Question

If $\mathit{f}\left(x\right)\mathbf{,}\mathit{g}\left(x\right)\mathbf{\in }\mathit{F}\left[x\right]$ prove

(a)${\left(f+g\right)}^{\mathbf{\text{'}}}\left(x\right)\mathbf{=}{\mathit{f}}^{\mathbf{\text{'}}}\left(x\right)\mathbf{+}{\mathit{g}}^{\mathbf{\text{'}}}\left(x\right)$

(b)If $\mathit{c}\mathbf{\in }\mathit{F}$ then ${\left(cf\right)}^{\mathbf{\text{'}}}\left(x\right)\mathbf{=}\mathit{c}{\mathit{f}}^{\mathbf{\text{'}}}\left(x\right)$.

Short answer

It is proved that:

(a)${\left(f+g\right)}^{\text{'}}\left(x\right)=f^{\text{'}}\left(x\right)+g^{\text{'}}\left(x\right)$

(b)${\left(cf\right)}^{\text{'}}\left(x\right)=c{f}^{\text{'}}\left(x\right)$

Step-by-step solution

Definition of irreducible polynomial

An irreducible polynomial is a non constant polynomial that cannot be factors into the product of two non constant polynomials.

(a)Step2: Showing that  f+g'x=f'x+g'x

Let $f\left(x\right),g\left(x\right)\in F\left[x\right]$and $f\left(x\right)=a_0+a_{1}x+a_2{x}^2+.....+a_n{x}^n$and $g\left(x\right)=b_0+b_{1}x+b_2{x}^2+......b_m{x}^m$where $m<n$. Then

$f+g=\left(a_0+b_0\right)+\left(a_1+b_1\right)x+.......+\left(a_n+b_m\right)x^{m+n}$

As ${\left(f+g\right)}^{\text{'}}\left(x\right)=f^{\text{'}}\left(x\right)+g^{\text{'}}\left(x\right)$take left hand side and rewrite as:

$\begin{array}{rcl}{\left(f+g\right)}^{\text{'}}\left(x\right)& =& \left(\left(a_0+a_{1}x+a_2{x}^2+....a_n{x}^n\right)+\left(b_0+b_{1}x+.....b_m{x}^m\right)\right)\\ & =& \left(a_0+b_0\right)+\left(a_1+b_1\right)x+.....\left(a_n+b_m\right)x^{m+n}\\ & =& \left(a_1+2x{a}_2+......+\left(n+m\right)a_n{x}^{n+m-1}\right)+\left(b_1+2x{b}_2+....\left(n+m\right)b_n{x}^{m+n-1}\right)\\ & & \end{array}$

Simplify further:

$\left(a_1+2x{a}_2+......+\left(n+m\right)a_n{x}^{n+m-1}\right)+\left(b_1+2x{b}_2+....\left(n+m\right)b_n{x}^{m+n-1}\right)=f^{\text{'}}\left(x\right)+g^{\text{'}}\left(x\right)$

This gives:

${\left(f+g\right)}^{\text{'}}\left(x\right)=f^{\text{'}}\left(x\right)+g^{\text{'}}\left(x\right)$

(b)Step 3: Showing that  cf'x=cf'x

Let $c\in F$and show that ${\left(cf\right)}^{\text{'}}\left(x\right)=c{f}^{\text{'}}\left(x\right)$

Now take left hand side and rewrite as:

$\begin{array}{rcl}{\left(cf\right)}^{\text{'}}& =& c{\left(a_0+a_{1}x+......+a_n{x}^n\right)}^{\text{'}}\\ & =& \left(c{a}_0+c{a}_{1}x+......+c{a}_n{x}^n\right)\\ & =& c\left(a_1+2x{a}_2+.......n{a}_n{x}^{n-1}\right)\\ & =& c{f}^{\text{'}}\left(x\right)\\ & & \end{array}$

This gives ${\left(cf\right)}^{\text{'}}\left(x\right)=c{f}^{\text{'}}\left(x\right)$.

It is proved that:

(a)${\left(f+g\right)}^{\text{'}}\left(x\right)=f^{\text{'}}\left(x\right)+g^{\text{'}}\left(x\right)$

(b)${\left(cf\right)}^{\text{'}}\left(x\right)=c{f}^{\text{'}}\left(x\right)$