- anonymous

Anyone good with large derivatives and want to help out?

- chestercat

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- anonymous

try me

- anonymous

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

- anonymous

V IS CONSTANT ?

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## More answers

- anonymous

no v is the variable for speed, it's the only non constant

- anonymous

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?

- anonymous

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?

- anonymous

just solve with c, I need to find the derivative of the equation above...

- anonymous

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

- anonymous

that is exactly what you do

- anonymous

yup...Im having trouble with simple algebra of fixing the variables up at the end

- anonymous

I can post the last step that I am sure of

- anonymous

MV squared over C use product rule, thenv squared over c squared product rule... yeah do that

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

the variables trip me up too man

- anonymous

ok I'm just going to throw out small parts of it that I'm unsure of

- anonymous

is (C^2(1-(v^2/c^2))^3/2)^3 the same as C^4*(1-(v^2/c^2)^3?

- anonymous

im pretty sure those are not the same, no

- anonymous

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

- anonymous

where are you getting C to the 4th up front from?

- anonymous

(C^2)^2....that last ^2 applies to both in the denominator doesnt it?

- anonymous

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)

- anonymous

yeah it does, I'm going to give up on this one for now, thanks for the help tho

- anonymous

sorry i couldnt help more, ill give it a try again after i make some food

- anonymous

you deriave with respect to what v ?

- anonymous

yes, v is independent, m(v) is dependent

- anonymous

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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