\({\rm{f}}\left( {\rm{x}} \right) = \left| {\begin{array}{*{20}{c}} {{{\rm{x}}^3}}&{\sin {\rm{x}}}&{\cos {\rm{x}}}\\ 6&{ - 1}&0\\ {\rm{p}}&{{{\rm{p}}^2}}&{{{\rm{p}}^3}} \end{array}} \right|\), where p is a constant

What is the value of f’(0)?

Concept:

Differentiation of the determinant of a matrix: Differentiate the first row (or column) and keep the entries of the other row untouched. Do this for all other rows (or columns). Then find the sum of all such determinants. Then you can get a determinant.

Calculation:

Given:

\({\rm{f}}\left( {\rm{x}} \right) = \left| {\begin{array}{*{20}{c}} {{{\rm{x}}^3}}&{\sin {\rm{x}}}&{\cos {\rm{x}}}\\ 6&{ - 1}&0\\ {\rm{p}}&{{{\rm{p}}^2}}&{{{\rm{p}}^3}} \end{array}} \right|\)

Differentiate with respect to x, we get

 \( \Rightarrow {\rm{f'}}\left( {\rm{x}} \right) = \left| {\begin{array}{*{20}{c}} {\frac{{{\rm{d}}\left( {{{\rm{x}}^3}} \right)}}{{{\rm{dx}}}}}&{\frac{{{\rm{d}}(\sin {\rm{x}})}}{{{\rm{dx}}}}}&{\frac{{{\rm{d}}(\cos {\rm{x}})}}{{{\rm{dx}}}}}\\ 6&{ - 1}&0\\ {\rm{p}}&{{{\rm{p}}^2}}&{{{\rm{p}}^3}} \end{array}} \right| + {\rm{\;}}\left| {\begin{array}{*{20}{c}} {{{\rm{x}}^3}}&{\sin {\rm{x}}}&{\cos {\rm{x}}}\\ {\frac{{{\rm{d}}\left( 6 \right)}}{{{\rm{dx}}}}}&{\frac{{{\rm{d}}\left( { - 1} \right)}}{{{\rm{dx}}}}}&{\frac{{{\rm{d}}\left( 0 \right)}}{{{\rm{dx}}}}}\\ {\rm{p}}&{{{\rm{p}}^2}}&{{{\rm{p}}^3}} \end{array}} \right| + {\rm{\;}}\left| {\begin{array}{*{20}{c}} {{{\rm{x}}^3}}&{\sin {\rm{x}}}&{\cos {\rm{x}}}\\ 6&{ - 1}&0\\ {\frac{{{\rm{d}}\left( {\rm{p}} \right)}}{{{\rm{dx}}}}}&{\frac{{{\rm{d}}\left( {{{\rm{p}}^2}} \right)}}{{{\rm{dx}}}}}&{\frac{{{\rm{d}}\left( {{{\rm{p}}^3}} \right)}}{{{\rm{dx}}}}} \end{array}} \right|\)

  \( \Rightarrow {\rm{f'}}\left( {\rm{x}} \right) = \left| {\begin{array}{*{20}{c}} {3{{\rm{x}}^2}}&{\cos {\rm{x}}}&{ - \sin {\rm{x}}}\\ 6&{ - 1}&0\\ {\rm{p}}&{{{\rm{p}}^2}}&{{{\rm{p}}^3}} \end{array}} \right| + 0 + 0 = \left| {\begin{array}{*{20}{c}} {3{{\rm{x}}^2}}&{\cos {\rm{x}}}&{ - \sin {\rm{x}}}\\ 6&{ - 1}&0\\ {\rm{p}}&{{{\rm{p}}^2}}&{{{\rm{p}}^3}} \end{array}} \right|\)

Put x = 0, we get

 \( \Rightarrow {\rm{f'}}\left( 0 \right) = \left| {\begin{array}{*{20}{c}} 0&{\cos 0}&{ - \sin 0}\\ 6&{ - 1}&0\\ {\rm{p}}&{{{\rm{p}}^2}}&{{{\rm{p}}^3}} \end{array}} \right| = {\rm{\;}}\left| {\begin{array}{*{20}{c}} 0&1&0\\ 6&{ - 1}&0\\ {\rm{p}}&{{{\rm{p}}^2}}&{{{\rm{p}}^3}} \end{array}} \right|\)

⇒ f’(0) = 0 – 1(6p³ – 0) + 0

∴ f’(0) = - 6p³