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 $A\mathit{}\mathrm{[}x\mathrm{]}$, then there exist unique polynomials $q$ and $A$ such that $p\mathit{}\mathrm{(}x\mathrm{+}y\mathrm{)}\mathrm{=}p\mathit{}\mathrm{(}x\mathrm{)}\mathrm{+}y\mathit{}q\mathit{}\mathrm{(}x\mathrm{)}\mathrm{+}{y}^{\mathrm{2}}\mathit{}r\mathit{}\mathrm{(}x\mathrm{,}y\mathrm{)}$.

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 $f(y)=yg(x,y)$ for some $g$ in $A[x,y]$. By definition of $f$, this means that $p(x+y)-p(x)=yg(x,y)$. 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 $h(y)=yr(x,y)$. Combining our definitions, we conclude that $p(x+y)=p(x)+yq(x)+y^{2}r(x,y)$.

We will now show uniqueness. Assume that there exists polymonomials $q,r,Q,R$ such that $p(x+y)=p(x)+yq(x)+y^{2}r(x,y)$ and $p(x+y)=p(x)+yQ(x)+y^{2}R(x,y)$. Subtracting and rearranging terms, $y(q(x)-Q(x))=y^2(R(x,y)-r(x,y))$. Cancelling $y$, 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 $A\mathit{}\mathrm{[}x\mathrm{]}$. Then $p^{\mathrm{\prime }}$ is the unique element of $A\mathit{}\mathrm{[}x\mathrm{]}$ such that $p\mathit{}\mathrm{(}x\mathrm{+}y\mathrm{)}\mathrm{=}p\mathit{}\mathrm{(}x\mathrm{)}\mathrm{+}y\mathit{}{p}^{\mathrm{\prime }}\mathit{}\mathrm{(}x\mathrm{)}\mathrm{+}{y}^{\mathrm{2}}\mathit{}r\mathit{}\mathrm{[}x\mathrm{,}y\mathrm{]}$ for some $r\mathrm{\in }A\mathit{}\mathrm{[}x\mathrm{,}y\mathrm{]}$

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

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

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

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

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

$\mathrm{\hspace{1em}}+y{p}^{\prime }(x)s(x,y)+y{q}^{\prime }(x)r(x,y)+y^{2}r(x,y)s(x,y)).$

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 $p\mathrm{,}q\mathrm{\in }A\mathit{}\mathrm{[}x\mathrm{]}$, then ${\mathrm{(}p\mathrm{\circ }q\mathrm{)}}^{\mathrm{\prime }}\mathrm{=}\mathrm{(}{p}^{\mathrm{\prime }}\mathrm{\circ }q\mathrm{)}\mathrm{\cdot }{q}^{\mathrm{\prime }}$.

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

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

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

$\mathrm{\hspace{1em}}+{\left(y{q}^{\prime }(y)+y^{2}s(x,y)\right)}^{2}r(q(x),y{q}^{\prime }(y)+y^{2}s(x,y))$

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

$\mathrm{\hspace{1em}}+y^2\left(s(x,y)p^{\prime }(q(x))+{\left(q^{\prime }(y)+ys(x,y)\right)}^{2}r(q(x),y{q}^{\prime }(y)+y^{2}s(x,y))\right)$

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

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