# Differentiation of Determinants

## Differentiation of Determinants

To differentiation a determinant, we differentiate one row (or column) at a time, keeping other unchanged. For example, if

$$\Delta \left( x \right)=\left| \begin{matrix} f(x) & g(x) \\ u\left( x \right) & v\left( x \right) \\\end{matrix} \right|$$.

Differentiation on both sides,

$$\frac{d}{dx}\left( \Delta x \right)=\left| \begin{matrix} f'(x) & g'(x) \\ u(x) & v(x) \\\end{matrix} \right|+\left| \begin{matrix} f(x) & g(x) \\ u'(x) & v'(x) \\\end{matrix} \right|$$.

(or)

$$\frac{d}{dx}\left( \Delta x \right)=\left| \begin{matrix} f'(x) & g(x) \\ u'(x) & v(x) \\\end{matrix} \right|+\left| \begin{matrix} f(x) & g'(x) \\ u(x) & v'(x) \\\end{matrix} \right|$$,

Similar results hold for the differentiation of determinants of higher order.

Example: If $$f\left( x \right)=\left| \begin{matrix} x+{{a}^{2}} & ab & ac \\ ab & x+{{b}^{2}} & bc \\ ac & bc & x+{{c}^{2}} \\\end{matrix} \right|$$, then prove that f’(x) = 3x² + 2x (a² + b² + c²) .

Solution: Given that  $$f\left( x \right)=\left| \begin{matrix} x+{{a}^{2}} & ab & ac \\ ab & x+{{b}^{2}} & bc \\ ac & bc & x+{{c}^{2}} \\\end{matrix} \right|$$,

Differentiation with respect x.

$$f’\left( x \right)=\left| \begin{matrix} (x+{{a}^{2}})’ & (ab)’ & (ac)’ \\ ab & x+{{b}^{2}} & bc \\ ac & bc & x+{{c}^{2}} \\\end{matrix} \right|+\left| \begin{matrix} x+{{a}^{2}} & ab & ac \\ (ab)’ & (x+{{b}^{2}})’ & (bc)’ \\ ac & bc & x+{{c}^{2}} \\\end{matrix} \right|+\left| \begin{matrix} x+{{a}^{2}} & ab & ac \\ ab & x+{{b}^{2}} & bc \\ (ac)’ & (bc)’ & (x+{{c}^{2}})’ \\\end{matrix} \right|$$,

$$f’\left( x \right)=\left| \begin{matrix} 1 & 0 & 0 \\ ab & x+{{b}^{2}} & bc \\ ac & bc & x+{{c}^{2}} \\\end{matrix} \right|+\left| \begin{matrix} x+{{a}^{2}} & ab & ac \\ 0 & 1 & 0 \\ ac & bc & x+{{c}^{2}} \\\end{matrix} \right|+\left| \begin{matrix} x+{{a}^{2}} & ab & ac \\ ab & x+{{b}^{2}} & bc \\ 0 & 0 & 1 \\\end{matrix} \right|$$.

$$=1\left( \left( x+{{b}^{2}} \right)\left( x+{{c}^{2}} \right)-{{\left( bc \right)}^{2}} \right)-0+0-0+1\left( \left( x+{{a}^{2}} \right)\left( x+{{c}^{2}} \right)-{{\left( ac \right)}^{2}} \right)$$$$+0-0+1\left( \left( x+{{a}^{2}} \right){{\left( x+b \right)}^{2}}-{{\left( ab \right)}^{2}} \right)$$,

$$=1\left( \left( x+{{b}^{2}} \right)\left( x+{{c}^{2}} \right)-{{b}^{2}}{{c}^{2}} \right)+1\left( \left( x+{{a}^{2}} \right)\left( x+{{c}^{2}} \right)-{{a}^{2}}{{c}^{2}} \right)+1\left( \left( x+{{a}^{2}} \right){{\left( x+b \right)}^{2}}-{{a}^{2}}{{b}^{2}} \right)$$,

f’(x) = 3x² + 2x (a² + b² + c²).