Basics of Differential Calculus
Professor Peter Cramton
Economics 300
Why differential calculus?
• Models explain economic behavior with system of
equations
• What happens if a variable changes?
– Comparative statics determines marginal change in
economic behavior
• How does change in tax rate alter consumption?
• How does change in NBA collective bargaining agreement
impact
– share of NBA revenues going to players?
– parity of teams across league?
Why differential calculus?
• Economic models assume rational optimizers
– Consumers maximize utility
– Producers maximize profits
– NBA owners maximize combination of wins and profits
• Optimization uses calculus to evaluate tradeoffs
– How much to consume?
• Consume until marginal utility = price
– How much to produce?
• Produce until marginal revenue = marginal cost
– Which free agents to go for?
Average rate of change over [x0,x1]
y f ( x1 )  f ( x0 )

, where
x
x1  x 0
x  x1  x0 and y  y1  y0
Average rate of change examples
y f ( x1 )  f ( x0 )

x
x1  x0
y bx1  bx0
y  a  bx :

b
x
x1  x0
2
2
x

x
( x1  x0 )( x1  x0 )

y
2
1
0
yx :


 x1  x0
x
x1  x0
x1  x0
yx
2
y
6
x
y
3
x
Average rate of change
and difference quotient
y f ( x1 )  f ( x) f ( x  x)  f ( x)


x
x1  x
x
y b( x  x)  bx
y  a  bx :

b
x
x
2
2
2

y
(
x


x
)

x
2
x

x


x
y  x2 :


 2 x  x
x
x
x
Some properties
• Rate of change of sum = sum of rates of change
– y, w, z are functions of x and y = w + z
– Then y
( w  z ) w
• Scaling:
z



x
x
x x
(ay )
y
a
x
x
Application: quadratic
( x 2 ) ( x  x) 2  x 2 2 xx  x 2


 2 x  x
x
x
x
y  ax  bx  c
2
y
( x )
x
1
a
b
c
x
x
x
x
 a (2 x  x)  b
2
Application: cubic
( x ) ( x  x)( x  2 xx  x )  x

x
x
3
2
2
3
3
( x  3 x x  3 xx  x )  x

x
2
2
 3 x  3 xx  x
3
2
2
3
y  gx  ax  bx  c
3
2
y
( x )
( x )
x
1
g
a
b
c
x
x
x
x
x
2
2
 g (3 x  3 xx  x )  a (2 x  x)  b
3
2
Exercise
• Find difference quotient for each function
y  5x
y  30  15 x
y
5
x
y
 15
x
y  6 x  2 x  9 y
 6(2 x  x)  2
x
2
y 1 x
2
y
 (2 x  x)
x
Exercise
• Total revenue: TR = P Q
• Price: P = 10  .5Q
• Difference quotient?
TR  (10  .5Q)Q  .5Q 2  10Q
TR
 .5(2Q  Q)  10
Q
• If Q = 5, what is impact of 1 unit increase in Q?
TR
 .5(2(5)  1)  10  4.5
Q
Derivative is difference quotient as x0
dy
f ( x  x)  f ( x)
 lim
dx x0
x
dy
b( x  x)  bx
y  a  bx :
 lim
b
dx x0
x
dy
( x  x)  x
yx :
 lim
 lim 2 x  x  2 x
x 0
dx x0
x
2
2
2
dy
2
2
2
y  x :  lim 3x  3xx  x  3x
dx x0
3
Some properties
• Derivative of sum = sum of derivatives
– y, w, z are functions of x and y = w + z
– Then
dy d ( w  z ) dw dz
dx
• Scaling:
• Application


dx
dx
d (ay )
dy
a
dx
dx

dx
y  gx  ax  bx  c
3
2
3
2
dy
d (x )
d (x )
dx
d1
g
a
b c
dx
dx
dx
dx
dx
2
 3 gx  2ax  b
Derivative is difference quotient as x0
average rate of change  difference quotient  derivative
y f ( x1 )  f ( x) f ( x  x)  f ( x)


x
x1  x
x
dy
f ( x  x)  f ( x)
 lim
dx x 0
x
Derivative is rate of change as x0
Derivative is instantaneous rate of change
Tangent line is limit of secant line
Derivative is slope of tangent line
Total tax revenue and marginal tax revenue
Exercise
• Cigarette tax yields revenue R(t) = 50 + 25t – 75t2
• What is marginal revenue?
dR
MR 
 25  75(2t )  25  150t
dt
• What tax rate maximizes revenues?
MR  25  150t  0
t  25 / 150  1/ 6
*
• Why is this a maximum?
concave
Functions not everywhere differentiable
Differentiable  Continuous
Demand and cost functions
Average vs. marginal
Difference quotient of a polynomial
(1)
( x)
0
1
x
x
2
( x )
 2 x  x
x
3
( x )
2
2
 3x  3xx  (x)
x
( x )
3
2
2
3
 4 x  6 x x  4 x(x)  (x)
x
4
Exercise: y = 4x2 – 8x + 3
• Find roots (x such that y = 0).
b  b  4ac 8  8  4(4)(3)
y

2a
2(4)
2
•
•
2
2 43

 1  .5
2
Derivative
dy
 8x  8
dx
Extreme value dy
*
 8x  8  0  x  1
dx
Exercise: y = 4x2 – 8x + 3
3
2
1
0.5
1
1.0
1.5
2.0
Exercise: y = 4x2 – 8x + 3
y’ = 8x - 8
5
y
0.5
5
1.0
1.5
2.0
Differential vs. derivative
• Derivative is rate of change as x  0
y
y dy
 lim

 f '( x)
x 0 x
x
dx
• Differential is change in y along tangent line
dy  f '( x)dx
Differential approximation and actual change
dy  f '( x0 )dx
Exercise: y = 16 – 4x + x3
y
2
2
• What is rate of change?
 4  3 x  3 xx  x
x
• What is derivative?
dy
2
 4  3x
dx
• What is differential?
dy  (3x  4)dx
2
2

y

(3(2
)  4)(8)  64
• Let x0 = 2; x = 8
y  (3(22 )  4  3(2)(8)  82 )(8)  960
• Let x0 = 2; x = .2 y  (3(22 )  4)(.2)  1.6
y  (3(22 )  4  3(2)(.2)  .22 )(.2)  1.848
Exercise
investment (I) and cost of borrowing (r)
2
I  f (r )  600  150r  400r
Compute: I  f (r )  r
r0  2%; r1  .5%; r2  1%
I1  (800r0  150) r1  (800(.02)  150)(.005)  .67
I 2  (800r0  150) r2  (800(.02)  150)(.01)  1.34
I 0  600  150(.02)  400(.02 )  597.16
2
I1  600  150(.025)  400(.025 )  596.5
2
I 2  600  150(.03)  400(.03 )  595.86
2
I1  596.5  597.16  .66
I 2  595.86  597.16  1.3