Economics 214
Lecture 19
Univariate Calculus
At the Margin


There is a direct correspondence between
the mathematical concept of the derivative
and the economic concept of considering
decisions made “at the margin,” the point
where the last unit is produced or
consumed.
This correspondence makes differentiation,
which is the evaluation of a derivative, one
of the core mathematical techniques in
economics.
Derivative Univariate Function
Recall that we difined the derivate of a univariate function
y  f(x) evaluated at x0 as
f x0  x   f x0 
dy
 lim
dx x0
x
Sum-Difference Rule
For any two functions f(x) and g(x)
d  f  x   g  x 
 f ' x   g ' x 
dx
Proof of Sum-Difference Rule
Let h( x)  f x   g x 
hx  x   hx 
x
x 0

f x0  x   g  x0  x    f  x0   g  x0 
 lim
x
x 0
f  x0  x   f  x0 
g  x0  x   g  x0 
 lim
 lim
x
x
x 0
x 0
 f ' x   g ' x 
h'  x   lim
Example Sum-Difference Rule


Let hx   3x 2  5 x  10  4  3x 
h' x   6 x  5   3  6 x  2
Note if we can rewrite h(x) as
h(x)  3x 2  2 x  6
Using our rule for derivate of quadratic function, we get
h'(x)  6 x-2
Scalar Rule
Let g(x)  k  f(x) where k is a real number. Then
g ' x   k  f ' x 
Proof :
k  f  x0  x   k  f  x0 


g ' x  lim
x
x 0
f  x0  x   f  x0 
 k  lim
 k  f ' x 
x
x 0
Example Scalar Rule

let g x   5  10 x  3x  2
2

g ' x   5  20 x  3  100 x  15
Rewrite g x  as
g x   50 x  15 x  10 then
2
g ' x   100 x  15
Product Rule
For f ( x)  g ( x)  h( x)
f ' ( x )  g ' ( x )  h( x )  h' ( x )  g ( x )
Proof :
f ( x  x)  f ( x)
f ' ( x)  lim
x
x 0
g  x  x h x  x   g  x h x 
 lim
x
x 0
Now add and substract h( x)  g  x  x  x
 g  x  x   g  x  
 hx  x   h( x) 


 lim h( x)  

g
x


x

 lim


x
x

x 0
x  0
 g ' ( x )  h( x )  h' ( x )  g ( x )
Example Product Rule
f ( x)  2 x  3  5 x  4 
f ' ( x)  2  5 x  4  5  2 x  3  10 x  8  10 x  15
 20 x  7
Note f ( x)  10 x 2  7 x  12
f ' ( x)  20 x  7
Extension of Product Rule
let f ( x)  g ( x)  h( x)  n( x)
also let z ( x)  h( x)  n( x) then
f ' ( x)  g ( x)  z ' x   z x   g ' ( x)
 g ( x)  h' ( x)  n( x)  h( x)  n' ( x)  h( x)  n( x) g ' ( x)
 g ' ( x )  h( x )  n( x )  h' ( x )  g ( x )  n( x )  n' ( x )  g ( x )  h( x )
Power Rule
Let
f ( x)  k  x n
then
f ' ( x)  n  k  x n 1
Proof : let g(x)  x n
g ' ( x)  lim
x  0
 lim
g ( x  x )  g ( x )
x
x  x n  x n
x
x  0
C0n x n  x   C1n x n 1  x  C2n x n  2  x     Cnn x 0 x   x n
 lim
x
x  0
0

2
 lim C1n x n 1  C2n x n  2  x     Cnn x 0 x 
x  0
 C1n x n 1  n  x n 1

f ' ( x)  k  g ' ( x)  n  k  x n 1
n 1
n

Examples of power Rule
Let f ( x)  k  k  x
f ' ( x)  n  k  x
n 1
0
1
 0k  x  0
k
Let f ( x)  n  k  x  n then
x
 nk
 n 1
f ' ( x)  n  k  x
 n 1
x
Examples of power Rule
Let f ( x)  6 x 5
then
f ' ( x)  5  6 x 51  30 x 4
Let f ( x)  4  3 x  4 x1 3 then
4 2 3
4
1 31
f ' ( x)  1 3  4  x
 x
 23
3
3x
Exponential Function Rule
Let f ( x)  e kx
then f ' ( x)  k  e kx
proof : We will furnish once we have
the chain rule.
Critical limit
lim
e
h
h 0

1
1
h
Proof
the above limit means that for very v ery small values of
e h  1 is approximat ely h
 e h is approximat ely h  1
 e is approximat ely 1  h 
1h
but we know
e  lim 1  h 
1h
h 0
Derivate of ex
Let f ( x)  e x
then


e x  x  e x
e x e x  1
f ' ( x)  lim
 lim
x
x
x 0
x 0
e
x
lim
x 0
e
 e x 1  e x
QED
x

1
By the previous slide we get
x
Growth Rate
Our continous growth model gave
X (t )  X (0)e rt
dX (t )
now
 r  X (0)e rt
dt
dX (t )
now
X (t )  r
dt
r is sometimes called the intantaneo us growth rate, since
it is the growth rate over a very small unit of time. Recall
proportial change
X t  X t 1 X X 1 X t



X t 1
X t 1
X t 1
X t 1
now as t  0, X t 1  X t hence the above ratio becomes
X t
X t
dX (t )




X (t )
t 0
dt
X t 1
Xt