rules of calculus for derivative of polynomial

rules of calculus for derivative of polynomial

In this entry, we will derive the properties of

derivatives
of polynomials

in a

rigorous

fashion. We begin by showing
that the

derivative

exists.

Theorem 1

.

If

A

is a

commutative ring

and

p

is a

polynomial

in

A

⁢

[

x

]

, then there exist unique polynomials

q

and

A

such that

p

⁢

(

x

+

y

)

=

p

⁢

(

x

)

+

y

⁢

q

⁢

(

x

)

+

y

2

⁢

r

⁢

(

x

,

y

)

.

Proof.

We will first show

existence

, then uniqueness. Define

f

⁢

(

y

)

=

p

⁢

(

x

+

y

)

-

p

⁢

(

x

)

. Since

f

is a polynomial in

y

with
coefficients in the ring

A

⁢

[

x

]

and

f

⁢

(

0

)

=

0

, we must have

y

be a factor of

f

⁢

(

y

)

, so

f

⁢

(

y

)

=

y

⁢

g

⁢

(

x

,

y

)

for some

g

in

A

⁢

[

x

,

y

]

. By definition of

f

, this means that

p

⁢

(

x

+

y

)

-

p

⁢

(

x

)

=

y

⁢

g

⁢

(

x

,

y

)

.

1

1

We are here making use of the
identification of

A

⁢

[

x

]

⁢

[

y

]

with

A

⁢

[

x

,

y

]

to write the
polynomial

g

either as a polynomial in

y

with coefficients
in

A

⁢

[

x

]

or as a polynomial in

x

and

y

with coefficients
in

A

.

Define

q

⁢

(

x

)

=

g

⁢

(

x

,

0

)

and

h

⁢

(

x

,

y

)

=

g

⁢

(

x

,

y

)

-

g

⁢

(

x

,

0

)

.
Regarding

h

as a polynomial in

y

with coefficients in

A

⁢

[

x

]

, we may, similiarly to what we did earlier, note that,
since

h

⁢

(

0

)

=

0

by construction,

y

must be a factor of

h

⁢

(

y

)

.
Hence there exists a polynomial

r

with coefficients in

A

⁢

[

x

,

y

]

such that

h

⁢

(

y

)

=

y

⁢

r

⁢

(

x

,

y

)

. Combining our definitions, we
conclude that

p

⁢

(

x

+

y

)

=

p

⁢

(

x

)

+

y

⁢

q

⁢

(

x

)

+

y

2

⁢

r

⁢

(

x

,

y

)

.

We will now show uniqueness. Assume that there
exists polymonomials

q

,

r

,

Q

,

R

such that

p

⁢

(

x

+

y

)

=

p

⁢

(

x

)

+

y

⁢

q

⁢

(

x

)

+

y

2

⁢

r

⁢

(

x

,

y

)

and

p

⁢

(

x

+

y

)

=

p

⁢

(

x

)

+

y

⁢

Q

⁢

(

x

)

+

y

2

⁢

R

⁢

(

x

,

y

)

. Subtracting and rearranging terms,

y

⁢

(

q

⁢

(

x

)

-

Q

⁢

(

x

)

)

=

y

2

⁢

(

R

⁢

(

x

,

y

)

-

r

⁢

(

x

,

y

)

)

. Cancelling

y

2

2

Note that, in
general, the

cancellation law

need not hold. However, even if

A

has

divisors

of zero, it still will be the case that the
polynomial

y

cannot divide zero, so we may cancel it.

,
we have

q

⁢

(

x

)

-

Q

⁢

(

x

)

=

y

⁢

(

R

⁢

(

x

,

y

)

-

r

⁢

(

x

,

y

)

)

. Substituting

0

for

y

, we have

q

⁢

(

x

)

-

Q

⁢

(

x

)

=

0

. Replacing this in our
equation,

y

⁢

(

R

⁢

(

x

,

y

)

-

r

⁢

(

x

,

y

)

)

=

0

. Cancelling another

y

,

R

⁢

(

x

,

y

)

-

r

⁢

(

x

,

y

)

=

0

. Hence, we conclude that

Q

=

q

and

R

=

r

, so our is unique.
∎

Hence, the following is well-defined:

Definition 1

.

Let

A

be a commutative ring and let

p

be polynomial in

A

⁢

[

x

]

. Then

p

′

is the unique element of

A

⁢

[

x

]

such that

p

⁢

(

x

+

y

)

=

p

⁢

(

x

)

+

y

⁢

p

′

⁢

(

x

)

+

y

2

⁢

r

⁢

[

x

,

y

]

for some

r

∈

A

⁢

[

x

,

y

]

We will now derive some of the rules for manipulating
derivatives familiar form calculus for polynomials using
purely algebraic operations with no limits involved.

Theorem 2

.

If

A

is a commutative ring and

p

,

q

∈

A

⁢

[

x

]

, then

(

p

+

q

)

′

=

p

′

+

q

′

.

Proof.

Let us write

p

⁢

(

x

+

y

)

=

p

⁢

(

x

)

+

y

⁢

p

′

⁢

(

x

)

+

y

2

⁢

r

⁢

(

x

,

y

)

and

q

⁢

(

x

+

y

)

=

q

⁢

(

x

)

+

y

⁢

q

′

⁢

(

y

)

+

y

2

⁢

s

⁢

(

x

,

y

)

. Adding,
we have

p

⁢

(

x

,

y

)

+

q

⁢

(

x

,

y

)

=

p

⁢

(

x

)

+

q

⁢

(

x

)

+

y

⁢

(

p

′

⁢

(

x

)

+

q

′

⁢

(

x

)

)

+

y

2

⁢

(

r

⁢

(

x

,

y

)

+

s

⁢

(

x

,

y

)

)

.

By definition of derivative, this means that

(

p

+

q

)

′

=

p

′

+

q

′

.
∎

Theorem 3

.

If

A

is a commutative ring and

p

,

q

∈

A

⁢

[

x

]

, then

(

p

⋅

q

)

′

=

p

′

⋅

q

+

p

⋅

q

′

.

Proof.

Let us write

p

⁢

(

x

+

y

)

=

p

⁢

(

x

)

+

y

⁢

p

′

⁢

(

x

)

+

y

2

⁢

r

⁢

(

x

,

y

)

and

q

⁢

(

x

+

y

)

=

q

⁢

(

x

)

+

y

⁢

q

′

⁢

(

y

)

+

y

2

⁢

s

⁢

(

x

,

y

)

. Multiplying,
grouping terms, and pulling out some common factors, we have

p

⁢

(

x

+

y

)

⁢

q

⁢

(

x

+

y

)

=

p

⁢

(

x

)

⁢

q

⁢

(

y

)

+

y

⁢

(

p

′

⁢

(

x

)

⁢

q

⁢

(

x

)

+

p

⁢

(

x

)

⁢

q

′

⁢

(

x

)

)

+

y

2

(

p

(

x

)

s

(

x

,

y

)

+

q

(

x

)

r

(

x

,

y

)

+

p

′

(

x

)

q

′

(

y

)

+

y

p

′

(

x

)

s

(

x

,

y

)

+

y

q

′

(

x

)

r

(

x

,

y

)

+

y

2

r

(

x

,

y

)

s

(

x

,

y

)

)

.

By definition of derivative, this means that

(

p

⋅

q

)

′

=

p

′

⋅

q

+

p

⋅

q

′

.
∎

Theorem 4

.

If

A

is a commutative ring and

p

,

q

∈

A

⁢

[

x

]

, then

(

p

∘

q

)

′

=

(

p

′

∘

q

)

⋅

q

′

.

Proof.

Let us write

p

⁢

(

x

+

y

)

=

p

⁢

(

x

)

+

y

⁢

p

′

⁢

(

x

)

+

y

2

⁢

r

⁢

(

x

,

y

)

and

q

⁢

(

x

+

y

)

=

q

⁢

(

x

)

+

y

⁢

q

′

⁢

(

y

)

+

y

2

⁢

s

⁢

(

x

,

y

)

. Composing,
grouping terms, and pulling out some common factors, we have

p

⁢

(

q

⁢

(

x

+

y

)

)

=

p

⁢

(

q

⁢

(

x

)

+

y

⁢

q

′

⁢

(

y

)

+

y

2

⁢

s

⁢

(

x

,

y

)

)

=

p

⁢

(

q

⁢

(

x

)

)

+

(

y

⁢

q

′

⁢

(

y

)

+

y

2

⁢

s

⁢

(

x

,

y

)

)

⁢

p

′

⁢

(

q

⁢

(

x

)

)

+

(

y

⁢

q

′

⁢

(

y

)

+

y

2

⁢

s

⁢

(

x

,

y

)

)

2

⁢

r

⁢

(

q

⁢

(

x

)

,

y

⁢

q

′

⁢

(

y

)

+

y

2

⁢

s

⁢

(

x

,

y

)

)

=

p

⁢

(

q

⁢

(

x

)

)

+

y

⁢

p

′

⁢

(

q

⁢

(

x

)

)

⁢

q

′

⁢

(

y

)

+

y

2

⁢

(

s

⁢

(

x

,

y

)

⁢

p

′

⁢

(

q

⁢

(

x

)

)

+

(

q

′

⁢

(

y

)

+

y

⁢

s

⁢

(

x

,

y

)

)

2

⁢

r

⁢

(

q

⁢

(

x

)

,

y

⁢

q

′

⁢

(

y

)

+

y

2

⁢

s

⁢

(

x

,

y

)

)

)

By definition of derivative, this means that

(

p

∘

q

)

′

=

(

p

′

∘

q

)

⋅

q

′

.
∎

Title

rules of calculus for derivative of polynomial

Canonical name

RulesOfCalculusForDerivativeOfPolynomial

Date of creation

2013-03-22 18:20:05

Last modified on

2013-03-22 18:20:05

Owner

rspuzio (6075)

Last modified by

rspuzio (6075)

Numerical id

10

Author

rspuzio (6075)

Entry type

Derivation

Classification

msc 13P05

Classification

msc 11C08

Classification

msc 12E05

Related topic

ProofOfPropertiesOfDerivativesByPureAlgebra

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