derivatives by pure algebra

derivatives by pure algebra

Let

R

be any commutative

unique factorization domain

(UFD) and

x

and

h

indeterminants. For

instance

, let

R

=

ℝ

the usual

real numbers

, or any other field. We treate

R

⁢

[

x

]

as a subring of

R

⁢

[

x

,

h

]

.

We derive a

definition

for

derivatives of polynomial

and

rational functions

over

R

along with the usual rules:

product rule

and

power rule

. Despite the abstract nature of the definitions, the mechanics

reflect

the general understanding of introductory

calculus

, without any appeal to the Cauchy style

ε

-

δ

limits of

analysis

.

Definition 1

.

Define

f

⁢

(

x

+

h

)

-

f

⁢

(

x

)

h

=

a

1

⁢

(

x

,

h

)

⁢

⋯

⁢

a

m

⁢

(

x

,

h

)

where

f

⁢

(

x

+

h

)

-

f

⁢

(

x

)

=

h

⁢

a

1

⁢

(

x

,

h

)

⁢

⋯

⁢

a

m

⁢

(

x

,

h

)

in the UFD

R

⁢

[

x

,

h

]

.
Furthermore, given

g

⁢

(

x

,

h

)

∈

R

⁢

[

x

,

h

]

define

lim

h

→

0

⁡

g

⁢

(

x

,

h

)

=

g

⁢

(

x

,

0

)

(which is simply the

evaluation homomorphism

at

h

=

0

.)
Finally define

d

⁢

f

d

⁢

x

:=

lim

h

→

0

⁡

f

⁢

(

x

+

h

)

-

f

⁢

(

x

)

h

=

a

1

⁢

(

x

,

0

)

⁢

⋯

⁢

a

m

⁢

(

x

,

0

)

.

We also denote

d

⁢

f

d

⁢

x

by

f

′

⁢

(

x

)

.

Example 1

.

d

d

⁢

x

⁢

(

5

⁢

x

2

-

7

⁢

x

+

9

)

=

10

⁢

x

-

7

.

Proof.

First we reduce the

fraction

in a manner identical to the usual methods of
calculus:

d

d

⁢

x

⁢

(

5

⁢

x

2

-

7

⁢

x

+

9

)

=

lim

h

→

0

⁡

(

5

⁢

(

x

+

h

)

2

-

7

⁢

(

x

+

h

)

+

9

)

-

(

5

⁢

x

2

-

7

⁢

x

+

9

)

h

=

lim

h

→

0

⁡

5

⁢

x

2

+

10

⁢

x

⁢

h

+

5

⁢

h

2

-

7

⁢

x

-

7

⁢

h

+

9

-

5

⁢

x

2

+

7

⁢

x

-

9

h

=

lim

h

→

0

⁡

10

⁢

x

⁢

h

+

5

⁢

h

2

-

7

⁢

h

h

=

lim

h

→

0

⁡

10

⁢

x

+

5

⁢

h

-

7

.

At this stage we must interpret the

lim

h

→

0

. Because the
limit notation simply means to evaluate this

polynomial

at

h

=

0

we find:

d

d

⁢

x

⁢

(

5

⁢

x

2

-

7

⁢

x

+

9

)

=

10

⁢

x

+

5

⁢

(

0

)

-

7

=

10

⁢

x

-

7

.

This is in contrast to the typical approach where

h

is said to “approach”

0

. However, no

difference

is found in the solution and almost no difference
is found in the method, only in the

interpretation

of the method.
∎

Proposition 2

.

The

derivative

formula

is

well-defined

. In particular,

h

divides

f

⁢

(

x

+

h

)

-

f

⁢

(

x

)

∈

R

⁢

[

x

,

h

]

for every

f

⁢

(

x

)

∈

R

⁢

[

x

]

,
and the

a

1

⁢

(

x

,

h

)

⁢

⋯

⁢

a

m

⁢

(

x

,

h

)

are unique to

f

⁢

(

x

)

.

Proof.

For all

f

⁢

(

x

)

,

g

⁢

(

x

)

∈

R

⁢

[

x

]

, it follows

(

f

+

g

)

⁢

(

x

+

h

)

-

(

f

+

g

)

⁢

(

x

)

=

f

⁢

(

x

+

h

)

+

g

⁢

(

x

+

h

)

-

f

⁢

(

x

)

-

g

⁢

(

x

)

⁢

(

f

⁢

(

x

+

h

)

-

f

⁢

(

x

)

)

+

(

g

⁢

(

x

+

h

)

-

g

⁢

(

x

)

)

.

Furthermore, for all

a

∈

R

(

a

⁢

f

)

⁢

(

x

+

h

)

-

(

a

⁢

f

)

⁢

(

x

)

=

a

⁢

f

⁢

(

x

+

h

)

-

a

⁢

f

⁢

(

x

)

=

a

⁢

(

f

⁢

(

x

+

h

)

-

f

⁢

(

x

)

)

.

So now if we take

f

⁢

(

x

)

=

a

0

+

a

1

⁢

x

+

⋯

+

a

n

⁢

x

n

,
then

h

|

(

f

(

x

+

h

)

-

f

(

x

)

)

if

h

|

(

(

x

+

h

)

i

-

x

i

)

for every

i

∈

ℕ

.
When

i

=

0

,

(

x

+

h

)

0

-

x

0

=

0

so

h

|

(

(

x

+

h

)

0

-

x

0

)

. Now take

i

>

0

and use of the

binomial theorem

to find:

(

x

+

h

)

i

-

x

i

=

∑

j

=

0

i

(

i

j

)

⁢

x

i

-

j

⁢

h

j

-

x

i

=

∑

j

=

1

i

(

i

j

)

⁢

x

i

-

j

⁢

h

j

=

h

⁢

∑

j

=

1

i

(

i

j

)

⁢

x

i

-

j

⁢

h

j

-

1

.

Hence

h

|

(

f

(

x

+

h

)

-

f

(

x

)

)

.

As

R

is a UFD, so is

R

⁢

[

x

,

h

]

. Also

h

is

irreducible

in

R

⁢

[

x

,

h

]

, and

h

|

(

f

(

x

+

h

)

-

f

(

x

)

)

, so

f

⁢

(

x

+

h

)

-

f

⁢

(

x

)

=

h

⁢

a

1

⁢

(

x

,

h

)

⁢

⋯

⁢

a

m

⁢

(

x

,

h

)

for some

a

i

⁢

(

x

,

h

)

∈

R

⁢

[

x

,

h

]

,

1

≤

i

≤

m

, with each

a

i

⁢

(

x

,

h

)

unique to

f

⁢

(

x

+

h

)

-

f

⁢

(

x

)

up to

multiplication

by a unit of

R

⁢

[

x

,

h

]

, that is, a unit of

R

. In particular,

a

1

⁢

(

x

,

h

)

⁢

⋯

⁢

a

m

⁢

(

x

,

h

)

is unique to

f

⁢

(

x

+

h

)

-

f

⁢

(

x

)

, and so unique to

f

⁢

(

x

)

.
∎

Remark 3

.

Although potentially obtuse, the notation

lim

h

→

0

is a

function

,

lim

h

→

0

:

R

⁢

[

x

,

h

]

→

R

⁢

[

x

]

and has kernel

R

⁢

h

=

(

h

)

. So
we have

R

⁢

[

x

,

h

]

/

(

h

)

≅

R

⁢

[

x

]

. Therefore we may also write:

d

⁢

f

d

⁢

x

≡

f

⁢

(

x

+

h

)

-

f

⁢

(

x

)

h

(

mod

h

)

.

It is important that we always reduce the fractions so that we are not encountering any

division

by 0 at any stage.

Theorem 4

.

Derivatives

satisfy

the following rules:

Linearity

For

f

⁢

(

x

)

,

g

⁢

(

x

)

∈

R

⁢

[

x

]

and

a

∈

R

d

d

⁢

x

⁢

(

f

⁢

(

x

)

+

g

⁢

(

x

)

)

=

d

⁢

f

d

⁢

x

+

d

⁢

g

d

⁢

x

,

d

d

⁢

x

⁢

(

a

⁢

f

⁢

(

x

)

)

=

a

⁢

d

⁢

f

d

⁢

x

,

Power Rule

d

d

⁢

x

⁢

(

x

n

)

=

n

⁢

x

n

-

1

.

Product Rule

d

d

⁢

x

⁢

(

f

⁢

(

x

)

⁢

g

⁢

(

x

)

)

=

d

⁢

f

d

⁢

x

⁢

g

⁢

(

x

)

+

f

⁢

(

x

)

⁢

d

⁢

g

d

⁢

x

.

This form of a formal derivative applies to any UFD and so it also applies to

ℝ

. Thus it is possible to express polynomial calculus in terms of algebraic theory without any proper use of limits. This obscures many of the geometric

properties

such as the slope of a

tangent line

to a graph. However, computationally this technique outlines how

ε

,

δ

-limits are not required for the computation of derivatives.

Although abstract algebra, such as quotients of rings, are required to properly understand

R

⁢

[

x

,

h

]

/

(

h

)

, this approach still provides

elementary proofs

of derivative rules like the product rule. Although it is not necessary, to draw a distinct between

R

⁢

[

x

]

and

R

⁢

[

x

,

h

]

/

(

h

)

one may use

≡

when we consider the

expressions

in

R

⁢

[

x

,

h

]

/

(

h

)

if the distinction is clarifying.

1

Derivatives of rational functions

One may also generalize the derivative to apply to general rational function

f

⁢

(

x

)

∈

R

⁢

(

x

)

by observing

1

=

x

n

⁢

x

-

n

. Therefore

d

d

⁢

x

⁢

(

1

)

=

d

d

⁢

x

⁢

(

x

n

⁢

x

-

n

)

0

=

d

d

⁢

x

⁢

(

x

n

)

⁢

x

-

n

+

x

n

⁢

d

d

⁢

x

⁢

(

x

-

n

)

0

=

n

⁢

x

n

-

1

⁢

x

-

n

+

x

n

⁢

d

d

⁢

x

⁢

(

x

-

n

)

=

n

x

+

x

n

⁢

d

d

⁢

x

⁢

(

x

-

n

)

.

Now solve for

d

d

⁢

x

⁢

(

x

-

n

)

.

d

d

⁢

x

⁢

(

x

-

n

)

=

-

n

x

⁢

1

x

n

=

(

-

n

)

⁢

x

(

-

n

)

-

1

.

Thus we also derive the usual

quotient rule

:

d

d

⁢

x

⁢

(

f

g

)

=

f

′

⁢

(

x

)

⁢

g

⁢

(

x

)

-

f

⁢

(

x

)

⁢

g

′

⁢

(

x

)

g

⁢

(

x

)

2

.

Title

derivatives by pure algebra

Canonical

name

DerivativesByPureAlgebra

Date of creation

2013-03-22 15:59:37

Last modified on

2013-03-22 15:59:37

Owner

Algeboy (12884)

Last modified by

Algeboy (12884)

Numerical id

16

Author

Algeboy (12884)

Entry type

Definition

Classification

msc 26B05

Classification

msc 46G05

Classification

msc 26A24

Related topic

DerivativeOfXn

Related topic

AlternativeProofOfDerivativeOfXn

Related topic

DerivativeOfPolynomial

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