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

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0keep moving the primes along

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.0Hmm 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 :)

dan815
 one year ago
Best ResponseYou've already chosen the best response.0lol did u know i wanted this because of arithmetic derivatives =.=

Empty
 one year ago
Best ResponseYou've already chosen the best response.2They're direct messages, not private messages :O

Empty
 one year ago
Best ResponseYou've already chosen the best response.2dan's being watched (and for good reason I might add)

Empty
 one year ago
Best ResponseYou've already chosen the best response.2ohhhh satellite is talking about like \[(fgh)'=f'gh+fg'h+fgh'\] moving these primes along dan haha

dan815
 one year ago
Best ResponseYou've already chosen the best response.0every permutations of the function  1 function must exist i see

Empty
 one year ago
Best ResponseYou've already chosen the best response.2\[(abc)'=(ab)'c+abc'=(a'b+ab')c+abc'=a'bc+ab'c+abc'\] 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)

Empty
 one year ago
Best ResponseYou've already chosen the best response.2Feel 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: \[\frac{(fgh)'}{fgh} = \frac{f'}{f}+\frac{g'}{g} + \frac{h'}{h}\]

dan815
 one year ago
Best ResponseYou've already chosen the best response.0hey i think i got it how about hte greatest GCD is all the primes 1 exponent

dan815
 one year ago
Best ResponseYou've already chosen the best response.0okay wait maybe we need another rule

dan815
 one year ago
Best ResponseYou've already chosen the best response.0it should be 4 so this isnt working lol

dan815
 one year ago
Best ResponseYou've already chosen the best response.0okay how about only for p1^p1 that is different

Empty
 one year ago
Best ResponseYou've already chosen the best response.2so you know the power rule, \[(p^k)' = kp^{k1}\] what numbers solve this differential equation: \[n'=n\]

dan815
 one year ago
Best ResponseYou've already chosen the best response.0ah so all the ones in form pn^pn

Empty
 one year ago
Best ResponseYou've already chosen the best response.2Yeah 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: \[(a^k)' = ka^{k1} a'\] one fun thing to prove; \[1'=?\]

Empty
 one year ago
Best ResponseYou've already chosen the best response.2\[1=(1*1)\] \[1'=1'*1+1*1'\] \[1'=2*1'\] \[1 \ne 2\] \[\implies\] \[1'=0\]
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