aarjaneiro/planetmath huggingface dataset:
product rule

product rule

The

product rule

states that if

f

:

ℝ

→

ℝ

and

g

:

ℝ

→

ℝ

are functions in one variable both

differentiable

at a point

x

0

, then the

derivative

of the product of the two functions, denoted

f

⋅

g

, at

x

0

is given by

d

d

⁢

x

⁢

(

f

⋅

g

)

⁢

(

x

0

)

=

f

⁢

(

x

0

)

⁢

g

′

⁢

(

x

0

)

+

f

′

⁢

(

x

0

)

⁢

g

⁢

(

x

0

)

.

Proof

See the proof of the product rule (

http://

planetmath

.org/ProofOfProductRule

).

0.1

Generalized Product Rule

More generally, for differentiable functions

f

1

,

f

2

,

…

,

f

n

in one variable, all differentiable at

x

0

, we have

D

⁢

(

f

1

⁢

⋯

⁢

f

n

)

⁢

(

x

0

)

=

∑

i

=

1

n

(

f

i

⁢

(

x

0

)

⁢

⋯

⁢

f

i

-

1

⁢

(

x

0

)

⋅

D

⁢

f

i

⁢

(

x

0

)

⋅

f

i

+

1

⁢

(

x

0

)

⁢

⋯

⁢

f

n

⁢

(

x

0

)

)

.

Also see Leibniz’ rule (

http://planetmath.org/LeibnizRule

).

Example

The derivative of

x

⁢

ln

⁡

|

x

|

can be found by application of this rule.
Let

f

⁢

(

x

)

=

x

,

g

⁢

(

x

)

=

ln

⁡

|

x

|

, so that

f

⁢

(

x

)

⁢

g

⁢

(

x

)

=

x

⁢

ln

⁡

|

x

|

. Then

f

′

⁢

(

x

)

=

1

and

g

′

⁢

(

x

)

=

1

x

. Therefore, by the product rule,

d

d

⁢

x

⁢

(

x

⁢

ln

⁡

|

x

|

)

=

f

⁢

(

x

)

⁢

g

′

⁢

(

x

)

+

f

′

⁢

(

x

)

⁢

g

⁢

(

x

)

=

x

x

+

1

⋅

ln

⁡

|

x

|

=

ln

⁡

|

x

|

+

1

Title

product rule

Canonical name

ProductRule

Date of creation

2013-03-22 12:27:57

Last modified on

2013-03-22 12:27:57

Owner

mathcam (2727)

Last modified by

mathcam (2727)

Numerical id

12

Author

mathcam (2727)

Entry type

Theorem

Classification

msc 26A06

Related topic

Derivative

Related topic

ProofOfProductRule

Related topic

ProductRule

Related topic

PowerRule

Related topic

ProofOfPowerRule

Related topic

SumRule

Related topic

ZeroesOfDerivativeOfComplexPolynomial

Generated on Fri Feb 9 19:45:05 2018 by

LaTeXML
proof of Barbalat’s lemma

proof of Barbalat’s lemma

We suppose that

y

⁢

(

t

)

↛

0

as

t

→

∞

. There exists a

sequence

(

t

n

)

in

ℝ

+

such that

t

n

→

∞

as

n

→

∞

and

|

y

⁢

(

t

n

)

|

≥

ε

for all

n

∈

ℕ

. By the

uniform continuity

of

y

, there exists a

δ

>

0

such that, for all

n

∈

ℕ

and all

t

∈

ℝ

,

|

t

n

-

t

|

≤

δ

⇒

|

y

⁢

(

t

n

)

-

y

⁢

(

t

)

|

≤

ε

2

.

So, for all

t

∈

[

t

n

,

t

n

+

δ

]

, and for all

n

∈

ℕ

we have

|

y

⁢

(

t

)

|

=

|

y

⁢

(

t

n

)

-

(

y

⁢

(

t

n

)

-

y

⁢

(

t

)

)

|

≥

|

y

⁢

(

t

n

)

|

-

|

y

⁢

(

t

n

)

-

y

⁢

(

t

)

|

≥

≥

ε

-

ε

2

=

ε

2

.

Therefore,

|

∫

0

t

n

+

δ

y

⁢

(

t

)

⁢

𝑑

t

-

∫

0

t

n

y

⁢

(

t

)

⁢

𝑑

t

|

=

|

∫

t

n

t

n

+

δ

y

⁢

(

t

)

⁢

𝑑

t

|

=

∫

t

n

t

n

+

δ

|

y

⁢

(

t

)

|

⁢

𝑑

t

≥

ε

⁢

δ

2

>

0

for each

n

∈

ℕ

. By the

hypothesis

, the improprer

Riemann integral

∫

0

∞

y

⁢

(

t

)

⁢

𝑑

t

exists, and thus the left hand side of the

inequality

converges

to 0 as

n

→

∞

, yielding a

contradiction

.

Title

proof of Barbalat’s lemma

Canonical

name

ProofOfBarbalatsLemma

Date of creation

2013-03-22 15:10:45

Last modified on

2013-03-22 15:10:45

Owner

ncrom (8997)

Last modified by

ncrom (8997)

Numerical id

7

Author

ncrom (8997)

Entry type

Proof

Classification

msc 26A06

Generated on Fri Feb 9 19:45:08 2018 by

LaTeXML
proof of binomial formula

proof of binomial formula

Let

p

∈

ℝ

and

x

∈

ℝ

,

|

x

|

<

1

be given. We wish to show
that

(

1

+

x

)

p

=

∑

n

=

0

∞

p

n

¯

⁢

x

n

n

!

,

where

p

n

¯

denotes the

n

th

falling factorial

of

p

.

The

convergence

of the series in the right-hand side of the above
equation is a straight-forward consequence of the

ratio test

. Set

f

⁢

(

x

)

=

(

1

+

x

)

p

.

and note that

f

(

n

)

⁢

(

x

)

=

p

n

¯

⁢

(

1

+

x

)

p

-

n

.

The desired equality now follows from Taylor’s Theorem. Q.E.D.

Title

proof of binomial formula

Canonical name

ProofOfBinomialFormula

Date of creation

2013-03-22 12:24:00

Last modified on

2013-03-22 12:24:00

Owner

rmilson (146)

Last modified by

rmilson (146)

Numerical id

6

Author

rmilson (146)

Entry type

Proof

Classification

msc 26A06

Generated on Fri Feb 9 19:45:10 2018 by

LaTeXML
proof of Bolzano’s theorem

proof of Bolzano’s theorem

Consider the compact

interval

[

a

,

b

]

,

a

<

b

and a

continuous

real valued

function

f

. If

f

⁢

(

a

)

.

f

⁢

(

b

)

<

0

then there exists

c

∈

(

a

,

b

)

such that

f

⁢

(

c

)

=

0

WLOG consider

f

⁢

(

a

)

<

0

and

f

⁢

(

b

)

>

0

. The other case can be proved using

-

f

⁢

(

x

)

which will also verify the theorem’s conditions.

consider

a

1

=

a

+

b

2

, three cases can occur:

•

f

⁢

(

a

1

)

=

0

, in this case the theorem is proved

c

=

a

1

•

f

⁢

(

a

1

)

>

0

, in this case consider the interval

I

1

=

(

a

,

a

1

)

•

f

⁢

(

a

1

)

<

0

, in this case consider the interval

I

1

=

(

a

1

,

b

)

so starting with an

open interval

I

0

=

(

a

,

b

)

we get another open interval

I

1

⊂

I

0

with length half of the original

|

I

1

|

=

|

I

0

|

2

.

Repeat the procedure to the interval

I

n

and get another interval

I

n

+

1

.

We can thus define a succession of open intervals

I

n

such that

I

n

+

1

⊂

I

n

,

|

I

n

|

=

2

-

n

⁢

|

I

0

|

, such that

I

n

=

(

a

n

,

b

n

)

and

f

⁢

(

a

n

)

<

0

<

f

⁢

(

b

n

)

.

The succession

c

2

⁢

n

=

a

n

,

c

2

⁢

n

+

1

=

b

n

is Cauchy by construction since

m

>

n

⟹

|

c

m

-

c

n

|

<

2

-

[

n

/

2

]

⁢

|

I

0

|

.

c

n

is therefore

convergent

c

n

→

c

∈

[

a

,

b

]

, and since

a

n

and

b

n

are sub-successions, they converge to the same limit.

f

is continuous in

[

a

,

b

]

so

x

n

→

x

⟹

f

⁢

(

x

n

)

→

f

⁢

(

x

)

By construction

f

⁢

(

a

n

)

<

0

and

f

⁢

(

b

n

)

>

0

so in the limit

lim

n

→

∞

⁡

f

⁢

(

a

n

)

=

f

⁢

(

lim

n

→

∞

⁡

a

n

)

=

f

⁢

(

c

)

≤

0

and

lim

n

→

∞

⁡

f

⁢

(

b

n

)

=

f

⁢

(

c

)

≥

0

.

So there exists

c

∈

[

a

,

b

]

such that

0

≤

f

⁢

(

c

)

≤

0

⟹

f

⁢

(

c

)

=

0

.

But since

f

⁢

(

a

)

.

f

⁢

(

b

)

<

0

, neither

f

⁢

(

a

)

=

0

nor

f

⁢

(

b

)

=

0

and since

f

⁢

(

c

)

=

0

,

c

∈

(

a

,

b

)

Title

proof of Bolzano’s theorem

Canonical name

ProofOfBolzanosTheorem

Date of creation

2013-03-22 15:43:29

Last modified on

2013-03-22 15:43:29

Owner

cvalente (11260)

Last modified by

cvalente (11260)

Numerical id

7

Author

cvalente (11260)

Entry type

Proof

Classification

msc 26A06

Generated on Fri Feb 9 19:45:12 2018 by

LaTeXML
proof of chain rule

proof of chain rule

Let’s say that

g

is

differentiable

in

x

0

and

f

is differentiable in

y

0

=

g

⁢

(

x

0

)

. We define:

φ

⁢

(

y

)

=

{

f

⁢

(

y

)

-

f

⁢

(

y

0

)

y

-

y

0

if

y

≠

y

0

f

′

⁢

(

y

0

)

if

y

=

y

0

Since

f

is differentiable in

y

0

,

φ

is

continuous

.
We observe that, for

x

≠

x

0

,

f

⁢

(

g

⁢

(

x

)

)

-

f

⁢

(

g

⁢

(

x

0

)

)

x

-

x

0

=

φ

⁢

(

g

⁢

(

x

)

)

⁢

g

⁢

(

x

)

-

g

⁢

(

x

0

)

x

-

x

0

,

in fact, if

g

⁢

(

x

)

≠

g

⁢

(

x

0

)

, it follows at once from the definition of

φ

, while if

g

⁢

(

x

)

=

g

⁢

(

x

0

)

, both members of the equation are 0.

Since

g

is continuous in

x

0

, and

φ

is continuous in

y

0

,

lim

x

→

x

0

⁡

φ

⁢

(

g

⁢

(

x

)

)

=

φ

⁢

(

g

⁢

(

x

0

)

)

=

f

′

⁢

(

g

⁢

(

x

0

)

)

,

hence

(

f

∘

g

)

′

⁢

(

x

0

)

=

lim

x

→

x

0

⁡

f

⁢

(

g

⁢

(

x

)

)

-

f

⁢

(

g

⁢

(

x

0

)

)

x

-

x

0

=

lim

x

→

x

0

⁡

φ

⁢

(

g

⁢

(

x

)

)

⁢

g

⁢

(

x

)

-

g

⁢

(

x

0

)

x

-

x

0

=

f

′

⁢

(

g

⁢

(

x

0

)

)

⁢

g

′

⁢

(

x

0

)

.

Title

proof of chain rule

Canonical name

ProofOfChainRule

Date of creation

2013-03-22 12:41:48

Last modified on

2013-03-22 12:41:48

Owner

n3o (216)

Last modified by

n3o (216)

Numerical id

6

Author

n3o (216)

Entry type

Proof

Classification

msc 26A06

Generated on Fri Feb 9 19:45:15 2018 by

LaTeXML
proof of complex mean-value theorem

proof of complex mean-value theorem

The

function

h

⁢

(

t

)

=

Re

⁡

f

⁢

(

a

+

t

⁢

(

b

-

a

)

)

-

f

⁢

(

a

)

b

-

a

is a function defined on [0,1].
We have

h

⁢

(

0

)

=

0

and

h

⁢

(

1

)

=

Re

⁡

f

⁢

(

b

)

-

f

⁢

(

a

)

b

-

a

.
By the ordinary

mean-value theorem

, there is a number

t

,

0

<

t

<

1

, such that

h

′

⁢

(

t

)

=

h

⁢

(

1

)

-

h

⁢

(

0

)

.
To evaluate

h

′

⁢

(

t

)

, we use the assumption that

f

is

complex differentiable

(holomorphic). The

derivative

of

f

⁢

(

a

+

t

⁢

(

b

-

a

)

)

-

f

⁢

(

a

)

b

-

a

is equal to

f

′

⁢

(

a

+

t

⁢

(

b

-

a

)

)

, then

h

′

⁢

(

t

)

=

Re

⁡

(

f

′

⁢

(

a

+

t

⁢

(

b

-

a

)

)

)

, so

u

=

a

+

t

⁢

(

b

-

a

)

satisfies the required equation.
The proof of the second assertion can be deduced from the result just proved by applying it to the function f multiplied by i.

Title

proof of complex mean-value theorem

Canonical name

ProofOfComplexMeanvalueTheorem

Date of creation

2013-03-22 14:34:39

Last modified on

2013-03-22 14:34:39

Owner

Wolfgang (5320)

Last modified by

Wolfgang (5320)

Numerical id

21

Author

Wolfgang (5320)

Entry type

Proof

Classification

msc 26A06

Generated on Fri Feb 9 19:45:17 2018 by

LaTeXML
proof of Darboux’s theorem

proof of Darboux’s theorem

Without loss of generality we migth and shall assume

f

+

′

⁢

(

a

)

>

t

>

f

-

′

⁢

(

b

)

. Let

g

⁢

(

x

)

:=

f

⁢

(

x

)

-

t

⁢

x

.
Then

g

′

⁢

(

x

)

=

f

′

⁢

(

x

)

-

t

,

g

+

′

⁢

(

a

)

>

0

>

g

-

′

⁢

(

b

)

, and we wish to find a zero of

g

′

.

Since

g

is a

continuous function

on

[

a

,

b

]

, it attains a

maximum

on

[

a

,

b

]

.
Since

g

+

′

⁢

(

a

)

>

0

and

g

+

′

⁢

(

b

)

<

0

Fermat’s

theorem

(

http://planetmath.org/FermatsTheoremStationaryPoints

) states that
neither

a

nor

b

can be points where

f

has a

local maximum

.
So a maximum is attained at some

c

∈

(

a

,

b

)

. But then

g

′

⁢

(

c

)

=

0

again by Fermat’s theorem (

http://planetmath.org/FermatsTheoremStationaryPoints

).

Title

proof of Darboux’s theorem

Canonical name

ProofOfDarbouxsTheorem

Date of creation

2013-03-22 12:45:04

Last modified on

2013-03-22 12:45:04

Owner

paolini (1187)

Last modified by

paolini (1187)

Numerical id

7

Author

paolini (1187)

Entry type

Proof

Classification

msc 26A06

Generated on Fri Feb 9 19:45:19 2018 by

LaTeXML
proof of extended mean-value theorem

proof of extended mean-value theorem

Let

f

:

[

a

,

b

]

→

ℝ

and

g

:

[

a

,

b

]

→

ℝ

be

continuous

on

[

a

,

b

]

and

differentiable

on

(

a

,

b

)

. Define the function

h

⁢

(

x

)

=

f

⁢

(

x

)

⁢

(

g

⁢

(

b

)

-

g

⁢

(

a

)

)

-

g

⁢

(

x

)

⁢

(

f

⁢

(

b

)

-

f

⁢

(

a

)

)

-

f

⁢

(

a

)

⁢

g

⁢

(

b

)

+

f

⁢

(

b

)

⁢

g

⁢

(

a

)

.

Because

f

and

g

are continuous on

[

a

,

b

]

and differentiable on

(

a

,

b

)

, so is

h

. Furthermore,

h

⁢

(

a

)

=

h

⁢

(

b

)

=

0

, so by Rolle’s theorem there exists a

ξ

∈

(

a

,

b

)

such that

h

′

⁢

(

ξ

)

=

0

. This implies that

f

′

⁢

(

ξ

)

⁢

(

g

⁢

(

b

)

-

g

⁢

(

a

)

)

-

g

′

⁢

(

ξ

)

⁢

(

f

⁢

(

b

)

-

f

⁢

(

a

)

)

=

0

and, if

g

⁢

(

b

)

≠

g

⁢

(

a

)

,

f

′

⁢

(

ξ

)

g

′

⁢

(

ξ

)

=

f

⁢

(

b

)

-

f

⁢

(

a

)

g

⁢

(

b

)

-

g

⁢

(

a

)

.

Title

proof of extended mean-value theorem

Canonical name

ProofOfExtendedMeanvalueTheorem

Date of creation

2013-03-22 13:09:00

Last modified on

2013-03-22 13:09:00

Owner

pbruin (1001)

Last modified by

pbruin (1001)

Numerical id

6

Author

pbruin (1001)

Entry type

Proof

Classification

msc 26A06

Synonym

proof of Cauchy’s

mean-value theorem

Related topic

MeanValueTheorem

Generated on Fri Feb 9 19:45:21 2018 by

LaTeXML
proof of Fermat’s Theorem (stationary points)

proof of Fermat’s Theorem (stationary points)

Suppose that

x

0

is a

local maximum

(a similar proof applies if

x

0

is a

local minimum

). Then there exists

δ

>

0

such that

(

x

0

-

δ

,

x

0

+

δ

)

⊂

(

a

,

b

)

and such that we have

f

⁢

(

x

0

)

≥

f

⁢

(

x

)

for all

x

with

|

x

-

x

0

|

<

δ

. Hence for

h

∈

(

0

,

δ

)

we notice that it holds

f

⁢

(

x

0

+

h

)

-

f

⁢

(

x

0

)

h

≤

0

.

Since the limit of this ratio as

h

→

0

+

exists and is equal to

f

′

⁢

(

x

0

)

we conclude that

f

′

⁢

(

x

0

)

≤

0

. On the other hand for

h

∈

(

-

δ

,

0

)

we notice that

f

⁢

(

x

0

+

h

)

-

f

⁢

(

x

0

)

h

≥

0

but again the limit as

h

→

0

+

exists and is equal to

f

′

⁢

(

x

0

)

so we also have

f

′

⁢

(

x

0

)

≥

0

.

Hence we conclude that

f

′

⁢

(

x

0

)

=

0

.

To prove the second part of the statement (when

x

0

is equal to

a

or

b

), just notice that in such
points we have only one of the two estimates written above.

Title

proof of Fermat’s Theorem (

stationary points

)

Canonical name

ProofOfFermatsTheoremstationaryPoints

Date of creation

2013-03-22 13:45:09

Last modified on

2013-03-22 13:45:09

Owner

paolini (1187)

Last modified by

paolini (1187)

Numerical id

5

Author

paolini (1187)

Entry type

Proof

Classification

msc 26A06

Generated on Fri Feb 9 19:45:23 2018 by

LaTeXML
proof of generalized Leibniz rule

proof of generalized Leibniz rule

The generalized

Leibniz rule

can be derived from the plain Leibniz rule by

induction

on

r

.

If

r

=

2

, the generalized Leibniz rule reduces to the plain Leibniz rule. This will be the starting point for the induction. To

complete

the induction, assume that the generalized Leibniz rule holds for a certain value of

r

; we shall now show that it holds for

r

+

1

.

Write

∏

i

=

1

r

+

1

f

i

⁢

(

t

)

=

(

f

r

+

1

⁢

(

t

)

)

⁢

(

∏

i

=

1

r

+

1

f

i

⁢

(

t

)

)

. Applying the plain Leibniz rule,

d

n

d

⁢

t

n

⁢

(

f

r

+

1

⁢

(

t

)

)

⁢

(

∏

i

=

1

r

+

1

f

i

⁢

(

t

)

)

=

∑

n

r

+

1

=

0

n

(

n

n

r

+

1

)

⁢

(

d

n

r

+

1

d

⁢

n

n

r

+

1

⁢

f

r

+

1

⁢

(

t

)

)

⁢

(

d

n

-

n

r

+

1

d

⁢

n

n

-

n

r

+

1

⁢

∏

i

=

1

r

+

1

f

i

⁢

(

t

)

)

By the generalized Leibniz rule for

r

(assumed to be true as the

induction hypothesis

), this equals

∑

n

r

+

1

=

0

n

∑

n

1

+

⋯

+

n

r

=

n

-

n

r

+

1

(

n

-

n

r

+

1

n

1

,

n

2

,

…

⁢

n

r

)

⁢

(

n

n

r

+

1

)

⁢

(

d

n

r

+

1

d

⁢

n

n

r

+

1

⁢

f

r

+

1

⁢

(

t

)

)

⁢

(

∏

i

=

1

r

d

n

i

d

⁢

t

n

i

⁢

f

i

⁢

(

t

)

)

Note that

(

n

-

n

r

+

1

n

1

,

n

2

,

…

⁢

n

r

)

⁢

(

n

n

r

+

1

)

=

(

n

-

n

r

+

1

n

1

,

n

2

,

…

⁢

n

r

,

n

r

+

1

)

This is an immediate

consequence

of the

expression

for

multinomial coefficients

as

quotients

of

factorials

. Using this

identity

, the quantity can be written as

∑

n

1

+

⋯

+

n

r

+

n

r

+

1

=

n

(

n

-

n

r

+

1

n

1

,

n

2

,

…

⁢

n

r

,

n

r

+

1

)

⁢

∏

i

=

1

r

+

1

d

n

i

d

⁢

t

n

i

⁢

f

i

⁢

(

t

)

which is the generalized Leibniz rule for the case of

r

+

1

.

Title

proof of generalized Leibniz rule

Canonical

name

ProofOfGeneralizedLeibnizRule

Date of creation

2013-03-22 14:34:14

Last modified on

2013-03-22 14:34:14

Owner

rspuzio (6075)

Last modified by

rspuzio (6075)

Numerical id

7

Author

rspuzio (6075)

Entry type

Proof

Classification

msc 26A06

Generated on Fri Feb 9 19:45:26 2018 by

LaTeXML
proof of intermediate value theorem

proof of intermediate value theorem

We first prove the following lemma.

If

f

:

[

a

,

b

]

→

ℝ

is a

continuous function

with

f

⁢

(

a

)

≤

0

≤

f

⁢

(

b

)

then there exists a

c

∈

[

a

,

b

]

such that

f

⁢

(

c

)

=

0

.

Define the

sequences

(

a

n

)

and

(

b

n

)

inductively, as follows.

a

0

=

a

b

0

=

b

c

n

=

a

n

+

b

n

2

(

a

n

,

b

n

)

=

{

(

a

n

-

1

,

c

n

-

1

)

f

⁢

(

c

n

-

1

)

≥

0

(

c

n

-

1

,

b

n

-

1

)

f

⁢

(

c

n

-

1

)

<

0

We note that

a

0

≤

a

1

≤

⋯

≤

a

n

≤

b

n

≤

⋯

≤

b

1

≤

b

0

(

b

n

-

a

n

)

=

2

-

n

⁢

(

b

0

-

a

0

)

(1)

f

⁢

(

a

n

)

≤

0

≤

f

⁢

(

b

n

)

(2)

By the fundamental axiom of

analysis

(

a

n

)

→

α

and

(

b

n

)

→

β

. But

(

b

n

-

a

n

)

→

0

so

α

=

β

.
By

continuity

of

f

(

f

⁢

(

a

n

)

)

→

f

⁢

(

α

)

(

f

⁢

(

b

n

)

)

→

f

⁢

(

α

)

But we have

f

⁢

(

α

)

≤

0

and

f

⁢

(

α

)

≥

0

so that

f

⁢

(

α

)

=

0

. Furthermore we have

a

≤

α

≤

b

, proving the assertion.

Set

g

⁢

(

x

)

=

f

⁢

(

x

)

-

k

where

f

⁢

(

a

)

≤

k

≤

f

⁢

(

b

)

.

g

satisfies

the same conditions as before, so there exists a

c

such that

f

⁢

(

c

)

=

k

. Thus proving the more general result.

Title

proof of intermediate value theorem

Canonical name

ProofOfIntermediateValueTheorem

Date of creation

2013-03-22 12:33:56

Last modified on

2013-03-22 12:33:56

Owner

yark (2760)

Last modified by

yark (2760)

Numerical id

9

Author

yark (2760)

Entry type

Proof

Classification

msc 26A06

Generated on Fri Feb 9 19:45:28 2018 by

LaTeXML
proof of mean value theorem

proof of mean value theorem

Define

h

⁢

(

x

)

on

[

a

,

b

]

by

h

⁢

(

x

)

=

f

⁢

(

x

)

-

f

⁢

(

a

)

-

(

f

⁢

(

b

)

-

f

⁢

(

a

)

b

-

a

)

⁢

(

x

-

a

)

Clearly,

h

is

continuous

on

[

a

,

b

]

,

differentiable

on

(

a

,

b

)

, and

h

⁢

(

a

)

=

f

⁢

(

a

)

-

f

⁢

(

a

)

=

0

h

⁢

(

b

)

=

f

⁢

(

b

)

-

f

⁢

(

a

)

-

(

f

⁢

(

b

)

-

f

⁢

(

a

)

b

-

a

)

⁢

(

b

-

a

)

=

0

Notice that

h

satisfies the conditions of Rolle’s Theorem. Therefore, by Rolle’s Theorem there exists

c

∈

(

a

,

b

)

such that

h

′

⁢

(

c

)

=

0

.

However, from the definition of

h

we obtain by

differentiation

that

h

′

⁢

(

x

)

=

f

′

⁢

(

x

)

-

f

⁢

(

b

)

-

f

⁢

(

a

)

b

-

a

Since

h

′

⁢

(

c

)

=

0

, we therefore have

f

′

⁢

(

c

)

=

f

⁢

(

b

)

-

f

⁢

(

a

)

b

-

a

as required.

References

1

Michael Spivak,

Calculus

, 3rd ed., Publish or Perish Inc., 1994.

Title

proof of mean value theorem

Canonical name

ProofOfMeanValueTheorem

Date of creation

2013-03-22 12:40:57

Last modified on

2013-03-22 12:40:57

Owner

Andrea Ambrosio (7332)

Last modified by

Andrea Ambrosio (7332)

Numerical id

5

Author

Andrea Ambrosio (7332)

Entry type

Proof

Classification

msc 26A06

Generated on Fri Feb 9 19:45:34 2018 by

LaTeXML
proof of monotonicity criterion

proof of monotonicity criterion

Let us start from the

implications

“

⇒

”.

Suppose that

f

′

⁢

(

x

)

≥

0

for all

x

∈

(

a

,

b

)

. We want to prove that therefore

f

is

increasing

. So take

x

1

,

x

2

∈

[

a

,

b

]

with

x

1

<

x

2

. Applying the

mean-value Theorem

on the

interval

[

x

1

,

x

2

]

we know that there exists a point

x

∈

(

x

1

,

x

2

)

such that

f

⁢

(

x

2

)

-

f

⁢

(

x

1

)

=

f

′

⁢

(

x

)

⁢

(

x

2

-

x

1

)

and being

f

′

⁢

(

x

)

≥

0

we conclude that

f

⁢

(

x

2

)

≥

f

⁢

(

x

1

)

.

This proves the first claim. The other three cases can be achieved with minor modifications: replace all “

≥

” respectively with

≤

,

>

and

<

.

Let us now prove the implication “

⇐

” for the first and second statement.

Given

x

∈

(

a

,

b

)

consider the ratio

f

⁢

(

x

+

h

)

-

f

⁢

(

x

)

h

.

If

f

is increasing the numerator of this ratio is

≥

0

when

h

>

0

and is

≤

0

when

h

<

0

. Anyway the ratio is

≥

0

since the denominator has the same sign of the numerator. Since we know by

hypothesis

that the

function

f

is

differentiable

in

x

we can pass to the limit to conclude that

f

′

⁢

(

x

)

=

lim

h

→

0

⁡

f

⁢

(

x

+

h

)

-

f

⁢

(

x

)

h

≥

0

.

If

f

is decreasing the ratio considered turns out to be

≤

0

hence the

conclusion

f

′

⁢

(

x

)

≤

0

.

Notice that if we suppose that

f

is strictly increasing we obtain the this ratio is

>

0

, but passing to the limit as

h

→

0

we cannot conclude that

f

′

⁢

(

x

)

>

0

but only (again)

f

′

⁢

(

x

)

≥

0

.

Title

proof of monotonicity criterion

Canonical name

ProofOfMonotonicityCriterion

Date of creation

2013-03-22 13:45:14

Last modified on

2013-03-22 13:45:14

Owner

paolini (1187)

Last modified by

paolini (1187)

Numerical id

6

Author

paolini (1187)

Entry type

Proof

Classification

msc 26A06

Generated on Fri Feb 9 19:45:36 2018 by

LaTeXML
proof of quotient rule

proof of quotient rule

Let

F

⁢

(

x

)

=

f

⁢

(

x

)

/

g

⁢

(

x

)

. Then

F

′

⁢

(

x

)

=

lim

h

→

0

⁡

F

⁢

(

x

+

h

)

-

F

⁢

(

x

)

h

=

lim

h

→

0

⁡

f

⁢

(

x

+

h

)

g

⁢

(

x

+

h

)

-

f

⁢

(

x

)

g

⁢

(

x

)

h

=

lim

h

→

0

⁡

f

⁢

(

x

+

h

)

⁢

g

⁢

(

x

)

-

f

⁢

(

x

)

⁢

g

⁢

(

x

+

h

)

h

⁢

g

⁢

(

x

+

h

)

⁢

g

⁢

(

x

)

Like the

product rule

, the key to this proof is subtracting and adding the same quantity. We separate

f

and

g

in the above

expression

by subtracting and adding the term

f

⁢

(

x

)

⁢

g

⁢

(

x

)

in the numerator.

F

′

⁢

(

x

)

=

lim

h

→

0

⁡

f

⁢

(

x

+

h

)

⁢

g

⁢

(

x

)

-

f

⁢

(

x

)

⁢

g

⁢

(

x

)

+

f

⁢

(

x

)

⁢

g

⁢

(

x

)

-

f

⁢

(

x

)

⁢

g

⁢

(

x

+

h

)

h

⁢

g

⁢

(

x

+

h

)

⁢

g

⁢

(

x

)

=

lim

h

→

0

⁡

g

⁢

(

x

)

⁢

f

⁢

(

x

+

h

)

-

f

⁢

(

x

)

h

-

f

⁢

(

x

)

⁢

g

⁢

(

x

+

h

)

-

g

⁢

(

x

)

h

g

⁢

(

x

+

h

)

⁢

g

⁢

(

x

)

=

lim

h

→

0

⁡

g

⁢

(

x

)

⋅

lim

h

→

0

⁡

f

⁢

(

x

+

h

)

-

f

⁢

(

x

)

h

-

lim

h

→

0

⁡

f

⁢

(

x

)

⋅

lim

h

→

0

⁡

g

⁢

(

x

+

h

)

-

g

⁢

(

x

)

h

lim

h

→

0

⁡

g

⁢

(

x

+

h

)

⋅

lim

h

→

0

⁡

g

⁢

(

x

)

=

g

⁢

(

x

)

⁢

f

′

⁢

(

x

)

-

f

⁢

(

x

)

⁢

g

′

⁢

(

x

)

[

g

⁢

(

x

)

]

2

Title

proof of quotient rule

Canonical

name

ProofOfQuotientRule

Date of creation

2013-03-22 12:38:58

Last modified on

2013-03-22 12:38:58

Owner

drini (3)

Last modified by

drini (3)

Numerical id

5

Author

drini (3)

Entry type

Proof

Classification

msc 26A06

Generated on Fri Feb 9 19:45:39 2018 by

LaTeXML
proof of quotient rule (using product rule)

proof of quotient rule (using product rule)

Suppose

f

and

g

are

differentiable functions

defined on some

interval

of

ℝ

, and

g

never

vanishes

. Let us prove that

(

f

g

)

′

=

f

′

⁢

g

-

f

⁢

g

′

g

2

.

Using the

product rule

(

f

⁢

g

)

′

=

f

′

⁢

g

+

f

⁢

g

′

, and

(

g

-

1

)

′

=

-

g

-

2

⁢

g

′

,
we have

(

f

g

)

′

=

(

f

⁢

g

-

1

)

′

=

f

′

⁢

g

-

1

+

f

⁢

(

g

-

1

)

′

=

f

′

⁢

g

-

1

+

f

⁢

(

-

1

)

⁢

g

-

2

⁢

g

′

=

f

′

g

-

f

⁢

g

′

g

2

=

f

′

⁢

g

-

f

⁢

g

′

g

2

.

Here

g

-

1

=

1

/

g

and

g

-

2

=

1

/

g

2

.

Title

proof of quotient rule

(using product rule)

Canonical name

ProofOfQuotientRuleusingProductRule

Date of creation

2013-03-22 15:00:45

Last modified on

2013-03-22 15:00:45

Owner

matte (1858)

Last modified by

matte (1858)

Numerical id

5

Author

matte (1858)

Entry type

Proof

Classification

msc 26A06

Generated on Fri Feb 9 19:45:41 2018 by

LaTeXML