CHAPTER 5
Differentiability
2. The derivative
Definition 5.2.1. Let f(x) be a function on a domain D, and c be an interior
point of D (there is an open interval containing c, which is contained in D). If
lim
x→c
f(x)− f(c)
x− c = limh→0
f(c+ h)− f(c)
h
exists, then we say f(x) is differentiable at x = c and denote this limit f ′(c).
Definition 5.2.2. We say f(x) is differentiable on an open interval (a, b) if f ′(c)
exists for every c ∈ (a, b).
Definition 5.2.3. We say f ∈ C1[a, b] if f(x) is differentiable on (a, b), both
f ′(a) = lim
x→a+
f(x)− f(a)
x− a and f
′(b) = lim
x→b−
f(x)− f(b)
x− b
exist and f ′(x) is a continuous function on [a, b].
Example 5.2.4. If γ(t) is the position of a particle at time t, then γ′(t) is the
velocity. If γ is not C1[0, T ], then the velocity is not continuous, which implies the
acceleration is infinite. This implies that there is an infinite amount of force acting
on the particle. This is not really realistic, and so we often require functions to be
C1, not simply differentiable when modeling physical or biological phenomenon.
1
2 5. DIFFERENTIABILITY
Can you write the definition of the derivative using the -δ definition of the limit?
∀ > 0, ∃δ > 0 so that 0 < |x− c| < δ implies that∣∣∣∣f(x)− f(c)x− c − f ′(c)
∣∣∣∣ < .
Example 5.2.5. Is the function
f(x) =
{
x sin( 1
x
), x 6= 0
0, x = 0,
continuous at 0? Differentiable at 0?
Yes, it is continuous at 0. If we fix > 0, and set δ = , then for |x| < δ, we have
|x sin(1/x)| < |x| · 1 < .
It is not differentiable at zero because limx→0
f(x)−f(0)
x−0 = limx→0
x sin(1/x)
x
= limx→0 sin(1/x)
does not exist. (We discussed this as Example 1.11 from Chapter 4.)
Example 5.2.6. Is the function
f(x) =
{
x2 sin( 1
x
), x 6= 0
0, x = 0,
continuous at 0? Differentiable at 0?
Yes, it is continuous at zero. It is also differentiable at zero because f ′(0) =
limx→0
x2 sin(1/x)
x
= limx→0 x sin(1/x) = 0.
2. THE DERIVATIVE 3
Question: Can you find a function which is n times differentiable, but not n + 1
times differentiable, for any n ∈ N?
The function f(x) = xn+1 sin(1/x) is n times differentiable, but not n + 1 times
differentiable.
Lemma 5.2.7. If f(x) is differentiable at c, then f(x) is continuous at c.
Proof: Assume that limx→c
f(x)−f(c)
x−c = L. Then we have
lim
x→c
f(x)− f(c) =
(
lim
x→c
f(x)− f(c)
x− c
)
·
(
lim
x→c
x− c
)
= L · 0 = 0.
Theorem 5.2.8 (Algebraic Differentiability Theorem). Let f and g be functions
defined on an interval A and assume both are differentiable at some point c ∈ A.
Then
(1) (f + g)′(c) = f ′(c) + g′(c)
(2) (kf)′(c) = kf ′(c) for all k ∈ R
(3) (fg)′(c) = f ′(c)g(c) + f(c)g′(c), and
(4) (f/g)′(c) = g(c)f
′(c)−f(c)g′(c)
[g(c)]2
, provided that g(c) 6= 0.
Proof of 3: We have
(fg)′(c) = lim
x→c
f(x)g(x)− f(c)g(c)
x− c = limx→c
f(x)g(x)− f(x)g(c) + f(x)g(c)− f(c)g(c)
x− c
= lim
x→c
f(x)
g(x)− g(c)
x− c + g(c)
f(x)− f(c)
x− c
= (lim
x→c
f(x)) ·
(
lim
x→c
g(x)− g(c)
x− c
)
+ g(c)
(
lim
x→c
f(x)− f(c)
x− c
)
= f(c)g′(c) + g(c)f ′(c) = f ′(c)g(c) + f(c)g′(c).
4 5. DIFFERENTIABILITY
Theorem 5.2.9 (Chain Rule). Let f : A → R and g : B → R satisfy f(A) ⊆ B
so that the composition g ◦ f : A → R is defined. If f is differentiable at c ∈ A
and g is differentiable at f(c) ∈ B, then g ◦ f is differentiable at c with (g ◦ f)′(c) =
g′(f(c)) · f ′(c).
Something that almost works but doesn’t quite:
lim
x→c
g(f(x))− g(f(c))
x− c = limx→c
g(f(x))− g(f(c))
f(x)− f(c) ·
f(x)− f(c)
x− c .
Set y = f(x). As x→ c, f(x)→ f(c). We get
lim
y→c
g(y)− g(f(c))
y − f(c) · limx→c
f(x)− f(c)
x− c = g
′(f(c)) · f ′(c).
Issue: We can’t divide by f(x)− f(c) if f(x) = f(c). And this might happen for
values of x with x 6= c.
Tweak: Define
d(y) =
{
g(y)−g(f(c))
y−f(c) if y 6= f(c)
g′(f(c)) if y = f(c).
Then d(y) is continuous on B and then we have
lim
x→c
g(f(x))− g(f(c))
x− c = limx→c d(f(x)) ·
f(x)− f(c)
x− c
=
(
lim
x→c
d(f(x))
)
·
(
lim
x→c
f(x)− f(c)
x− c
)
= d(f(c))f ′(c) = g′(f(c))f ′(c).
2. THE DERIVATIVE 5
Theorem 5.2.10 (How to solve max/min problems). Let f be differentiable on
(a, b). If f attains a maximum value at some point c ∈ (a, b), then f ′(c) = 0. The
same is true if f(c) is a minimum value.
Assume that f has a maximum value at c and let f(c) = M . Let xn be an
increasing sequence of real numbers that converges to c. We then have that
f ′(c) = lim
n→∞
f(xn)− f(c)
xn − c .
Since xn ≤ c for all n, xn − c ≤ 0. Also, since f(c) is a maximum, f(xn)− f(c) ≤ 0.
Thus, f(xn)−f(c)
xn−c ≥ 0 for all n and by the order limit theorem, f ′(c) ≥ 0.
Now, let yn be a decreasing sequence of real numbers that converges to c. We
have
f ′(c) = lim
n→∞
f(yn)− f(c)
yn − c .
Now, yn ≥ c for all n and f(yn) − f(c) ≤ 0 (since f(c) is a maximum). Thus,
f(yn)−f(c)
yn−c ≤ 0 and so f ′(c) ≤ 0 by the order limit theorem.
6 5. DIFFERENTIABILITY
3. Mean value theorems
Theorem 5.3.1 (Rolle’s Theorem). If f(x) is continuous on [a, b], differentiable
on (a, b) and f(b) = f(a), then there exists a c ∈ (a, b) such that f ′(c) = 0.
Since f is continuous on [a, b], there are x1, x2 ∈ [a, b] so that f(x1) = min{f(x) :
x ∈ [a, b]} and f(x2) = max{f(x) : x ∈ [a, b]}. If x1 ∈ (a, b) then by the previous
theorem f ′(x1) = 0. Similarly if x2 ∈ (a, b) then f ′(x2) = 0.
So suppose that x1, x2 ∈ {a, b}. Since f(a) = f(b), this implies that the min and
the max values of f on [a, b] are the same. So f(x) = f(a) for all x ∈ [a, b]. It is easy
to verify that f ′(c) = 0 for all c ∈ (a, b).
Theorem 5.3.2 (Mean Value Theorem). If f(x) is continuous on [a, b] and dif-
ferentiable on (a, b) then there exists some c ∈ (a, b) such that
f ′(c) =
f(b)− f(a)
b− a .
Let g(x) = f(x)−
[(
f(b)−f(a)
b−a
)
(x− a) + f(a)
]
. Then g(a) = f(a)−
[(
f(b)−f(a)
b−a
)
· 0 + f(a)
]
=
0. Also, g(b) = f(b) − [f(b)− f(a) + f(a)] = 0. Thus, by Rolle’s theorem, there
is some c ∈ (a, b) so that g′(c) = 0. But g′(x) = f ′(x) − f(b)−f(a)
b−a . So we have
f ′(c) = f(b)−f(a)
b−a .
3. MEAN VALUE THEOREMS 7
Theorem 5.3.3 (Generalized Mean Value Theorem). If f and g are continuous
on [a, b] and differentiable on (a, b), then there exists a point c ∈ (a, b) where
[f(b)− f(a)]g′(c) = [g(b)− g(a)]f ′(c).
If g′ is never zero on (a, b), then we can state the conclusion as
f ′(c)
g′(c)
=
f(b)− f(a)
g(b)− g(a) .
Apply the mean value theorem to h(x) = [f(b)− f(a)]g(x)− [g(b)− g(a)]f(x).