{"id":5139,"date":"2021-09-08T20:44:55","date_gmt":"2021-09-08T15:14:55","guid":{"rendered":"https:\/\/mathemerize.com\/?p=5139"},"modified":"2021-11-22T20:40:09","modified_gmt":"2021-11-22T15:10:09","slug":"differentiation-of-determinant","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/differentiation-of-determinant\/","title":{"rendered":"Differentiation of Determinant"},"content":{"rendered":"
Here you will learn differentiation of determinant with example.<\/p>\n
Let’s begin –<\/p>\n
To differentiate a determinant, we differerentiate one row (or column) at a time, keeping others unchanged.<\/p>\n
for example, if<\/p>\n
D(x) = \\(\\begin{vmatrix} f(x) & g(x) \\\\ u(x) & v(x) \\end{vmatrix}\\) , then<\/p>\n
\\(d\\over dx\\){D(x)} = \\(\\begin{vmatrix} f'(x) & g'(x) \\\\ u(x) & v(x) \\end{vmatrix}\\) + \\(\\begin{vmatrix} f(x) & g(x) \\\\ u'(x) & v'(x) \\end{vmatrix}\\)<\/p>\n
Also,<\/p>\n
\\(d\\over dx\\){D(x)} = \\(\\begin{vmatrix} f'(x) & g(x) \\\\ u'(x) & v(x) \\end{vmatrix}\\) + \\(\\begin{vmatrix} f(x) & g'(x) \\\\ u(x) & v'(x) \\end{vmatrix}\\)<\/p>\n
Similar results holds for the differentiation of determinant of higher order. <\/p>\n
Example<\/span><\/strong> : If f(x) = \\(\\begin{vmatrix} x^2 + a^2 & a \\\\ x^2 + b^2 & b \\end{vmatrix}\\), find f'(x).<\/p>\n Solution<\/span><\/strong> : We have,<\/p>\n f(x) = \\(\\begin{vmatrix} x^2 + a^2 & a \\\\ x^2 + b^2 & b \\end{vmatrix}\\)<\/p>\n \\(\\implies\\) f'(x) = \\(\\begin{vmatrix} 2x & 0 \\\\ x^2 + b^2 & b \\end{vmatrix}\\) + \\(\\begin{vmatrix} x^2 + a^2 & a \\\\ 2x & 0 \\end{vmatrix}\\)<\/p>\n \\(\\implies\\) f'(x) = {2bx} + {2ax}<\/p>\n f'(x) = 2x(a +b)<\/p>\n Example<\/span><\/strong> : If f(x) = \\(\\begin{vmatrix} x + a^2 & ab & ac \\\\ ab & x + b^2 & bc \\\\ ac & bc & x + c^2 \\end{vmatrix}\\), find f'(x).<\/p>\n Solution<\/span><\/strong> : We have,<\/p>\n f(x) = \\(\\begin{vmatrix} x + a^2 & ab & ac \\\\ ab & x + b^2 & bc \\\\ ac & bc & x + c^2 \\end{vmatrix}\\)<\/p>\n \\(\\implies\\) f'(x) = \\(\\begin{vmatrix} 1 & 0 & 0 \\\\ ab & x + b^2 & bc \\\\ ac & bc & x + c^2 \\end{vmatrix}\\) + \\(\\begin{vmatrix} x + a^2 & ab & ac \\\\ 0 & 1 & 0 \\\\ ac & bc & x + c^2 \\end{vmatrix}\\) + \\(\\begin{vmatrix} x + a^2 & ab & ac \\\\ ab & x + b^2 & bc \\\\ 0 & 0 & 1 \\end{vmatrix}\\)<\/p>\n \\(\\implies\\) f'(x) = \\(\\begin{vmatrix} x + b^2 & bc \\\\ bc & x + c^2 \\end{vmatrix}\\) + \\(\\begin{vmatrix} x + a^2 & ac \\\\ ac & x + c^2 \\end{vmatrix}\\) + \\(\\begin{vmatrix} x + a^2 & ab \\\\ ab & x + b^2 \\end{vmatrix}\\)<\/p>\n \\(\\implies\\) f'(x) = {\\((x+b^2)(x+c^2) – b^2c^2\\)} + {\\((x+a^2)(x+c^2) – a^2c^2\\)} + {\\((x+a^2)(x+b^2) – a^2b^2\\)}<\/p>\n \\(\\implies\\) f'(x) = \\(x^2\\) + x\\(b^2+c^2\\) + \\(x^2\\) + x\\(a^2+c^2\\) + \\(x^2\\) + x\\(a^2+b^2\\)<\/p>\n f'(x) = \\(3x^2\\) + 2x\\(a^2 + b^2 + c^2\\).<\/p>\n\n\n