dan815
  • dan815
Can you write a general expression for the derivative of the product of n functions
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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dan815
  • dan815
@Empty
anonymous
  • anonymous
keep moving the primes along
zepdrix
  • 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 :)

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dan815
  • dan815
lol did u know i wanted this because of arithmetic derivatives =.=
Empty
  • Empty
They're direct messages, not private messages :O
Empty
  • Empty
dan's being watched (and for good reason I might add)
Empty
  • Empty
ohhhh satellite is talking about like \[(fgh)'=f'gh+fg'h+fgh'\] moving these primes along dan haha
dan815
  • dan815
every permutations of the function - 1 function must exist i see
Empty
  • 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)
anonymous
  • anonymous
oops
Empty
  • 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: \[\frac{(fgh)'}{fgh} = \frac{f'}{f}+\frac{g'}{g} + \frac{h'}{h}\]
dan815
  • dan815
hey i think i got it how about hte greatest GCD is all the primes -1 exponent
Empty
  • Empty
\[\gcd(4,4')=?\]
dan815
  • dan815
(4,(2*2)')=2
Empty
  • Empty
do it again slower
dan815
  • dan815
okay wait maybe we need another rule
dan815
  • dan815
it should be 4 so this isnt working lol
dan815
  • dan815
okay how about only for p1^p1 that is different
Empty
  • Empty
so you know the power rule, \[(p^k)' = kp^{k-1}\] what numbers solve this differential equation: \[n'=n\]
dan815
  • dan815
ah so all the ones in form pn^pn
dan815
  • dan815
when k=p
Empty
  • 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; \[1'=?\]
Empty
  • Empty
\[1=(1*1)\] \[1'=1'*1+1*1'\] \[1'=2*1'\] \[1 \ne 2\] \[\implies\] \[1'=0\]

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