Math 341 { Lecture Notes on Chapter 5 { The Derivativemath341.cardon.byu.edu/Math-341-Lecture-Notes/341_Lectures_Cha… · Proof. By hypothesis, f0(c) = lim x!c g(x) g(c) x c exits.

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