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Math 341 – Lecture Notes on Chapter 5 – The Derivative
Lecture 26
§5.2: Derivative and the Intermediate Value Property
Definition of the Derivative
Let g : A → R be a function defined on an interval A. Given c ∈ A, the
derivative of g at c is defined by
g′(c) = limx→c
g(x)− g(c)
x− c,
provided this limit exists.
An interpretation of g′(c) as a tangent line to the curve y = g(x) is depicted in the
figure below:
Note: Alternative Definitions of g′(c)
In Exercise 5.2.6, you verify that
g′(c) = limt→c
g(t)− g(c)
t− c= lim
h→0
g(c+ h)− g(c)
h= lim
h→0
g(c+ h)− g(c− h)
2h.
1
Lipschitz Functions
Let A be an interval. A function f : A → R is said to be Lipschitz if there
exists a positive constant K such that
|f(x)− f(y)| < K|x− y| for all x, y ∈ A.
This implies that the difference quotients are bounded:
f(x)− f(y)
x− y< K,
which comes close to differentiability. In fact, a Lipschitz function (although
not necessarily differentiable) is uniformly continuous on A. If 0 < c < 1 and
|g(x)− g(y)| < c|x− y| for all x, y ∈ R,
then g is called a contraction mapping. In this case (see Exercise 4.3.11),
there will exist some x0 ∈ A such that g(x0) = x0. The point x0 is called a
fixed point of g.
Examples
(a) f(x) = xn where n ∈ N. Recall that
xn − cn = (x− c)(xn−1 + c1xn−2 + c2xn−3 + · · ·+ cn−2x1 + cn−1
).
So,
f ′(c) = limx→c
xn − cn
x− c= lim
x→c
(xn−1+c1xn−2+c2xn−3+· · ·+cn−2x1+cn−1
)= ncn−1
(b) f(x) = |x|. Then
f ′(c) =
1 if c > 0,
−1 if c < 0,
DNE if c = 0.
So, continuity at a point does NOT imply differentiability.
Theorem (differentiability implies continuity)
If g : A→ R is differentiable at c ∈ A, then g is continuous at c.
2
Proof. By hypothesis,
f ′(c) = limx→c
g(x)− g(c)
x− cexits. Also, we know that limx→c(x− c) = 0. By the Algebraic Limit Theorem,
limx→c
(f(x)− f(c)
)= lim
x→c
[f(x)− f(c)
x− c· (x− c)
]ALT= lim
x→c
g(x)− g(c)
x− c· limx→c
(x− c)
= f ′(c) · 0 = 0.
So, limx→c f(x) = f(c), which implies that f is continuous at c.
Algebraic Differentiability Theorem
Let f and g be functions defined on an interval A. Assume f and g are differ-
entiable at c ∈ A. Then
1. (f + g)′(c) = f ′(c) + g′(c).
2. (kf)′(c) = kf ′(c) for all k ∈ R.
3. (Product Rule) (fg)′(c) = f ′(c)g(c) + f(c)g′(c).
4. (Quotient Rule)(fg
)′(c) =
f ′(c)g(c)− f(c)g′(c)
[g(c)]2.
Proof. The proofs are all fairly basic. As an illustration we’ll only go through the
proof of the product rule.
(fg)(x)− (fg)(c)
x− c=f(x)g(x)− f(c)g(c)
x− c
=f(x)g(x)− f(c)g(x) + f(c)g(x)− f(c)g(c)
x− c
=(f(x)− f(c)
x− c
)g(x) + f(c)
(g(x)− g(c)
x− c
)Taking the limit as x→ c and using the continuity of g give
(fg)′(c) = f ′(c)g(c) + f(c)g′(c).
3
Chain Rule
Let f : A → R and g : B → R satisfy f(A) ⊆ B so that g ◦ f : A → R. If f is
differentiable at c and g is differentiable at f(c), then g ◦ f is differentiable at
c and
(g ◦ f)′(c) = g′(f(c)) f ′(c).
Proof. Since g is differentiable at f(c),
g′(f(c)) = limy→f(c)
g(y)− g(f(c))
y − f(c).
Define the function d : B → R by
d(y) =
{g(y)−g(f(c))y−f(c) if y ∈ B and y 6= f(c),
g′(f(c)) if y = f(c).(∗)
Then
limy→f(c)
d(y) = g′(f(c))
and so d is continuous at f(c). Now rewrite equation (∗) as
g(y)− g(f(c)) = d(y)[y − f(c)].
The last equation is true for all y. Substitute f(t) for y and divide by t − c for
t 6= c to get
g(f(t))− g(f(c))
t− c= d(f(t))
(f(t)− f(c)
t− c
)(t 6= c).
Apply the Algebraic Limit Theorem to take the limit as t→ c to get
(g ◦ f)′(c) = g′(f(c)) f ′(c).
4
Lecture 27
§5.2: Derivative and the Intermediate Value Property (Con-
tinued)
Let’s look at another proof that differentiability implies continuity. This time we’ll
use the ε-δ definition directly without using the Algebraic Limit Theorem.
Theorem (Differentiability Implies Continuity)
Let g : A→ R be differentiable at c ∈ A, where A is an interval. Then g is also
continuous at c.
Proof. Let ε > 0 be given. We wish to show that there exists δ > 0 such that∣∣f(x)− f(c)∣∣ < ε
whenever |x− c| < δ and x ∈ A.
Since g′(c) = limx→cf(x)−f(c)
x−c , there exists α > 0 such that if 0 < |x − c| < α and
x ∈ A, then ∣∣∣f(x)− f(c)
x− c− f ′(c)
∣∣∣ < ε,
which implies
f ′(c)− ε < f(x)− f(c)
x− c< f ′(c) + ε
and so ∣∣f(x)− f(c)∣∣ ≤ (|f ′(c)|+ ε
)|x− c|. (∗)
Set δ = min(α, ε|f ′(c)|+ε
). Then if |x− c| < δ and x ∈ A, we have∣∣f(x)− f(c)
∣∣ ≤ (|f ′(c)|+ ε)|x− c| by inequality (∗) since |x− c| < δ ≤ α
<(|f ′(c)|+ ε
)( ε
|f ′(c)|+ ε
)since |x− c| < δ ≤ ε
|f ′(c)|+ε
≤ ε.
This proves that g is continuous at c.
5
Example (Derivative exists everywhere but is not continuous every-
where)
Consider the function
f(x) =
{x2 sin(1/x) if x 6= 0,
0 if x = 0.
If x 6= 0,
f ′(x) = 2x sin(1/x) + x2 cos(1/x)(−1/x2) = 2x sin(1/x)− cos(1/x).
If x = 0,
f ′(0) = limt→0
f(t)− f(0)
t− 0= lim
t→0
t2 sin(1/t)− 0
t− 0= lim
t→0t sin(1/t) = 0.
Thus, f is everywhere differentiable and
f ′(x) =
{2x sin(1/x)− cos(1/x) if x 6= 0,
0 if x = 0.
Note that f ′(x) exists at all x ∈ R, however f ′(x) is not everywhere continuous
since limx→0 f′(x) does not exist, but f ′(0) does exists. So f ′(x) is discontinuous
at x = 0. To verify the nonexistence of the limit, we can use the two sequences
(xn) =( 1
2nπ
)→ 0 f ′(xn) = −1,
(yn) =( 1
(2n+ 1)π
)→ 0 f ′(yn) = 1.
Graphs, created with Mathematica, of the functions in this example are included
on the following two pages.
6
Derivative Example f (x) = x2 sin(1 /x).In[1]:= f[x_] := x2 * Sin
1
x
For large x, the curve y = x2 sin(1 /x) asymptotically approaches the line y = x.
In[2]:= Plot[{f[x], x}, {x, -2, 2}]
Out[2]=-2 -1 1 2
-2
-1
1
2
For small x, the curve y = x2 sin(1 /x) oscillates between the curves y = -x2 and y = x2.
In[3]:= Plotx2 * Sin1
x, x2, -x2, x,
-1
4,1
4
Out[3]=-0.2 -0.1 0.1 0.2
-0.06
-0.04
-0.02
0.02
0.04
0.06
In[4]:= D[f[x], x]
Out[4]= -Cos1
x + 2 x Sin
1
x
For large x, the derivative curve y = f ' (x) asymptotically approaches the line y = 1.
In[5]:= Plot-Cos1
x + 2 x Sin
1
x, 1, {x, -5, 5}, PlotRange → All, PlotPoints → 200
Out[5]=
-4 -2 2 4
-1.0
-0.5
0.5
1.0
For small x, the curve y = f ' (x) oscillates approximately between the lines y = -1 and y = 1.
In[6]:= Plot-Cos1
x + 2 x Sin
1
x, -1, 1, {x, -1, 1}, PlotRange → All, PlotPoints → 200
Out[6]=
-1.0 -0.5 0.5 1.0
-1.0
-0.5
0.5
1.0
2 Graph of derivative example in Section 5.2.nb
Interior Extremum Theorem
Suppose f is differentiable on an open interval (a, b) and attains a maximum
(or a minimum) at c ∈ (a, b). Then f ′(c) = 0.
Proof. Suppose f attains a maximum at c ∈ (a, b). Let (xn) and (yn) be sequence
in (a, b) with
a < xn < c < yn < b
and lim xn = lim yn = c. Then
f ′(c) = limn→∞
f(xn)− f(c)
xn − c≥ 0 since f(xn)− f(c) ≤ 0 and xn − c < 0,
and
f ′(c) = limn→∞
f(yn)− f(c)
yn − c≤ 0 since f(yn)− f(c) ≤ 0 and yn − c > 0.
So, f ′(c) = 0.
The proof in the case when f attains a minimum at c ∈ (a, b) is entirely similar.
Next we will see that there is an intermediate value property for derivatives.
Darboux’s Theorem
Assume f is differentiable on a closed interval [a, b] and f ′(a) < α < f ′(b) [or
f ′(a) > α > f ′(b)], then there exists c ∈ (a, b) such that f ′(c) = α.
Note: As we saw in an earlier example, the derivative can exist everywhere on
[a, b] yet not be continuous everywhere on [a, b]. So, the proof cannot assume that
f ′ is continuous. However, the proof will use both the differentiability of f and the
continuity of f .
Proof. Assume f ′(a) < α < f ′(b). Consider the new function
g(x) := f(x)− αx.
Then
g′(x) = f ′(x)− α
9
and
g′(a) < 0 < g′(b).
For the positive number ε0 = |g′(a)|2 , there exists a δ > 0 (with 0 < δ < b− a) such
that if a < x < a+ δ, then
g(x)− g(a)
x− a< g′(a) + ε0 = g′(a) +
|g′(a)|2
=g′(a)
2< 0,
which implies
g(x)− g(a) <g′(a)
2
(x− a) < 0.
Thus there exists at least one point x ∈ (a, b) where
g(a) > g(x).
By a similar argument, there exists at least one y ∈ (a, b) such that
g(y) < g(b).
The continuous function g : [a, b]→ R has a minimum by the Extreme Value The-
orem, but this minimum is not at x = a or x = b. Thus there exists c ∈ (a, b) such
that g has a minimum at c. But then
g′(c) = f ′(c)− α = 0
So f ′(c) = α.
The proof in the case f ′(a) > α > f ′(b) is entirely similar.
10
Lecture 28
§5.3 Mean Value Theorems
The most important version of the Mean Value Theorem is the following:
Theorem (Mean Value Theorem)
If f : [a, b] → R is continuous on [a, b] and differentiable on (a, b), then there
exists a point c ∈ (a, b) where
f ′(c) =f(b)− f(a)
b− a.
Here is a picture depicting the theorem:
To prove the Mean Value Theorem, we need a lemma:
Lemma (Rolle’s Theorem)
Let f : [a, b] → R be continuous on [a, b] and differentiable on (a, b). If f(a) =
f(b), then there exists a point c ∈ (a, b) where f ′(c) = 0.
11
Proof. Since f is continuous on [a, b], which is compact, f attains a maximum and
a minumum on [a, b]. If the maximum and minimum occur at the endpoints, then
f is a constant and f ′(x) = 0 for all x ∈ (a, b). In this case we may choose any
c ∈ (a, b) and then f ′(c) = 0.
If either the maximum or minimum occurs at an interior point c ∈ (a, b), then by
the Interior Extremum Theorem, f ′(c) = 0.
Now we’ll prove the Mean Value Theorem. The following figure will be helpful for
understanding the proof.
Proof of Mean Value Theorem. The equation of the line through the points (a, f(a))
and (b, f(b)) is
y =
(f(b)− f(a)
b− a
)(x− a) + f(a).
12
Consider the difference between this line and the curve y = f(x). Set
d(x) := f(x)−[(
f(b)− f(a)
b− a
)(x− a) + f(a)
].
Then d is continuous on [a, b], differentiable on (a, b), and d(a) = d(b). By Rolle’s
Theorem, there exists c ∈ (a, b) where d′(c) = 0. Since
d′(x) = f ′(x)−(f(b)− f(a)
b− a
),
we get
0 = f ′(c)−(f(b)− f(a)
b− a
).
Corollary
If g : A → R is differentiable on an interval A and satisfies g′(x) = 0 for all
x ∈ A, then g(x) = k for some constant k ∈ R.
Proof. Let x, y ∈ A where x < y. By the Mean Value Theorem applied to g on the
interval [x, y], there exists c ∈ (x, y) such that
0 = g′(c) =f(y)− f(x)
y − x
which implies g(x) = g(y). Set k to this common value. Since x and y were
arbitrary, g(x) = k for all x ∈ A.
Corollary
Suppose f and g are differentiable functions on an interval A and f ′(x) = g′(x)
for all x ∈ A. Then f(x) = g(x) + k for some constant k.
Proof. Let h(x) = f(x) − g(x). Then h′(x) = f ′(x) − g′(x) = 0 for all x ∈ A. By
the previous corollary, h(x) = k for some constant k. So, f(x) = g(x) + k for some
constant k ∈ R and all x ∈ A.
13
Lecture 29
First, recall the Mean Value Theorem that we studied last time:
Theorem (Mean Value Theorem)
If f : [a, b] → R is continuous on [a, b] and differentiable on (a, b), then there
exists a point c ∈ (a, b) where
f ′(c) =f(b)− f(a)
b− a.
Let’s study an example.
Example
Recall that a function f : A → R is Lipschitz on A if there exists an M > 0
such that ∣∣∣∣f(x)− f(y)
x− y
∣∣∣∣ ≤M
for all x, y ∈ A with x 6= y.
Assume f is twice-differentiable on [a, b]. Thus f ′ is differentiable on [a, b] and
so f ′ is also continuous on the compact set [a, b]. Hence f ′ is bounded on [a, b]
and so there exists some M > 0 such that |f ′(x)| ≤M for all x ∈ [a, b].
Let x, y ∈ [a, b] where x < y. By the Mean Value Theorem, there exists
c ∈ (a, b) such thatf(x)− f(y)
x− y= f ′(c).
Then ∣∣∣∣f(x)− f(y)
x− y
∣∣∣∣ = |f ′(c)| ≤M.
Since x and y were arbitrary, f is Lipschitz on [a, b]. Thus, we have shown that
twice-differentiable functions defined on a finite closed interval are Lipschitz on
that interval.
Now consider a pair of function f : A→ R and g : A→ R that are both continuous
on [a, b] and differentiable on (a, b). By the Mean Value Theorem, there exist
14
c1, c2 ∈ (a, b) such that
f ′(c1) =f(b)− f(a)
b− aand g′(c2) =
g(b)− g(a)
b− a.
If g(a) 6= g(b), we can divide to obtain
f ′(c1)
g′(c2)=f(b)− f(a)
g(b)− g(a).
In fact, a stronger statement is true. It turns out to be possible to choose c1 = c2in the last equation (but not in the earlier equations).
Generalized Mean Value Theorem
If f and g are continuous on the closed interval [a, b] and differentiable on the
open interval (a, b), then there exists a point c ∈ (a, b) such that(f(b)− f(a)
)g′(c) =
(g(b)− g(a)
)f ′(c).
If g′ is never zero on (a, b), this can be stated as
f ′(c)
g′(c)=f(b)− f(a)
g(b)− g(a).
Proof. Consider the function
h(x) = [f(b)− f(a)]g(x)− [g(b)− g(a)]f(x).
Then h is continuous on [a, b] and differentiable on (a, b) and
h(a) = f(b)g(a)− f(a)g(b) = h(b).
By Rolle’s Theorem, there exists c ∈ (a, b) such that
0 = h′(c) = [f(b)− f(a)]g′(c)− [g(b)− g(a)]f ′(c)
and so
[f(b)− f(a)]g′(c) = [g(b)− g(a)]f ′(c).
15
L’Hospital’s Rule : 0/0 Case
Let f and g be continuous on an interval containing a. Assume f and g are
differentiable on this initerval with the possible exception of the point a. If
f(a) = g(a) = 0 and g′(x) 6= 0 for all x 6= a, then
limx→a
f ′(x)
g′(x)= L implies lim
x→a
f(x)
g(x)= L.
Proof. (The proof of this version of L’Hospital’s book is an exercise in the textbook.)
Assume
limx→a
f ′(x)
g′(x)= L.
For every ε > 0 there exists δ such that if 0 < |x− a| < δ, then∣∣∣∣f ′(x)
g′(x)− L
∣∣∣∣ < ε.
Now for any particular x satisfying 0 < |x−a| < δ, by the Generalized Mean Value
Theorem, there exists c between a and x such that
f ′(c)
g′(c)=f(x)− f(a)
g(x)− g(a)=f(x)
g(x).
But then∣∣∣∣f(x)
g(x)− L
∣∣∣∣ =
∣∣∣∣f ′(c)g′(c)− L
∣∣∣∣ < ε (since 0 < |c− a| < |x− a| < δ.)
So,
limx→a
f(x)
g(x)= L.
Examples
limx→0
sinx
x= lim
x→0
cosx
1=
1
1.
limx→0
x− sinx
x3= lim
x→0
1− cosx
3x2= lim
x→0
sinx
6x= lim
x→0
cosx
6=
1
6.
We’ll state the next L’Hospital Rule as a one-sided limit.
16
L’Hospital’s Rule: ∞/∞ Case
Assume f and g are differntiable on (a, b) and that g′(x) 6= 0 for all x ∈ (a, b).
If limx→a g(x) =∞ (or −∞), then
limx→a
f ′(x)
g′(x)= L implies lim
x→a
f(x)
g(x)= L.
Proof. The proof is given in the textbook.
Exercise 5.3.6
(a) Let g : [0, a] → R be differentiable, g(0) = 0, and |g′(x)| ≤ M for all
x ∈ [0, a]. Show that |g(x)| ≤Mx for all x ∈ [0, a].
(b) Let h : [0, a] → R be twice differentiable, h′(0) = h(0) = 0, and |h′′(x)| ≤M for all x ∈ [0, a]. Show that |h(x)| ≤ Mx2
2 for all x ∈ [0, a].
(c) Let f : [0, a]→ R be three times differentiable,
0 = f(0) = f ′(0) = f ′′(0) and |f ′′′(x)| ≤M for all x ∈ [0, a].
Show that |f(x)| ≤ Mx3
3! for all x ∈ [0, a].
Proof of (a). Let x ∈ (0, a]. By the Mean Value Theorem there exists c1 ∈ (0, x)
such that
g′(c1) =g(x)− g(0)
x− 0=g(x)
x.
Then
g(x) = xg′(c1)
|g(x)| = Mx since |g′(c1)| ≤M .
By the continuity of g at 0, the inequality holds for all x ∈ [0, a].
Proof of (b). Let x ∈ (0, a]. We apply the Generalized Mean Value Theorem to the
pair of functions h(x) and x2. There exists c1 ∈ (0, x) such that
h′(c1)
2c1=h(x)− h(0)
x2 − 02=h(x)
x2⇒ h(x) =
h′(c1)x2
2c1.
17
By the Mean Value Theorem there exists c2 ∈ (0, c1) such that
h′′(c2) =h′(c1)− h′(0)
c1 − 0=h′(c1)
c1⇒ h′(c1) = c1h
′′(c2).
Combining these results gives
h(x) =h′(c1)x
2
2c1=c1h′′(c2)x
2
2c1=h′′(c2)x
2
2
|h(x)| ≤ Mx2
2for x ∈ (0, a].
By the continuity of h at 0, the inequality holds for all x ∈ [0, a].
Proof of (c). Let x ∈ (0, a]. We apply the Generalized Mean Value Theorem to the
pair of functions f(x) and x3. There exists c1 ∈ (0, x) such that
f ′(c1)
3c21=f(x)− f(0)
x3 − 03⇒ f(x) =
f ′(c1)x3
3c21.
Again apply the Generalized Mean Value Theorem to the pair of functions f ′(x)
and x2. There exists c2 ∈ (0, c1) such that
f ′′(c2)
2c2=f ′(c1)− f ′(0)
c21 − 02=f ′(c1)
c21⇒ f ′(c1) =
f ′′(c2)c21
2c2.
Finally, apply the Mean Value Theorem to the function f ′′ on the interval (0, c2).
There exists c3 ∈ (0, c2) such that
f ′′′(c3) =f ′′(c2)− f ′′(0)
c2 − 0=f ′′(c2)
c2⇒ f ′′(c2) = c2f
′′′(c3).
Putting the pieces together gives
f(x) = f ′(c1) ·x3
3c21= f ′′(c2) ·
c212c2· x
3
3c21= f ′′′(c2)c2 ·
c212c2· x
3
3c21= f ′′′(c3) ·
x3
3!.
Since |f ′′′(c3)| ≤M , this gives
|f(x)| ≤ Mx3
3!
for x ∈ (0, a]. By continuity at 0, the inequality holds for all x ∈ [0, a].
18
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