In Problems$\mathbf{8}\mathbf{}\mathbf{to}\mathbf{}\mathbf{15}$,use $\mathbf{(}\mathbf{8}\mathbf{.}\mathbf{5}\mathbf{)}$ to show that the given functions are linearly independent.

$\mathbf{1}\mathbf{,}{\mathbf{x}}^{\mathbf{2}}\mathbf{,}{\mathbf{x}}^{\mathbf{4}}\mathbf{,}{\mathbf{x}}^{\mathbf{6}}$

The function${\mathbf{f}}_{\mathbf{1}}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{,}\mathbf{}{\mathbf{f}}_{\mathbf{2}}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{,}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{,}{\mathbf{f}}_{\mathbf{n}}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{}$are linearly independent if the determinant

$\mathbf{W}\mathbf{=}\mathbf{\left|}\begin{array}{cccc}{\mathbf{f}}_{\mathbf{1}}\mathbf{\left(}\mathbf{x}\mathbf{\right)}& {\mathbf{f}}_{\mathbf{2}}\mathbf{\left(}\mathbf{x}\mathbf{\right)}& \mathbf{\cdots }& {\mathbf{f}}_{\mathbf{n}}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\\ {\mathbf{f}}_{\mathbf{1}}^{\mathbf{\prime }}\mathbf{\left(}\mathbf{x}\mathbf{\right)}& {\mathbf{f}}_{\mathbf{2}}^{\mathbf{\prime }}\mathbf{\left(}\mathbf{x}\mathbf{\right)}& \mathbf{\cdots }& {\mathbf{f}}_{\mathbf{n}}^{\mathbf{\prime }}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\\ \mathbf{⋮}& \mathbf{⋮}& \mathbf{\ddots }& \mathbf{⋮}\\ {\mathbf{f}}_{\mathbf{1}}^{\mathbf{(}\mathbf{n}\mathbf{-}\mathbf{1}\mathbf{)}}\mathbf{\left(}\mathbf{x}\mathbf{\right)}& {\mathbf{f}}_{\mathbf{2}}^{\mathbf{(}\mathbf{n}\mathbf{-}\mathbf{1}\mathbf{)}}\mathbf{\left(}\mathbf{x}\mathbf{\right)}& \mathbf{\cdots }& {\mathbf{f}}_{\mathbf{n}}^{\mathbf{(}\mathbf{n}\mathbf{-}\mathbf{1}\mathbf{)}}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\end{array}\mathbf{\right|}\mathbf{\ne }\mathbf{0}$

Here, W is called the Wronksian of functions.

Find the derivatives of the function ${\mathrm{f}}_1\left(\mathrm{x}\right)=1$of order $3.$

${\mathrm{f}}_1\left(\mathrm{x}\right)=1$

${{\mathrm{f}}_1}^{\text{'}}\left(\mathrm{x}\right)=0$

${{\mathrm{f}}_1}^{\text{'}\text{'}}\left(\mathrm{x}\right)=0$

${{\mathrm{f}}_1}^{\text{'}\text{'}\text{'}}\left(\mathrm{x}\right)=0$

Find the derivatives of the function ${\mathrm{f}}_2\left(\mathrm{x}\right)={\mathrm{x}}^2$of order $3.$

${\mathrm{f}}_2\left(\mathrm{x}\right)={\mathrm{x}}^2$

${{\mathrm{f}}_2}^{\text{'}}\left(\mathrm{x}\right)=2\mathrm{x}$

${{\mathrm{f}}_2}^{\text{'}\text{'}}\left(\mathrm{x}\right)=2$

${{\mathrm{f}}_2}^{\text{'}\text{'}\text{'}}\left(\mathrm{x}\right)=0$

Find the derivatives of the function ${\mathrm{f}}_3\left(\mathrm{x}\right)={\mathrm{x}}^4$of order $3.$

${\mathrm{f}}_3\left(\mathrm{x}\right)={\mathrm{x}}^4$

${{\mathrm{f}}_3}^{\text{'}}\left(\mathrm{x}\right)=4{\mathrm{x}}^3$

${{\mathrm{f}}_3}^{\text{'}\text{'}}\left(\mathrm{x}\right)=12{\mathrm{x}}^2$

${{\mathrm{f}}_3}^{\text{'}\text{'}\text{'}}\left(\mathrm{x}\right)=24\mathrm{x}$

Find the derivatives of the function ${\mathrm{f}}_4\left(\mathrm{x}\right)={\mathrm{x}}^6$of order $3.$

${\mathrm{f}}_4\left(\mathrm{x}\right)={\mathrm{x}}^6$

${{\mathrm{f}}_4}^{\text{'}}\left(\mathrm{x}\right)=6{\mathrm{x}}^5$

${{\mathrm{f}}_4}^{\text{'}\text{'}}\left(\mathrm{x}\right)=30{\mathrm{x}}^4$

${{\mathrm{f}}_4}^{\text{'}\text{'}\text{'}}\left(\mathrm{x}\right)=120{\mathrm{x}}^3$

Substitute the derivatives in the Wronksian formula and simplify as follows:

$\mathrm{W}=\left|\begin{array}{cccc}1& {\mathrm{x}}^2& {\mathrm{x}}^4& {\mathrm{x}}^5\\ 0& 2\mathrm{x}& 4{\mathrm{x}}^3& 6{\mathrm{x}}^5\\ 0& 2& 12{\mathrm{x}}^2& 30{\mathrm{x}}^4\\ 0& 0& 24\mathrm{x}& 120{\mathrm{x}}^3\end{array}\right|$

$=\left|\begin{array}{ccc}2x& 4x^3& 6x^5\\ 2& 12x^2& 30x^4\\ 0& 24x& 120x^3\end{array}\right|$

$=2\mathrm{x}\left|\begin{array}{cc}12{\mathrm{x}}^2& 30{\mathrm{x}}^4\\ 24\mathrm{x}& 120{\mathrm{x}}^3\end{array}\right|-2\left|\begin{array}{cc}4{\mathrm{x}}^3& 6{\mathrm{x}}^5\\ 24\mathrm{x}& 120{\mathrm{x}}^3\end{array}\right|$

$=2\mathrm{x}(1440{\mathrm{x}}^5-720{\mathrm{x}}^5)-2(480{\mathrm{x}}^6-144{\mathrm{x}}^6)$

Solve the determinant as follows:

$\mathrm{W}=1440{\mathrm{x}}^6-672{\mathrm{x}}^6$

$=768{\mathrm{x}}^6$

Her, $\mathrm{W}\ne 0.$Therefore, the functions$1,{\mathrm{x}}^2,{\mathrm{x}}^4,{\mathrm{x}}^6$are linearly independent.