{"id":5575,"date":"2021-09-21T01:19:03","date_gmt":"2021-09-20T19:49:03","guid":{"rendered":"https:\/\/mathemerize.com\/?p=5575"},"modified":"2022-01-16T17:03:35","modified_gmt":"2022-01-16T11:33:35","slug":"higher-order-derivatives","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/higher-order-derivatives\/","title":{"rendered":"Higher Order Derivatives – Definition and Example"},"content":{"rendered":"
Here you will learn higher order derivatives of functions with examples.<\/p>\n
Let’s begin –<\/p>\n
If y = f(x), then \\(dy\\over dx\\), the derivative of y with respect to x, is itself, in general, a function of x and can be differentiated gain.<\/p>\n
To fix up the idea, we shall call \\(dy\\over dx\\) as the first order derivative of y with respect to x and the derivative of \\(dy\\over dx\\) with respect to x as the second order derivative of y with respect to x and will be denoted by \\(d^2y\\over dx^2\\).<\/p>\n
Similarly the derivative of \\(d^2y\\over dx^2\\) with respect to x will be termed as the third order derivative of y with respect to x and will be denoted by \\(d^3y\\over dx^3\\) and so on. The \\(n^{th}\\) order derivative of y with respect to x will be denoted by \\(d^ny\\over dx^n\\).<\/p>\n
If y = f(x), then the other alternative notations for<\/p>\n
\n\\(dy\\over dx\\), \\(d^2y\\over dx^2\\), \\(d^3y\\over dx^3\\), …… , \\(d^ny\\over dx^n\\) are<\/p>\n
\\(y_1\\), \\(y_2\\), \\(y^3\\), …… , \\(y_n\\)<\/p>\n
y’ , y” , y”’ , ……. , \\(y^(n)\\)<\/p>\n
Dy, \\(D^2\\)y , \\(D^3\\)y , ….. , \\(D^n\\)y<\/p>\n
f'(x) , f”(x) , f”'(x) , …… , \\(f^{n}\\) (x)<\/p>\n<\/blockquote>\n
The value of these derivatives at x = a are denoted by \\(y_n\\) (a), \\(y^n\\) (a) , \\(D^n\\)y (a) or, \\(({d^ny\\over dx^n})_{x = a}\\).<\/p>\n
Example<\/strong><\/span> : If y = \\(sin^{-1}x\\), show that \\(d^2y\\over dx^2\\) = \\(x\\over {(1 – x^2)^{3\/2}}\\)<\/p>\n
Solution<\/strong><\/span> : We have, y = \\(sin^{-1}x\\).<\/p>\n
On differentiating with respect to x, we get<\/p>\n
\\(dy\\over dx\\) = \\(1\\over \\sqrt{1 – x^2}\\)<\/p>\n
On differentiating again with respect to x, we get<\/p>\n
\\(d^2y\\over dx^2\\) = \\(d\\over dx\\)\\(({1\\over \\sqrt{1 – x^2}})\\) <\/p>\n
= \\(d\\over dx\\)\\([{(1 – x^2)^{-1\/2}}]\\) <\/p>\n
= \\(-1\\over 2\\) \\((1 – x^2)^{-3\/2}\\) \\(\\times\\) \\(d\\over dx\\) (\\(1 – x^2\\))<\/p>\n
\\(\\implies\\) \\(d^2y\\over dx^2\\) = -\\(1\\over 2(1 – x^2)^{3\/2}\\)(-2x) = \\(x\\over {(1 – x^2)^{3\/2}}\\)<\/p>\n\n\n