dan815 one year ago Can you write a general expression for the derivative of the product of n functions

1. dan815

@Empty

2. anonymous

keep moving the primes along

3. zepdrix

Hmm I know you can write the nth derivative in a fancy way. it follows the same type of format as the binomial theorem. But n functions? Hmm interesting idea :)

4. dan815

lol did u know i wanted this because of arithmetic derivatives =.=

5. Empty

They're direct messages, not private messages :O

6. Empty

dan's being watched (and for good reason I might add)

7. Empty

ohhhh satellite is talking about like $(fgh)'=f'gh+fg'h+fgh'$ moving these primes along dan haha

8. dan815

every permutations of the function - 1 function must exist i see

9. Empty

$(abc)'=(ab)'c+abc'=(a'b+ab')c+abc'=a'bc+ab'c+abc'$ Actually the derivative for the arithmetic functions will have the same general form I think FOR THOSE WHO AREN'T ME OR DAN LISTEN UP! We're playing around with the arithmetic derivative itself found here: https://projecteuler.net/problem=484 (We're not attempting the problem though, maybe another day haha)

10. anonymous

oops

11. Empty

Feel free to join in it's super fun to play with lol. dan, here's my hint to you, divide by fgh and you get this: $\frac{(fgh)'}{fgh} = \frac{f'}{f}+\frac{g'}{g} + \frac{h'}{h}$

12. dan815

hey i think i got it how about hte greatest GCD is all the primes -1 exponent

13. Empty

$\gcd(4,4')=?$

14. dan815

(4,(2*2)')=2

15. Empty

do it again slower

16. dan815

okay wait maybe we need another rule

17. dan815

it should be 4 so this isnt working lol

18. dan815

okay how about only for p1^p1 that is different

19. Empty

so you know the power rule, $(p^k)' = kp^{k-1}$ what numbers solve this differential equation: $n'=n$

20. dan815

ah so all the ones in form pn^pn

21. dan815

when k=p

22. Empty

Yeah so it's pretty cool we have like: $(p^{kp})' = k p^{kp}$ kinda serving like $(e^{kx})' =ke^{kx}$ haha Also here's the chain rule: $(a^k)' = ka^{k-1} a'$ one fun thing to prove; $1'=?$

23. Empty

$1=(1*1)$ $1'=1'*1+1*1'$ $1'=2*1'$ $1 \ne 2$ $\implies$ $1'=0$