Understanding the Problem: Differentiating a Determinant Function

The question asks us to find the value of the derivative of a given function \({\rm{f}}\left( {\rm{x}} \right)\) at \({\rm{x}} = 0\). The function \({\rm{f}}\left( {\rm{x}} \right)\) is defined as the determinant of a 3x3 matrix whose elements are functions of \({\rm{x}}\).

The given function is:

Here, \({\rm{p}}\) is a constant.

Calculating the Derivative of a Determinant

To find the derivative of a determinant where the elements are functions of \({\rm{x}}\), we can use a property that is similar to the product rule of differentiation. The derivative of the determinant is the sum of determinants obtained by differentiating one row (or column) at a time while keeping the other rows (or columns) unchanged.

For a 3x3 determinant, the derivative \({\rm{f}}'\left( {\rm{x}} \right)\) is given by:

Let's find the derivatives of the elements:

• \(\frac{{\rm{d}}}{{{\rm{dx}}}}\left( {{{\rm{x}}^3}} \right) = 3{{\rm{x}}^2}\)
• \(\frac{{\rm{d}}}{{{\rm{dx}}}}\left( {\sin {\rm{x}}} \right) = \cos {\rm{x}}\)
• \(\frac{{\rm{d}}}{{{\rm{dx}}}}\left( {\cos {\rm{x}}} \right) = - \sin {\rm{x}}\)
• \(\frac{{\rm{d}}}{{{\rm{dx}}}}\left( 6 \right) = 0\) (derivative of a constant is 0)
• \(\frac{{\rm{d}}}{{{\rm{dx}}}}\left( { - 1} \right) = 0\)
• \(\frac{{\rm{d}}}{{{\rm{dx}}}}\left( 0 \right) = 0\)
• \(\frac{{\rm{d}}}{{{\rm{dx}}}}\left( {\rm{p}} \right) = 0\) (derivative of a constant \({\rm{p}}\) is 0)
• \(\frac{{\rm{d}}}{{{\rm{dx}}}}\left( {{{\rm{p}}^2}} \right) = 0\)
• \(\frac{{\rm{d}}}{{{\rm{dx}}}}\left( {{{\rm{p}}^3}} \right) = 0\)

Substitute these derivatives back into the expression for \({\rm{f}}'\left( {\rm{x}} \right)\):

Evaluating f'(0)

Now we need to find the value of \({\rm{f}}'\left( {\rm{x}} \right)\) at \({\rm{x}} = 0\). We substitute \({\rm{x}} = 0\) into the expression for \({\rm{f}}'\left( {\rm{x}} \right)\).

Recall the values of the terms at \({\rm{x}} = 0\):

• \(3{{\rm{x}}^2}\) becomes \(3(0)^2 = 0\)
• \({{\rm{x}}^3}\) becomes \(0^3 = 0\)
• \(\cos {\rm{x}}\) becomes \(\cos(0) = 1\)
• \(\sin {\rm{x}}\) becomes \(\sin(0) = 0\)

Substitute these values:

Let's evaluate each determinant:

First Determinant: \(\left| {\begin{array}{*{20}{c}} 0&1&0\\ 6&{ - 1}&0\\ {\rm{p}}&{{{\rm{p}}^2}}&{{{\rm{p}}^3}} \end{array}} \right|\)

Expand along the first row:

Alternatively, expand along the third column (since it has two zeros):

The value of the first determinant is \(-6{\rm{p}}^3\).

Second Determinant: \(\left| {\begin{array}{*{20}{c}} 0&0&1\\ 0&0&0\\ {\rm{p}}&{{{\rm{p}}^2}}&{{{\rm{p}}^3}} \end{array}} \right|\)

This determinant has a row of zeros (the second row). A property of determinants states that if a matrix has a row or column consisting entirely of zeros, its determinant is 0.

The value of the second determinant is 0.

Third Determinant: \(\left| {\begin{array}{*{20}{c}} 0&0&1\\ 6&{ - 1}&0\\ 0&0&0 \end{array}} \right|\)

This determinant also has a row of zeros (the third row). Therefore, its determinant is 0.

The value of the third determinant is 0.

Now, sum the values of the three determinants to find \({\rm{f}}'\left( 0 \right)\):

The value of f'(0) is \(-6p^3\).

Revision Table: Key Calculus and Matrix Concepts

Additional Information: Differentiation of Matrix Functions

While the problem specifically deals with the determinant of a matrix function, the concept of differentiating matrices can be broader. If you have a matrix \({\bf{A}}({\rm{x}})\) whose elements are functions of \({\rm{x}}\), the derivative of the matrix \({\bf{A}}'({\rm{x}})\) is simply the matrix obtained by differentiating each element with respect to \({\rm{x}}\). For example, if \({\bf{A}}({\rm{x}}) = \left[ {\begin{array}{*{20}{c}} {{{\rm{a}}_{11}}\left( {\rm{x}} \right)}&{{{\rm{a}}_{12}}\left( {\rm{x}} \right)}\\ {{{\rm{a}}_{21}}\left( {\rm{x}} \right)}&{{{\rm{a}}_{22}}\left( {\rm{x}} \right)} \end{array}} \right]\), then \({\bf{A}}'({\rm{x}}) = \left[ {\begin{array}{*{20}{c}} {{{\rm{a}}'_{11}}\left( {\rm{x}} \right)}&{{{\rm{a}}'_{12}}\left( {\rm{x}} \right)}\\ {{{\rm{a}}'_{21}}\left( {\rm{x}} \right)}&{{{\rm{a}}'_{22}}\left( {\rm{x}} \right)} \end{array}} \right]\). However, the derivative of the determinant is calculated using the specific rule involving the sum of determinants, as shown in this problem.