Derivative for Product rule a new one?

[Baharx]

i found a formula but I'm not sure is and maybe this is exist i don't know.

we know it F(x)= f(x).g(x) → F'(x)= f'(x).g(x)+g'(x).f(x)

and i want to calculate nth order.


n>1
fⁿ(x).g(x)+ $\sum^{n}_{k=1}$ binomial coefficient
$\left(n k\right)$. f^(n-k)(x).g^k(x)+ gⁿ.f(x)


is it true? and i found myself but I'm sure this formula is exist.:uhh:

 

[lurflurf]

That is called Leibniz (product) rule it is in most good calculus books.

 

[Baharx]

oh, i didnt know that. i studying Calculus and that was an exercise. Leibniz? anyway.

first i found on wikipedia this : http://en.wikipedia.org/wiki/Difference_quotient

but i didn't understand then 6 hours i did get a fix on this exercise.
Thank you for reply.

but I'm seeing just sigma, where is fⁿ(x).g(x) and gⁿ(x).f(x) i added these.

and i started k=1 and Mr.Leibniz started k=0.

 

[Mute]

Your "extra terms" are contained in the sigma notation of the Leibniz formula. (In fact, since your summation goes all the way up to n, you have actually counted the last term twice. Also, there is an error in your first post - you wrote g^n(x), not g^n(x)f(x).) Even though you're not the first to derive this result, good on you for figuring it out yourself.

 

[Baharx]

Your "extra terms" are contained in the sigma notation of the Leibniz formula. (In fact, since your summation goes all the way up to n, you have actually counted the last term twice. Also, there is an error in your first post - you wrote g^n(x), not g^n(x)f(x).) Even though you're not the first to derive this result, good on you for figuring it out yourself.

Thank you, i understand but besides i wrote n>1 condition. This is why first derivative can't be find this formula. but i tried it's working 3th order and nth order :uhh:

Thank you again.

 

[Baharx]

but i understand. Thanks for the help. i want to study for master degree department of math. but i think i must study work hard.