Function Posted February 27, 2014 Share Posted February 27, 2014 (edited) Hello everyone I was wondering if there was a general formula (sum, product, ...) to define the n-th derivative of the m-th power of a function f(x) So far, I've found that for [math]n[/math] up to 3, the following should be right (if I made no mistakes): [math]\frac{d^{n}\left[f(x)\right]^m}{dx^n}=\frac{d^{n}f}{dx^n}m\left[f(x)\right]^{m-1}+\left(\frac{df}{dx}\right)^n\left[f(x)\right]^{m-n}\cdot\prod_{i=0}^{m-1}{(m-i)}+n\left[f(x)\right]^{m-n+1}\cdot\prod_{i=1}^{m-1}{\frac{d^{i}f}{dx^i}}\cdot\prod_{i=0}^{m-2}{(m-i)}[/math] Can anyone tell me if it's even possible to make such formula, and if yes, what that formula is exactly? Thanks. Function Edited February 27, 2014 by Function Link to comment Share on other sites More sharing options...
studiot Posted February 27, 2014 Share Posted February 27, 2014 (edited) Indeed there are. You have uncovered Leibnitz rules. Well done. This develops the nth derivative of a product of two functions u and v via the binomial coefficients of lower derivatives, For example [math]\frac{{{d^2}}}{{d{x^2}}}\left( {uv} \right) = u\frac{{{d^2}v}}{{d{x^2}}} + 2\frac{{du}}{{dx}}\frac{{dv}}{{dx}} + v\frac{{{d^2}u}}{{d{x^2}}}[/math] [math]\frac{{{d^3}}}{{d{x^3}}}\left( {uv} \right) = u\frac{{{d^3}v}}{{d{x^3}}} + 3\frac{{du}}{{dx}}\frac{{{d^2}v}}{{d{x^2}}} + 3\frac{{{d^2}u}}{{d{x^2}}}\frac{{dv}}{{dx}} + v\frac{{{d^3}u}}{{d{x^3}}}[/math] http://en.wikipedia.org/wiki/General_Leibniz_rule Edited February 27, 2014 by studiot 1 Link to comment Share on other sites More sharing options...
Function Posted February 28, 2014 Author Share Posted February 28, 2014 (edited) Indeed there are. You have uncovered Leibnitz rules. Well done. This develops the nth derivative of a product of two functions u and v via the binomial coefficients of lower derivatives, For example [math]\frac{{{d^2}}}{{d{x^2}}}\left( {uv} \right) = u\frac{{{d^2}v}}{{d{x^2}}} + 2\frac{{du}}{{dx}}\frac{{dv}}{{dx}} + v\frac{{{d^2}u}}{{d{x^2}}}[/math] [math]\frac{{{d^3}}}{{d{x^3}}}\left( {uv} \right) = u\frac{{{d^3}v}}{{d{x^3}}} + 3\frac{{du}}{{dx}}\frac{{{d^2}v}}{{d{x^2}}} + 3\frac{{{d^2}u}}{{d{x^2}}}\frac{{dv}}{{dx}} + v\frac{{{d^3}u}}{{d{x^3}}}[/math] http://en.wikipedia.org/wiki/General_Leibniz_rule Ah.. Those binomials are everywhere Thanks. But how about m-numbers of functions? E.g. 3 functions instead of 2, or maybe 4, 5, ...? Edited February 28, 2014 by Function Link to comment Share on other sites More sharing options...
Endercreeper01 Posted February 28, 2014 Share Posted February 28, 2014 If you have m functions, and you all raise the functions to a power of a, and then add them together, the nth derivative would be [latex]\frac{d^n}{dx^n} \sum_{i=1}^m (f_i (x))^a [/latex] Using the chain rule, this is given by [latex]\frac{d^n}{dx^n} {\sum_{i=1}^m (f_i (x))^a} = {\sum_{i=1}^m \frac{a!}{(a-n)!} (f_i (x))^{a-n}}[/latex] This is derived from the fact that the nth derivative of f(x) to the power of a is [latex] \frac{a!}{(a - n)!} (f(x))^{a-n} [/latex] Which was found using the chain rule. The nth derivative of the sum of a set of these functions would then be given by [latex]{\sum_{i=1}^m \frac{a!}{(a-n)!} (f_i (x))^{a-n}}[/latex] Which gives us our final answer for this problem. Link to comment Share on other sites More sharing options...
Bignose Posted February 28, 2014 Share Posted February 28, 2014 Using the chain rule, this is given by [latex]\frac{d^n}{dx^n} {\sum_{i=1}^m (f_i (x))^a} = {\sum_{i=1}^m \frac{a!}{(a-n)!} (f_i (x))^{a-n}}[/latex] This isn't right. The RHS doesn't have any df/dx terms. What you've posted here isn't the chain rule. 1 Link to comment Share on other sites More sharing options...
studiot Posted February 28, 2014 Share Posted February 28, 2014 (edited) But how about m-numbers of functions? E.g. 3 functions instead of 2, or maybe 4, 5, ...? You really are a glutton for the extreme. I haven't ever had need to think about it but the usual thing in those circumstances would be to generalise the binomial expansion with the multinomial expansion. http://en.wikipedia.org/wiki/Multinomial_theorem Edited February 28, 2014 by studiot 1 Link to comment Share on other sites More sharing options...
Function Posted February 28, 2014 Author Share Posted February 28, 2014 You really are a glutton for the extreme. I haven't ever had need to think about it but the usual thing in those circumstances would be to generalise the binomial expansion with the multinomial expansion. http://en.wikipedia.org/wiki/Multinomial_theorem I'm a glutton for the most general formulas Thanks for the link. Link to comment Share on other sites More sharing options...
Endercreeper01 Posted February 28, 2014 Share Posted February 28, 2014 This isn't right. The RHS doesn't have any df/dx terms. What you've posted here isn't the chain rule. Oh yes, I wasn't paying attention. My formula would not be correct then. Link to comment Share on other sites More sharing options...
Function Posted May 29, 2014 Author Share Posted May 29, 2014 (edited) I have reviewed this; is this correct: [math]\frac{d^n\left(\left[f(x)\right]^{m}\right)}{dx^n}=\prod_{i=0}^{n-1}{\left((m-i)\cdot\left[f(x)\right]^{m-i-1}\cdot\frac{df}{dx}\right)}[/math] Nope.. Nvm.. Went too quickly on this one. Edited May 29, 2014 by Function Link to comment Share on other sites More sharing options...
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