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anonymous
 5 years ago
Anyone good with large derivatives and want to help out?
anonymous
 5 years ago
Anyone good with large derivatives and want to help out?

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok, 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

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0no v is the variable for speed, it's the only non constant

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I 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?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0what 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?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0just solve with c, I need to find the derivative of the equation above...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so 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

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0that is exactly what you do

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yup...Im having trouble with simple algebra of fixing the variables up at the end

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I can post the last step that I am sure of

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0MV squared over C use product rule, thenv squared over c squared product rule... yeah do that

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i 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

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok 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

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0haha, if that isn't clear let me know...it's the [C^2...+ 3MoV...] divided by the C^2...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the variables trip me up too man

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok I'm just going to throw out small parts of it that I'm unsure of

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0is (C^2(1(v^2/c^2))^3/2)^3 the same as C^4*(1(v^2/c^2)^3?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0im pretty sure those are not the same, no

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0right 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

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0where are you getting C to the 4th up front from?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0(C^2)^2....that last ^2 applies to both in the denominator doesnt it?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0which 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)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yeah it does, I'm going to give up on this one for now, thanks for the help tho

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0sorry i couldnt help more, ill give it a try again after i make some food

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you deriave with respect to what v ?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yes, v is independent, m(v) is dependent

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0if 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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