If y = sin px and y_n is the n^{th} derivativ | vedclass

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(B) Given $y = \sin px$. The derivatives are:$y_1 = p \cos px$$y_2 = -p^2 \sin px$$y_3 = -p^3 \cos px$$y_4 = p^4 \sin px$$y_5 = p^5 \cos px$$y_6 = -p^

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$0$

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(B) Given $y = \sin px$. The derivatives are:$y_1 = p \cos px$$y_2 = -p^2 \sin px$$y_3 = -p^3 \cos px$$y_4 = p^4 \sin px$$y_5 = p^5 \cos px$$y_6 = -p^6 \sin px$$y_7 = -p^7 \cos px$$y_8 = p^8 \sin px$Substituting these into the determinant $D$:$D = \left| \begin{array}{ccc} \sin px & p \cos px & -p^2 \sin px \ -p^3 \cos px & p^4 \sin px & p^5 \cos px \ -p^6 \sin px & -p^7 \cos px & p^8 \sin px \end{array} ight|$Factoring out $p$ from the second column and $p^2$ from the third column:$D = p^3 \left| \begin{array}{ccc} \sin px & \cos px & -\sin px \ -p^3 \cos px & p^3 \sin px & p^3 \cos px \ -p^6 \sin px & -p^6 \cos px & p^6 \sin px \end{array} ight|$Factoring out $p^3$ from the second row and $p^6$ from the third row:$D = p^3 \cdot p^3 \cdot p^6 \left| \begin{array}{ccc} \sin px & \cos px & -\sin px \ -\cos px & \sin px & \cos px \ -\sin px & -\cos px & \sin px \end{array} ight|$Notice that the third row is the negative of the first row $(R_3 = -R_1)$.Since two rows are proportional,the value of the determinant is $0$.

If $y = \sin px$ and $y_n$ is the $n^{th}$ derivative of $y$,then $\left| \begin{array}{ccc} y & y_1 & y_2 \\ y_3 & y_4 & y_5 \\ y_6 & y_7 & y_8 \end{array} \right|$ is equal to

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