Differentiation

1
Chapter 11

• Differentiation

2
Slope of a Line

• Suppose that we have a straight line
• y mx b
• Earlier in the class, we found that the slope of
  the line can be found by making the calculation
• m (y2 y1)/(x2 x1)
• This is illustrated on the following slide

3
Graph of a Line
m (y2 y1)/(x2 x1)
y
Y mx b
y2
y1
x
x1
x2
4
Limits and the Slope

• Again, suppose that y f(x) mx b
• Suppose that we wish to find the limit as h ? 0
  of the difference quotient
• f(x h) f(x)/h
• m(x h) b mx b/h
• mh/h m

5
Interpretation

• Suppose that we choose to evaluate the slope of
  the line between two x-points, x and x h.
• Then, the values of y f(x) associated with
  these values of x are f(x) and f(xh)
• So, the slope is f(x h) f(x)/h
• When we take the limit as h ? 0, we are simply
  moving the two x-points ever closer together
  until they are one and the same.

6
The Graph Revisited
m (y2 y1)/(x2 x1)
y
m f(x h) f(x)/h
y2
y1
h (x2 x1)
x
x1
x2
7
Nonlinear Functions

• When we compute the slope of a linear function,
  it does not matter whether the x-points are close
  together or far apartthe slope is always the
  same.
• If the line is y mx b, then the slope is m
  everywhere.
• But this is not so with a nonlinear function.
  Consider y x2

8
Graph of y x2
y
The curve becomes steeper as X gets bigger
x
9
Slope of y x2

• We can find the slope of y f(x) x2 by taking
  the limit as h ? 0 of
• f(x h) f(x)/h
• So, we would want to look at
• (x h)2 x2/h (x2 2xh h2 x2)/h
• (2xh h2)/h 2x h
• This shows that the limit of the difference
  quotient as h ? 0 2x
• So, when x 1, the slope is 2. When x 2, the
  slope is 4. When x4, the slope is 8 etc., in
  other words the slope increases with x

10
Interpretation

• The slope of a line or a curve is interpreted as
  the change in y for a (small) change in x
• In calculus, the change in y for a (small) change
  in x is called the derivative denoted as either
  (more possible ways to denote the derivative, see
  p. 541 of text)
• dy/dx
• f (x)

11
Uses of Derivatives

• In business and economics, there are many
  instances when we want to be able to find a rate
  of change
• Marginal revenuethe change in total revenue from
  selling one more unit
• Marginal costthe change in total cost from
  producing one more unit
• Marginal productthe change in output from
  applying one more unit of a factor, such as labor

12
Finding Derivatives

• If we want to find the slope of a function (or
  the rate of change in y for a given change in x),
  we can use the difference quotient.
• But, a number of rules based on the difference
  quotient already have been established to speed
  things up.
• Constant function rule
• Constant factor rule
• Power function rule
• Sum difference rule

13
Constant Function Rule

• Suppose y f(x) c, where c is a constant.
  Then dy/dx 0
• Interpretation if y equals a constant, then
  there is no change in y when if x changes.
• Example let y denote fixed cost and let x denote
  output. Then, as output changes, fixed cost does
  not change the derivative of fixed cost with
  respect to output is zero.

14
Constant Factor Rule

• Suppose that y cf(x). Then dy/dx cf (x)
• Example Suppose that we have a linear demand
  curve p 30 0.5q
• dp/dq -0.5 is the slope of the demand slope
  (Note that the demand schedule is linear so it is
  in the form y mx b)

15
Constant Factor Rule (cont.)

• Now suppose that the demand curve shifts by a
  constant factor so that
• p 2(30 0.5q) 60 q
• dp/dq 2(-0.5) -1.0
• The constant factor (2) doubles the absolute
  value of the slope. The slope goes from 0.5 to
  1.0, so dy/dx goes from 0.5 to 1.0

16
Power Function Rule

• Suppose that y f(x) xn. Then dy/dx nxn-1
• Example Suppose that a total cost function can
  be expressed as C Q2
• Then dC/dQ 2Q
• Interpretation dC/dQ is the slope of the total
  cost curve (marginal cost) the change in total
  cost with respect to a small change in output
• In this situation, MC 2Q, MC increases linearly
  with Q.

17
Illustration
C Q2
C
MC
dC/dQ 2Q
Q
Q
18
Sum Difference Rule

• Suppose that f(x) and g(x) are differentiable
  functions
• Then d/dxf(x) g(x) f (x) g (x)
• And d/dxf(x) - g(x) f (x) - g (x)
• Example Suppose that demand for a product is
  given by p 30 q and suppose that total cost
  is given by C 20 10q 4q2. Find the
  addition to profit from making and selling one
  more unit of output

19
Solution

• If demand is given by p 30 q, then total
  revenue is pq 30q q2
• Profit is total revenue total cost or
• Profit ? 30q q2 (20 10q 4q2)
• d?/dq (30 2q) (10 8q)
• So, marginal profit is just marginal revenue
  minus marginal cost, or
• d?/dq 20 10q
• Suppose that we set d?/dq 20 10q 0
• In this case q 2

20
Interpretation
p
MC
28
Demand
q
MR
2
21
Another View
?
d?/dq
q
2
22
Homework

• Pp 551-552 1,3,7,17,75,79. Also, do a number
  of other odd numbered problems. Taking
  derivatives requires a lot of practice.