Question

Can you write a general expression for
the derivative of the product of n functions

Answer 1

@Empty

Answer 2

keep moving the primes along

Answer 3

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 :)

Answer 4

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

Answer 5

They're direct messages, not private messages :O

Answer 6

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

Answer 7

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

Answer 8

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

Answer 9

\[(abc)'=(ab)'c+abc'=(a'b+ab')c+abc'=a'bc+ab'c+abc'\]

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)

Answer 10

oops

Answer 11

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:
\[\frac{(fgh)'}{fgh} = \frac{f'}{f}+\frac{g'}{g} + \frac{h'}{h}\]

Answer 12

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

Answer 13

\[\gcd(4,4')=?\]

Answer 14

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

Answer 15

do it again slower

Answer 16

okay wait maybe we need another rule

Answer 17

it should be 4 so this isnt working lol

Answer 18

okay how about only for p1^p1 that is different

Answer 19

so you know the power rule, \[(p^k)' = kp^{k-1}\]

what numbers solve this differential equation: \[n'=n\]

Answer 20

ah so all the ones in form pn^pn

Answer 21

when k=p

Answer 22

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; \[1'=?\]

Answer 23

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