q3 − 6Q2 + 15Q. Find marginal and average cost functions.
➠ Example (Example 54A in Book):
Suppose the demand function of a product supplied by a monopolist is p
= 9 − 1.5Q.
➛ Find the total revenue and marginal revenue.
➛ Plot the three functions on one pair of axes.
Banda–UKZN
42
Mathematical Sciences - 2006
MATH134 - Quantitative methods: Calculus.
Economic applications of the derivative:
➠ Price Elasticity of demand (measuring how demand responds to price changes):
Price elasticity of demand is loosely defined to be:
η
=
−
percentage change in QD
percentage change in p
=
−
∆QD
QD × 100
∆p
p × 100
=
−
∆QD
p
×
.
∆p
QD
➠ More precisely:
η
=
lim −
∆p→0
p
p
∆QD
dQD
×
×
=−
.
∆p
QD
dp
QD
³
dp ´ p
or η = −1/
.
dQD QD
Banda–UKZN
43
MATH134 - Quantitative methods: Calculus.
Mathematical Sciences - 2006
Price inelastic:
if price changes substantially people continue to buy item since there is no substitute.
e.g. Bread.
Price elastic:
if price changes substantially people will buy less of the item.
e.g. CDs.
➛ Example (Example 57A in Book): Demand function: QD = 12/p. Find price elasticity of
demand.
➛ Example (Example 58B in Book): Demand function: p = 8 − 3QD . Find price elasticity of
demand.
Banda–UKZN
44
Mathematical Sciences - 2006
MATH134 - Quantitative methods: Calculus.
1.7 Welcome to the World of e
➠ Consider the expansion of (a + b)n . Can multiply out but tedious and prone to mistakes.
➠ Clever method: Binomial Theorem:
➠
n
(a + b) =

where 
k=0

n
k
=
n
X



n
k
 an−k bk
n!
and n! = n × (n − 1) × (n − 2) × · · · × 1
(n − k)!k!
➠ Building Block:
0! = 1; 1! = 1
➠ Examples: (a + b)2 = a2 + 2ab + b2 ;
(a + b)3 = a3 + 3a2 b + 3ab2 + b3
➠ Check out Pascal’s Triangle for small n.
Banda–UKZN
45
MATH134 - Quantitative methods: Calculus.
Mathematical Sciences - 2006
Welcome to the World of e contd....
➠ Sequences: Collection of numbers: a1 , a2 , a3 , . . . , an , . . . if the nth number has form:
an = f (n).
➠ Example:
➛ an = n2 , for n = 1, 2, 3, . . .
➛ an = (n − 1)!, for n = 1, 2, 3, . . .
➛ an = (−1)n (n − 1), for n = 1, 2, 3, . . .
➠ Psychologists use sequences to design IQ tests: the better you are at spotting obscure
relationships between numbers, the higher is your IQ supposed to be.
Banda–UKZN
46
Mathematical Sciences - 2006
MATH134 - Quantitative methods: Calculus.
Welcome to the World of e contd....
➠ e a limit of a special sequence: Consider a sequence an = (1 +
a1
a2
a3
a4
a5
a6
a7
2
2.25
2.370
2.411
2.488
2.522
2.546
···
a20
···
a100
···
a1000
···
2.653
2.704814
2.716923
1 n
) .
n
a8
a9
a10
2.566
2.581
2.594
a1,000,000
···
2.718260
➠ As n becomes bigger and bigger or as n tends to infinity, an = (1 + n1 )n tends to
2.718260 . . .
³
1 ´n
➠ Notation: lim an = lim 1 +
.
n→∞
n→∞
n
➠ Thus successive values of an stop increasing and converge to e.
³
1 ´n
➠ We write lim 1 +
= e = 2.718281828
n→∞
n
Banda–UKZN
47
Mathematical Sciences - 2006
MATH134 - Quantitative methods: Calculus.
Welcome to the World of e contd....
➠ Expansions of e and ex
a
➠ Note lim
=0
n→∞ n
³
´n
➠ What is 1 + 1/n ? Use Binomial Theorem for (a + b)n with a = 1 and b = 1/n:
➠
³
1 ´n
1+
n


= 1+




n
n
n
 1 +
 1 +
 1 + ···
n
n2
n3
1
2
3
1
n(n − 1)
1
n(n − 1)(n − 2) 1
= 1+n× +
× 2+
+ ···
n
2!
n
3!
n3
n n−1
1
n n−1 n−2
1
+ × ×
×
+ ···
= 1+1+ × ×
2! n
n
3! n
n
n
³
´
³
´³
´
1
1
1
1
2
= 1+1+
1−
+
1−
1−
+
2!
n
3!
n
n
³
´³
´³
´
1
1
2
3
1−
1−
1−
+ ···
4!
n
n
n
Banda–UKZN
48
Mathematical Sciences - 2006
MATH134 - Quantitative methods: Calculus.
Welcome to the World of e contd....
➠ Expansions of e and ex
➠ Consider n → ∞:
➠
³
1 ´n
1+
n
=
1³
1´ 1 ³
1 ´³
2´
1+1+
1−
+
1−
1−
+
2!
n
3!
n
n
³
´³
´³
´
1
1
2
3
1−
1−
1−
+ ···
4!
n
n
n
³
e =
=
=
Banda–UKZN
1 ´n
1
1
1
lim 1 +
= 1 + 1 + + + + ···
n→∞
n
2! 3! 4!
1
1
1
1
1
+ + + + + ···
0! 1! 2! 3! 4!
∞
X
1
m!
m=0
49
MATH134 - Quantitative methods: Calculus.
Mathematical Sciences - 2006
➠ Example (Example
³ 1Axin´nBook):
Show that lim 1 +
= ex .
n→∞
n
Put m = n/x, then clearly as n tends to infinity so does m.
Thus
³
³
³
x ´n
1 ´mx h
1 ´m i x
lim 1 +
= lim 1 +
= lim 1 +
= ex
n→∞
m→∞
m→∞
n
m
m
³ 1 ³ x − 2 ´2 ´
2
− 12 ( xx−2
x
2 −3 )
➠ e is written exp(x) or exp x e.g. exp −
=e
2
2 x −3
Banda–UKZN
50
Mathematical Sciences - 2006
MATH134 - Quantitative methods: Calculus.
Rule 9: The derivative of the logarithm function:
➠ If y = loge x, then
dy
1
= .
dx
x
➠ The derivation of this rule shows that base e is essential. That means for other bases one must
change the base to e using change of base rule on Page 14:
y = log10 x can be rewritten as:
y=
1
loge x
=
loge x
loge 10
loge 10
➠ Example (Example 6B in Book): Find the derivative of y = x2 log x; x3 / log x; S = log u.
Banda–UKZN
51
Mathematical Sciences - 2006
MATH134 - Quantitative methods: Calculus.
Rule 10: The derivative of the logarithm function:
g 0 (x)
dy
=
.
➠ If y = loge g(x), then
dx
g(x)
➠ Example (Example 9B in Book): Differentiate y = log(4x3 − 1);
y = log(2x)/(1 + 2x log x3 ); S = log u.
(2x4 + 1)2 (5x2 − 4x + 2)
➠ Example (Example 11A in Book): Find g (x) if g(x) =
(1 + 3x2 )3
0
➠ Example (Example 13B in Book): Differentiate y = ax where a is a positive constant.
➠ Example (Example 14B in Book): Differentiate y = xr where r is any nonzero real number.
Banda–UKZN
52
Mathematical Sciences - 2006
MATH134 - Quantitative methods: Calculus.
(2x4 + 1)2 (5x2 − 4x + 2)
➠ Example (Example 11A in Book): Find g (x) if g(x) =
(1 + 3x2 )3
0
➠ Use a trick: Take log to base e:
➠
(2x4 + 1)2 (5x2 − 4x + 2)
loge g(x) = loge
;
2
3
(1 + 3x )
= 2 loge (2x4 + 1) + loge (5x2 − 4x + 2) − 3 loge (1 + 3x2 ).
➠ Differentiate both sides to obtain:
➠
g 0 (x)
g(x)
Banda–UKZN
=
2 × 8x3
10x − 4
3 × 6x
+
−
;
2x4 + 1 5x2 − 4x + 2 1 + 3x2
=
10x − 4
18x
16x3
+
−
.
2x4 + 1 5x2 − 4x + 2 1 + 3x2
53
MATH134 - Quantitative methods: Calculus.
Mathematical Sciences - 2006
➠ To find g 0 (x) multiply through by g(x):
➠
g 0 (x) =
=
³ 16x3
10x − 4
18x ´
+
−
g(x);
2x4 + 1 5x2 − 4x + 2 1 + 3x2
³ 16x3
10x − 4
18x ´³ (2x4 + 1)2 (5x2 − 4x + 2) ´
+
−
.
2x4 + 1 5x2 − 4x + 2 1 + 3x2
(1 + 3x2 )3
➠ Similarly, you do example 13B and example 14B.
Banda–UKZN
54
MATH134 - Quantitative methods: Calculus.
Mathematical Sciences - 2006
Rule 11: The derivative of an exponential function:
dy
➠ If y = e , then
= ex .
dx
➠ ➛ Since y = ex then take logarithms: loge y = x loge e = x.
x
➛ y 0 /y = 1. Hence y 0 = y = ex
Banda–UKZN
55
MATH134 - Quantitative methods: Calculus.
Mathematical Sciences - 2006
Rule 12: The derivative of an exponential function:
➠ If y = eg(x) , then
dy
= g 0 (x)eg(x) .
dx
➠ ➛ Consider y = eg(x) = exp(g(x)) and take logarithms:
➛ log y = g(x) then differentiate: y 0 /y = g 0 (x).
➛ Hence y 0 = g 0 (x)y = g 0 (x) exp(g(x)).
➠ Example (Example 19B in Book): Find derivatives of y = e
Banda–UKZN
x4 +1
;y
= x2 e−x .
56