# Differentiation of Determinants

## Differentiation of Determinants

If $$y=\left| \begin{matrix} f(x) & g(x) & h(x) \\ p(x) & q(x) & r(x) \\ u(x) & v(x) & w(x) \\\end{matrix} \right|$$

Then, $$\frac{dy}{dx}=\left| \begin{matrix} f'(x) & g'(x) & h'(x) \\ p(x) & q(x) & r(x) \\ u(x) & v(x) & w(x) \\\end{matrix} \right|+\left| \begin{matrix} f(x) & g(x) & h(x) \\ p'(x) & q'(x) & r'(x) \\ u(x) & v(x) & w(x) \\\end{matrix} \right|+\left| \begin{matrix} f(x) & g(x) & h(x) \\ p(x) & q(x) & r(x) \\ u'(x) & v'(x) & w'(x) \\\end{matrix} \right|$$

i.e., to differentiate a determinant, differentiate one row (or column) at a time, keeping other unchanged.

Example: If $$f(x)=\left| \begin{matrix} x & {{x}^{2}} & {{x}^{3}} \\ 1 & 2x & 3{{x}^{2}} \\ 0 & 2 & 6x \\\end{matrix} \right|$$, then f’(x) is equal to

Solution:

∵ $$f(x)=\left| \begin{matrix} x & {{x}^{2}} & {{x}^{3}} \\ 1 & 2x & 3{{x}^{2}} \\ 0 & 2 & 6x \\\end{matrix} \right|$$

∴ $$f'(x)=\left| \begin{matrix} x & {{x}^{2}} & {{x}^{3}} \\ 0 & 2x & 3{{x}^{2}} \\ 0 & 2 & 6x \\\end{matrix} \right|+\left| \begin{matrix} x & 2x & {{x}^{3}} \\ 1 & 2 & 3{{x}^{2}} \\ 0 & 0 & 6x \\\end{matrix} \right|+\left| \begin{matrix} x & {{x}^{2}} & 3{{x}^{2}} \\ 1 & 2x & 6x \\0 & 2 & 6 \\\end{matrix} \right|$$

= 1(12x² – 6x²) + 6x (2x – 2x) – 2(6x² – 3x²) + 6 (2x² – x²)

= 6x² + 0 – 6x² + 6x² = 6x².

If $$y=\left| \begin{matrix} f(x) & g(x) & h(x) \\ p(x) & q(x) & r(x) \\ u(x) & v(x) & w(x) \\\end{matrix} \right|$$

Then, $$\frac{dy}{dx}=\left| \begin{matrix} f'(x) & g'(x) & h'(x) \\ p(x) & q(x) & r(x) \\ u(x) & v(x) & w(x) \\\end{matrix} \right|+\left| \begin{matrix} f(x) & g(x) & h(x) \\ p'(x) & q'(x) & r'(x) \\ u(x) & v(x) & w(x) \\\end{matrix} \right|+\left| \begin{matrix} f(x) & g(x) & h(x) \\ p(x) & q(x) & r(x) \\ u'(x) & v'(x) & w'(x) \\\end{matrix} \right|$$

i.e., to differentiate a determinant, differentiate one row (or column) at a time, keeping other unchanged.

Example: If $$f(x)=\left| \begin{matrix} x & {{x}^{2}} & {{x}^{3}} \\ 1 & 2x & 3{{x}^{2}} \\ 0 & 2 & 6x \\\end{matrix} \right|$$, then f’(x) is equal to

Solution:

∵ $$f(x)=\left| \begin{matrix} x & {{x}^{2}} & {{x}^{3}} \\ 1 & 2x & 3{{x}^{2}} \\ 0 & 2 & 6x \\\end{matrix} \right|$$

∴ $$f'(x)=\left| \begin{matrix} x & {{x}^{2}} & {{x}^{3}} \\ 0 & 2x & 3{{x}^{2}} \\ 0 & 2 & 6x \\\end{matrix} \right|+\left| \begin{matrix} x & 2x & {{x}^{3}} \\ 1 & 2 & 3{{x}^{2}} \\ 0 & 0 & 6x \\\end{matrix} \right|+\left| \begin{matrix} x & {{x}^{2}} & 3{{x}^{2}} \\ 1 & 2x & 6x \\0 & 2 & 6 \\\end{matrix} \right|$$

= 1(12x² – 6x²) + 6x (2x – 2x) – 2(6x² – 3x²) + 6 (2x² – x²)

= 6x² + 0 – 6x² + 6x² = 6x².