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, 201411 yr by 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

Edited February 27, 201411 yr by 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

 

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, 201411 yr by Function

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.

 

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.

 

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, 201411 yr by studiot

 

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.

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.

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, 201411 yr by Function