n-th derivative of [f(x)]^m

[Function]

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

[studiot]

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

[Function]

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

[Endercreeper01]

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.

 

[Bignose]

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.

[studiot]

 

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

[Function]

 

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.

[Endercreeper01]

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.

[Function]

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