Lecture # 8 - Derivatives of Functions of One Variable (cont.)
Continuity of a Function
Suppose we evaluate a limit lim g (v) = L.
v!N
If and only if N is in the domain of the function, we can get g(N ):
If N is in the domain AND lim g (v) = g (N ) ; then we say that the function g (:) is
v!N
continuous at N:
In other words, continuity needs at least three requirements:
1. N is in the domain of the function
2. The limit of the function at N , lim g (v) ; exists
v!N
3. lim g (v) = g (N )
v!N
Examples
– Step function violates (2)
– Hyperbole function violates (1)
Notes
– Any polynomial function is continuous.
– Rational functions are not continuous IF the denominator equals 0 for some value of
x
1
Di¤erentiabilty of a Function
y
x
Di¤erentiation means whether the limit lim
x!0
exists
Note that di¤erentiation implies continuity. In other words
1. Any function that is di¤erentiable is also continuous
2. Not all continuous functions are di¤erentiable
To prove (1):
– Suppose f (x) is di¤erentiable, i.e., f 0 (x) = lim
x!0
– Consider the di¤erence f (x +
x)
– We can rewirte as follows: f (x +
f (x+ x) f (x)
x
holds.
f (x)
x)
f (x)
f (x+ x) f (x)
x
– Let’s take the limits to both sides of the last expression as
x
x ! 0:
Left side:
lim [f (x +
x!0
x)
f (x)]
=
lim f (x +
x)
x!0
lim f (x)
=
x!0
lim f (x +
x)
x!0
f (x)
Right side:
lim
f (x +
x!0
x)
x
f (x)
x
=
lim
f (x +
x!0
– Then the expression can be rewritten as: lim f (x +
x!0
niently, lim f (x +
x!0
– lim f (x +
x!0
x)
x
x)
f (x)
lim
x!0
x
= f 0 (x) 0 = 0
f (x) = 0 or more conve-
x) = f (x)
x) = f (x) is just another way to write continuity (for the particular case
x ! 0)
– Hence, if a function is di¤erentiable, then it is also continuous.
For (2), let’s use a counterexample: Consider the function f (x) = jxj
– It is continuous.
x!0
f (0+ x) f (0)
x
j xj
x
= lim
– It is not di¤erentiable at x = 0, since the limit lim
lim j xj
does
x!0 x
= lim
x!0
j0+ xj j0j
x
=
not exist. Why?
Approach the limit from the left side:
Approach the limit from the left side:
2
lim
x!0
lim
x!0+
j xj
x
x!0
= lim
x!0+
x
x
x
x
= lim
x!0
1=
= lim 1 = 1
x!0+
1
Rules of Di¤erentiation
Derivative of constant function:
y = k =)
dy
dk
=
=0
dx
dx
Example 1 What is the derivative of the …xed cost when output changes?
Proof.
dy
y
f (x0 +
= lim
= lim
x!0 x
x!0
dx
x)
x
f (x0 )
= lim
x!0
Derivative of a power function:
y = xn =)
dy
= f 0 [x] = nxn
dx
1
– Rule valid for any value of n
Example 2 y = x0 =)
dy
dx
Example 3 y =
p
3
x =)
Example 4 y =
1
x
=)
= :::
dy
dx
dy
dx
= :::
= :::
Generalized power function rule
y = kxn =)
Example 5 y = 21 x2 =)
1
Example 6 y = 6 p
3 x =)
dy
dx
dy
dx
dy
= f 0 [x] = knxn
dx
= :::
= :::
3
1
k
k
=0
x
Derivative of a sum (di¤erence) of two functions:
d
[f (x)
dx
Example 7 C (Q) = Q3
g (x)] =
d
f (x)
dx
d
g (x) = f 0 (x) + g 0 (x)
dx
4Q2 + 10Q + 1200 =)
Example 8 y = (x + 2)2 = x2 + 4x + 4 =)
dy
dx
d
dQ C
(Q) = : : :
= :::
Derivative of a product of two functions:
d
d
d
[f (x) g (x)] =
f (x) g (x) + f (x)
g (x) = f 0 (x) g (x) + f (x) g 0 (x)
dx
dx
dx
Example 9 h (x) = 2x2
2x2
1
5x2 + 2 : Instead of multiplying …rst, you can de…ine f (x) =
1 and g (x) = 5x2 + 2
Example 10 De…ne total revenue T R = P (Q) Q; where P (Q) is the inverse demand.
Then marginal revenue is M R = P 0 (Q) Q+P (Q) : Notice that M R P (Q) = P 0 (Q) Q
0:
Derivative of a quotient of two functions:
f 0 (x) g (x) f (x) g 0 (x)
d f (x)
=
dx g (x)
[g (x)]2
Example 11 h (x) =
3x 5
x 2
Example 12 Consider the total cost function C (Q) : Then AC =
C(Q)
Q :
Further, the rate
of change of the AC with respect to Q is
d
AC =
dQ
=
=
=
In other words, if M C > AC, then
d C (Q)
dQ
Q
0
C (Q) Q C (Q)
Q2
1
C (Q)
C 0 (Q)
Q
Q
1
(M C AC)
Q
d
dQ AC
> 0: If M C < AC then
4
d
dQ AC
<0
Chain Rule
– Suppose we have that
y is a function of u;
in turn, u is a function of x
=) y is a composite function of x
– If x changes, then u changes and then y also changes =) chain reaction
x!
u!
y
– Chain rule:
dy
dy du
=
dx
du dx
Example 13 y =
x 5
x+2
Example 14 y = 5 1 +
1
2
p
: De…ne u =
x3 + 1
25
x 5
x+2
=)
(we are using the chain rule twice)
5
dy
dx
so that y = u
= 125 1 +
p
1
2
x3 + 1
24
1
2
x3 + 1
1
2
3x2