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anonymous

  • 5 years ago

Anyone good with large derivatives and want to help out?

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  1. anonymous
    • 5 years ago
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    try me

  2. anonymous
    • 5 years ago
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    ok, the derivative is Mo*V/C^2(1-(v^2/c^2))^3/2 where Mo=mass of object at rest, c=speed of light and v=speed

  3. anonymous
    • 5 years ago
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    V IS CONSTANT ?

  4. anonymous
    • 5 years ago
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    no v is the variable for speed, it's the only non constant

  5. anonymous
    • 5 years ago
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    I wish I could represent this better, but I can't use the equation editor worth a darn, is there any other complications you see with my equations?

  6. anonymous
    • 5 years ago
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    what value do they want you to use for speed of light, or are they having you solve the function in terms of speed of light as a constant C?

  7. anonymous
    • 5 years ago
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    just solve with c, I need to find the derivative of the equation above...

  8. anonymous
    • 5 years ago
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    so far I see you use the product rule for the numerator.....the product rule for the denominator, followed by the chain rule then the quotient rule....then quotient rule for the whole thing

  9. anonymous
    • 5 years ago
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    that is exactly what you do

  10. anonymous
    • 5 years ago
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    yup...Im having trouble with simple algebra of fixing the variables up at the end

  11. anonymous
    • 5 years ago
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    I can post the last step that I am sure of

  12. anonymous
    • 5 years ago
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    MV squared over C use product rule, thenv squared over c squared product rule... yeah do that

  13. anonymous
    • 5 years ago
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    i keep tripping myself up on the variables, funny i can solve one of these big guys when its full of decimals and garbage but simple variables trip me up haha

  14. anonymous
    • 5 years ago
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    ok here's what I got..... {[C^2(1-(v^2/c^2))^3/2]+[3MoV(1-(v^2/c^2))^1/2]}/ (C^2(1-(v^2/c^2))^3/2)^3

  15. anonymous
    • 5 years ago
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    haha, if that isn't clear let me know...it's the [C^2...+ 3MoV...] divided by the C^2...

  16. anonymous
    • 5 years ago
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    the variables trip me up too man

  17. anonymous
    • 5 years ago
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    ok I'm just going to throw out small parts of it that I'm unsure of

  18. anonymous
    • 5 years ago
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    is (C^2(1-(v^2/c^2))^3/2)^3 the same as C^4*(1-(v^2/c^2)^3?

  19. anonymous
    • 5 years ago
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    im pretty sure those are not the same, no

  20. anonymous
    • 5 years ago
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    right sorry I meant (C^2(1-(v^2/c^2))^3/2)^3 the same as C^4*(1-(v^2/c^2)^2...last number should have been a 2

  21. anonymous
    • 5 years ago
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    where are you getting C to the 4th up front from?

  22. anonymous
    • 5 years ago
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    (C^2)^2....that last ^2 applies to both in the denominator doesnt it?

  23. anonymous
    • 5 years ago
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    which last ^2 are you talking about? theres a ^3 at the end and the v squared over c squared is in parenthesis so im not sure what your saying (text based math formulas suck)

  24. anonymous
    • 5 years ago
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    yeah it does, I'm going to give up on this one for now, thanks for the help tho

  25. anonymous
    • 5 years ago
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    sorry i couldnt help more, ill give it a try again after i make some food

  26. anonymous
    • 5 years ago
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    you deriave with respect to what v ?

  27. anonymous
    • 5 years ago
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    yes, v is independent, m(v) is dependent

  28. anonymous
    • 5 years ago
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    if my equation confuses you, it's already the derivative of Einstein's equation m(v) mass of a moving particle, now we have to derive it again

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