If y= sin px and `y_(n)` is the nth derivative of y, then
`|{:(y,y_(1),y_(2)),(y_(3),y_(4),y_(5)),(y_(6),y_(7),y_(8)):}|`is

Options

A. `m^(9)`
B. `m^(2)`
C. `m^(3)`
D. none of these

Answer

To solve the given problem, we need to find the determinant of the matrix formed by \(y\) and its derivatives up to the 8th order, where \(y = \sin(px)\) and \(y_n\) is the nth derivative of \(y\).

### Step-by-Step Solution:

1. **Define the function and its derivatives**:
- Let \(y = \sin(px)\).
- The derivatives of \(y\) can be computed as follows:
- \(y' = \frac{dy}{dx} = p \cos(px)\)
- \(y'' = \frac{d^2y}{dx^2} = -p^2 \sin(px)\)
- \(y^{(3)} = \frac{d^3y}{dx^3} = -p^3 \cos(px)\)
- \(y^{(4)} = \frac{d^4y}{dx^4} = p^4 \sin(px)\)
- \(y^{(5)} = \frac{d^5y}{dx^5} = p^5 \cos(px)\)
- \(y^{(6)} = \frac{d^6y}{dx^6} = -p^6 \sin(px)\)
- \(y^{(7)} = \frac{d^7y}{dx^7} = -p^7 \cos(px)\)
- \(y^{(8)} = \frac{d^8y}{dx^8} = p^8 \sin(px)\)

2. **Construct the determinant**:
- We need to evaluate the determinant:
\[ D = \begin{vmatrix} y & y' & y'' \\ y^{(3)} & y^{(4)} & y^{(5)} \\ y^{(6)} & y^{(7)} & y^{(8)} \end{vmatrix} \]
Substituting the derivatives we calculated:
\[ D = \begin{vmatrix} \sin(px) & p \cos(px) & -p^2 \sin(px) \\ -p^3 \cos(px) & p^4 \sin(px) & p^5 \cos(px) \\ -p^6 \sin(px) & -p^7 \cos(px) & p^8 \sin(px) \end{vmatrix} \]

3. **Factor out common terms**:
- From the second row, we can factor out \(p^3\):
\[ D = p^3 \begin{vmatrix} \sin(px) & p \cos(px) & -p^2 \sin(px) \\ -\cos(px) & p \sin(px) & p^2 \cos(px) \\ -p^3 \sin(px) & -p^4 \cos(px) & p^5 \sin(px) \end{vmatrix} \]
- From the third row, we can factor out \(-p^6\):
\[ D = -p^9 \begin{vmatrix} \sin(px) & p \cos(px) & -p^2 \sin(px) \\ -\cos(px) & p \sin(px) & p^2 \cos(px) \\ \sin(px) & p \cos(px) & -p^2 \sin(px) \end{vmatrix} \]

4. **Row transformation**:
- Now, we can perform a row operation \(R_1 + R_3\) to simplify the determinant:
\[ D = -p^9 \begin{vmatrix} 0 & 0 & 0 \\ -\cos(px) & p \sin(px) & p^2 \cos(px) \\ \sin(px) & p \cos(px) & -p^2 \sin(px) \end{vmatrix} \]

5. **Evaluate the determinant**:
- Since the first row is all zeros, the determinant evaluates to zero:
\[ D = 0 \]

### Final Result:
Thus, the value of the determinant is:
\[ \boxed{0} \]