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(D) Given $y = \sin(mx)$. The $n$-th derivative is $y_n = m^n \sin(mx + \frac{n\pi}{2})$.Let the determinant be $\Delta = \left| \begin{array}{ccc} y

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independent of $m$ and $x$

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(D) Given $y = \sin(mx)$. The $n$-th derivative is $y_n = m^n \sin(mx + \frac{n\pi}{2})$.Let the determinant be $\Delta = \left| \begin{array}{ccc} y & y_1 & y_2 \ y_3 & y_4 & y_5 \ y_6 & y_7 & y_8 \end{array} ight|$.We observe the derivatives:$y_0 = \sin(mx)$$y_1 = m \cos(mx)$$y_2 = -m^2 \sin(mx)$$y_3 = -m^3 \cos(mx)$$y_4 = m^4 \sin(mx)$$y_5 = m^5 \cos(mx)$$y_6 = -m^6 \sin(mx)$$y_7 = -m^7 \cos(mx)$$y_8 = m^8 \sin(mx)$Notice that $y_2 = -m^2 y_0$,$y_3 = -m^2 y_1$,$y_4 = -m^2 y_2$,$y_5 = -m^2 y_3$,$y_6 = -m^2 y_4$,$y_7 = -m^2 y_5$,$y_8 = -m^2 y_6$.Since each column is a scalar multiple of the previous column (or related by a constant factor $-m^2$),the columns are linearly dependent.Specifically,$C_2 = \frac{y_1}{y} C_1$ is not constant,but the ratio of consecutive derivatives $y_{n+1}/y_n$ is constant for specific patterns. More simply,the determinant of any matrix where rows/columns are linearly dependent is $0$.Since the value is $0$,it is independent of both $m$ and $x$.

If $y = \sin(mx)$,then the value of $\left| \begin{array}{ccc} y & y_1 & y_2 \\ y_3 & y_4 & y_5 \\ y_6 & y_7 & y_8 \end{array} \right|$ (where subscripts of $y$ denote the order of derivative) is:

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