Question

(a) If $\mathit{f}\left(x\right)\mathbf{=}\mathit{c}{\mathit{x}}^{\mathbf{n}}\mathbf{\in }\mathit{F}\left[x\right]$ and $\mathit{g}\left(x\right)\mathbf{=}{\mathit{b}}_{\mathbf{0}}\mathbf{+}{\mathit{b}}_{\mathbf{1}}\mathit{x}\mathbf{+}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}{\mathit{b}}_{\mathbf{k}}{\mathit{x}}^{\mathbf{k}}\mathbf{\in }\mathit{F}\left[x\right]$ prove that ${\left(fg\right)}^{\mathbf{\text{'}}}\left(x\right)\mathbf{=}\mathit{f}\left(x\right){\mathit{g}}^{\mathbf{\text{'}}}\left(x\right)\mathbf{+}{\mathit{f}}^{\mathbf{\text{'}}}\left(x\right)\mathit{g}\left(x\right)$.

(b)If $\mathit{f}\left(x\right)\mathbf{,}\mathit{g}\left(x\right)$ are any polynomial in $F\left[x\right]$. Prove that ${\left(fg\right)}^{\mathbf{\text{'}}}\left(x\right)\mathbf{=}\mathit{f}\left(x\right){\mathit{g}}^{\mathbf{\text{'}}}\left(x\right)\mathbf{+}{\mathit{f}}^{\mathbf{\text{'}}}\left(x\right)\mathit{g}\left(x\right)$.

Short answer

It is proved that:

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

(b)${\left(fg\right)}^{\text{'}}\left(x\right)=f\left(x\right)g^{\text{'}}\left(x\right)+f^{\text{'}}\left(x\right)g\left(x\right)$ if $f\left(x\right),g\left(x\right)$ are polynomial

Step-by-step solution

Definition of polynomial 

A polynomial $p\left(x\right)$ over a given field K is separable if its root are distinct in an algebraic closure of K .

(a)Step2: Showing that fg'x=fxg'x+f'xgx

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

$\begin{array}{rcl}\left(fg\right)& =& c{x}^n+b_0+b_{1}x+.......b_n{x}^k\\ & =& \left(b_0\right)+\left(b_1\right)x+.......\left(c+b_k\right)x^{n+k}\\ & & \end{array}$

$$\begin{gathered} f^{\text{'}}\left(x\right)=nc{x}^{n-1} \\ {g}^{\text{'}}\left(x\right)=b_1+2b_{2}x+.......+k{b}_k{x}^{k-1} \end{gathered}$$

Show that${\left(fg\right)}^{\text{'}}\left(x\right)=f\left(x\right)g^{\text{'}}\left(x\right)+f^{\text{'}}\left(x\right)g\left(x\right)$,take left hand side and simplify as:

$\begin{array}{rcl}\left(fg\right)\left(x\right)& =& c{x}^n\left(b_0+b_{1}x+.......+b_k{x}^k\right)\\ & =& c{x}^n{b}_0+c{b}_{1}x{x}^n+c{b}_2{x}^2{x}^n+....c{b}_k{x}^{n+k}\\ & & \end{array}$

Take derivatives of the above:

$\begin{array}{rcl}{\left(fg\right)}^{\text{'}}\left(x\right)& =& \left(\left(c{b}_0\right)x^n+\left(c{b}_1\right)x^{n+1}+.....c{b}_k{x}^{n+k}\right)\\ & =& nc{b}_0{x}^{n-1}+\left(n+1\right)c{b}_1{x}^n+..........\left(n+k\right)c{b}_k{x}^{n+k-1}\\ & & \end{array}$

Consider right hand side of the claim:

$\begin{array}{rcl}f\left(x\right)g^{\text{'}}\left(x\right)+f^{\text{'}}\left(x\right)g\left(x\right)& =& c{x}^n\left(b_1+2b_{2}x+......k{b}_k{x}^{k-1}\right)+nc{x}^{n-1}\left(b_0+b_{1}x+b_2{x}^2+....+b_k{x}^k\right)\\ & =& c{b}_1{x}^n+2c{b}_2{x}^{n-1}+.......+kc{b}_k{x}^{n+k-1}+nc{b}_0{x}^{n-1}+nc{b}_1{x}^n+........+nc{b}_k{x}^{n+k-1}\\ & =& n\left(c{b}_0\right)x^{n-1}+..........+\left(n+k\right)\left(c{b}_k\right)x^{n+k-1}\\ & & \end{array}$

Since both the side of the claim are equal. Therefore,

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

 (b)Step3: Showing that  fg'x=fxg'x+f'xgx if  fx,gx are polynomial

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 $g\left(x\right)=b_0+b_{1}x+b_2{x}^2+.....+b_k{x}^k$..

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

$\begin{array}{rcl}\left(fg\right)\left(x\right)& =& \left(a_0+a_{1}x+.....+a_n{x}^n\right)\left(b_0+b_{1}x+.....+b_k{x}^k\right)\\ & =& \left(a_0{b}_0+a_0{b}_{1}x+.....+a_0{b}_k{x}^k\right)+\left(a_1{b}_{0}x+a_1{b}_1{x}^2+.....a_1{b}_1{x}^{k+1}\right)+...+\left(a_n{b}_k{x}^{k+1}\right)\\ & =& a_0{b}_0+\left(a_0{b}_1+a_1{b}_0\right)x+......+\left(a_n{b}_k{x}^{n+k}\right)\\ & & \end{array}$

Taking derivative of the above:

$f^{\text{'}}\left(x\right)=a_1+2a_{2}x+.....+n{a}_n{x}^{n-1}$

$\begin{array}{rcl}{\left(fg\right)}^{\text{'}}\left(x\right)& =& a_0{b}_0+\left(a_0{b}_1+a_1{b}_0\right)x+......+{\left(a_n{b}_k{x}^{n+k}\right)}^{\text{'}}\\ & =& \left(a_0{b}_1+a_1{b}_0\right)x+......+\left(a_n{b}_k{x}^{n+k-1}\right)\\ & & \end{array}$

Consider:

$\begin{array}{rcl}f\left(x\right)g^{\text{'}}\left(x\right)& =& \left(a_0+a_{1}x+....a_n{x}^n\right)\left(b_1+2b_{2}x+........k{b}_k{x}^{k-1}\right)\\ & =& a_0\left(b_1+2b_{2}x+......k{b}_k{x}^{k-1}\right)+a_{1}x\left(b_1+2b_{2}x+........+k{b}_k{x}^{k-1}\right)+...+a_n{x}^n\left(b_1+......+k{b}_k{x}^{k-1}\right)\\ & =& \left(a_0{b}_1+2\left(a_0{b}_2+a_1{b}_1\right)x+.....+k{a}_n{b}_n{x}^{n+k-1}\right)\\ & & \end{array}$

Now

$f^{\text{'}}\left(x\right)g\left(x\right)=\left(a_1+2a_{2}x+.....+n{a}_n{x}^{n-1}\right)\left(b_0+b_{1}x+.....+b_k{x}^k\right)$

Adding value of both$f\left(x\right)g^{\text{'}}\left(x\right)$and.$f^{\text{'}}\left(x\right)g\left(x\right)$ Therefore,

$\begin{array}{rcl}f\left(x\right)g^{\text{'}}\left(x\right)+f^{\text{'}}\left(x\right)g\left(x\right)& =& \left(a_0{b}_1+....+2\left(a_0{b}_2+a_1{b}_1\right)x+.....\right)+\left(a_1{b}_0+2\left(a_2{b}_0\right)x+.....+n{a}_n{b}_k{x}^{n+k-1}\right)\\ & =& a_0{b}_1+a_1{b}_0+2\left(a_2{b}_0+a_0{b}_2+a_1{b}_1\right)x+........+\left(n+k\right)a_n{b}_k{x}^{n+k-1}\\ & & \end{array}$

Thus both the sides of claim are equal. Therefore,

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