If `f(x)=|{:(x^(3),x^(4),3x^(2)),(1,-6,4),(p,p^(2),p^(3)):}|`, where p is a constant, then `(d^(3))/(dx^(3))(f(x))`, is

Options

A. proportional to `x^(2)`
B. proportional to x
C. proportional to `x^(3)`
D. a constant

Answer

To solve the problem, we need to find the third derivative of the function \( f(x) \) defined as the determinant of the following matrix:

\[ f(x) = \begin{vmatrix} x^3 & 1 & p \\ x^4 & -6 & p^2 \\ 3x^2 & 4 & p^3 \end{vmatrix} \]

### Step 1: Calculate the Determinant

To compute the determinant, we can use the method of cofactor expansion. We can expand along the first row:

\[ f(x) = x^3 \begin{vmatrix} -6 & p^2 \\ 4 & p^3 \end{vmatrix} - 1 \begin{vmatrix} x^4 & p^2 \\ 3x^2 & p^3 \end{vmatrix} + p \begin{vmatrix} x^4 & -6 \\ 3x^2 & 4 \end{vmatrix} \]

Calculating each of these 2x2 determinants:

1. For the first determinant:
\[ \begin{vmatrix} -6 & p^2 \\ 4 & p^3 \end{vmatrix} = (-6)(p^3) - (p^2)(4) = -6p^3 - 4p^2 \]

2. For the second determinant:
\[ \begin{vmatrix} x^4 & p^2 \\ 3x^2 & p^3 \end{vmatrix} = (x^4)(p^3) - (p^2)(3x^2) = x^4p^3 - 3p^2x^2 \]

3. For the third determinant:
\[ \begin{vmatrix} x^4 & -6 \\ 3x^2 & 4 \end{vmatrix} = (x^4)(4) - (-6)(3x^2) = 4x^4 + 18x^2 \]

Now substituting back into the determinant:

\[ f(x) = x^3(-6p^3 - 4p^2) - (x^4p^3 - 3p^2x^2) + p(4x^4 + 18x^2) \]

### Step 2: Simplify \( f(x) \)

Combining the terms:

\[ f(x) = -6p^3x^3 - 4p^2x^3 - x^4p^3 + 3p^2x^2 + 4px^4 + 18px^2 \]

Rearranging gives:

\[ f(x) = (-x^4p^3 + 4px^4) + (-6p^3x^3 - 4p^2x^3) + (3p^2x^2 + 18px^2) \]

### Step 3: Differentiate \( f(x) \)

Now we differentiate \( f(x) \) three times.

1. **First Derivative \( f'(x) \)**:
Using the power rule:
\[ f'(x) = \frac{d}{dx}(-p^3x^4 + 4px^4) + \frac{d}{dx}(-6p^3x^3 - 4p^2x^3) + \frac{d}{dx}(3p^2x^2 + 18px^2) \]
Calculating gives:
\[ f'(x) = (-4p^3 + 4p)x^3 + (-18p^3 - 12p^2)x^2 + (6p^2 + 36p)x \]

2. **Second Derivative \( f''(x) \)**:
Differentiating \( f'(x) \):
\[ f''(x) = (-12p^3 + 12p)x^2 + (-36p^3 - 24p^2)x + (6p^2 + 36p) \]

3. **Third Derivative \( f'''(x) \)**:
Differentiating \( f''(x) \):
\[ f'''(x) = (-24p^3 + 24p)x + (-36p^3 - 24p^2) \]

### Final Result

Thus, the third derivative \( f'''(x) \) is:

\[ f'''(x) = (-24p^3 + 24p)x - (36p^3 + 24p^2) \]

### Conclusion

The final expression shows that \( f'''(x) \) is a linear function in \( x \), indicating that it is proportional to \( x \).