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Homerules of calculus for derivative of polynomial

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# 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 representation 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 $p^{{\prime}}$ is the unique element of $A[x]$ such that $p(x+y)=p(x)+y\,p^{{\prime}}(x)+y^{2}\,r[x,y]$ for some $r\in A[x,y]$

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 $p,q\in A[x]$, then $(p+q)^{{\prime}}=p^{{\prime}}+q^{{\prime}}$.

###### Proof.

Let us write $p(x+y)=p(x)+y\,p^{{\prime}}(x)+y^{2}\,r(x,y)$ and $q(x+y)=q(x)+y\,q^{{\prime}}(y)+y^{2}\,s(x,y)$. Adding, we have

$p(x,y)+q(x,y)=p(x)+q(x)+y(p^{{\prime}}(x)+q^{{\prime}}(x))+y^{2}(r(x,y)+s(x,y)).$ |

By definition of derivative, this means that $(p+q)^{{\prime}}=p^{{\prime}}+q^{{\prime}}$. ∎

###### Theorem 3.

If $A$ is a commutative ring and $p,q\in A[x]$, then $(p\cdot q)^{{\prime}}=p^{{\prime}}\cdot q+p\cdot q^{{\prime}}$.

###### Proof.

Let us write $p(x+y)=p(x)+y\,p^{{\prime}}(x)+y^{2}\,r(x,y)$ and $q(x+y)=q(x)+y\,q^{{\prime}}(y)+y^{2}\,s(x,y)$. Multiplying, grouping terms, and pulling out some common factors, we have

$\displaystyle p(x+y)q(x+y)$ | $\displaystyle=p(x)q(y)+y(p^{{\prime}}(x)q(x)+p(x)q^{{\prime}}(x))$ | ||

$\displaystyle+y^{2}(p(x)s(x,y)+q(x)r(x,y)+p^{{\prime}}(x)q^{{\prime}}(y)$ | |||

$\displaystyle\quad+y\,p^{{\prime}}(x)s(x,y)+y\,q^{{\prime}}(x)r(x,y)+y^{2}\,r(% x,y)s(x,y)).$ |

By definition of derivative, this means that $(p\cdot q)^{{\prime}}=p^{{\prime}}\cdot q+p\cdot q^{{\prime}}$. ∎

###### Theorem 4.

If $A$ is a commutative ring and $p,q\in A[x]$, then $(p\circ q)^{{\prime}}=(p^{{\prime}}\circ q)\cdot q^{{\prime}}$.

###### Proof.

Let us write $p(x+y)=p(x)+y\,p^{{\prime}}(x)+y^{2}\,r(x,y)$ and $q(x+y)=q(x)+y\,q^{{\prime}}(y)+y^{2}\,s(x,y)$. Composing, grouping terms, and pulling out some common factors, we have

$\displaystyle p(q(x+y))$ | $\displaystyle=p\left(q(x)+y\,q^{{\prime}}(y)+y^{2}\,s(x,y)\right)$ | ||

$\displaystyle=p(q(x))+\left(y\,q^{{\prime}}(y)+y^{2}\,s(x,y)\right)p^{{\prime}% }(q(x))$ | |||

$\displaystyle\quad+\left(y\,q^{{\prime}}(y)+y^{2}\,s(x,y)\right)^{2}r\left(q(x% ),y\,q^{{\prime}}(y)+y^{2}\,s(x,y)\right)$ | |||

$\displaystyle=p(q(x))+y\,p^{{\prime}}(q(x))q^{{\prime}}(y)$ | |||

$\displaystyle\quad+y^{2}\left(s(x,y)p^{{\prime}}(q(x))+\left(q^{{\prime}}(y)+y% \,s(x,y)\right)^{2}r\left(q(x),y\,q^{{\prime}}(y)+y^{2}\,s(x,y)\right)\right)$ |

By definition of derivative, this means that $(p\circ q)^{{\prime}}=(p^{{\prime}}\circ q)\cdot q^{{\prime}}$. ∎

## Mathematics Subject Classification

13P05*no label found*11C08

*no label found*12E05

*no label found*

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