A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

dan815

  • one year ago

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

  • This Question is Closed
  1. dan815
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    @Empty

  2. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    keep moving the primes along

  3. zepdrix
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  5. Empty
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    They're direct messages, not private messages :O

  6. Empty
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

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

  7. Empty
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

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

  8. dan815
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  9. Empty
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 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)

  10. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    oops

  11. Empty
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    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
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  13. Empty
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

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

  14. dan815
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  15. Empty
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    do it again slower

  16. dan815
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    okay wait maybe we need another rule

  17. dan815
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    it should be 4 so this isnt working lol

  18. dan815
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    okay how about only for p1^p1 that is different

  19. Empty
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

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

  20. dan815
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    ah so all the ones in form pn^pn

  21. dan815
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    when k=p

  22. Empty
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    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
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

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

  24. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.