anonymous
  • anonymous
Anyone good with large derivatives and want to help out?
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
try me
anonymous
  • 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
  • anonymous
V IS CONSTANT ?

Looking for something else?

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

More answers

anonymous
  • anonymous
no v is the variable for speed, it's the only non constant
anonymous
  • 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
  • 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
  • anonymous
just solve with c, I need to find the derivative of the equation above...
anonymous
  • 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
  • anonymous
that is exactly what you do
anonymous
  • anonymous
yup...Im having trouble with simple algebra of fixing the variables up at the end
anonymous
  • anonymous
I can post the last step that I am sure of
anonymous
  • anonymous
MV squared over C use product rule, thenv squared over c squared product rule... yeah do that
anonymous
  • 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
  • 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
  • anonymous
haha, if that isn't clear let me know...it's the [C^2...+ 3MoV...] divided by the C^2...
anonymous
  • anonymous
the variables trip me up too man
anonymous
  • anonymous
ok I'm just going to throw out small parts of it that I'm unsure of
anonymous
  • anonymous
is (C^2(1-(v^2/c^2))^3/2)^3 the same as C^4*(1-(v^2/c^2)^3?
anonymous
  • anonymous
im pretty sure those are not the same, no
anonymous
  • 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
  • anonymous
where are you getting C to the 4th up front from?
anonymous
  • anonymous
(C^2)^2....that last ^2 applies to both in the denominator doesnt it?
anonymous
  • 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
  • anonymous
yeah it does, I'm going to give up on this one for now, thanks for the help tho
anonymous
  • anonymous
sorry i couldnt help more, ill give it a try again after i make some food
anonymous
  • anonymous
you deriave with respect to what v ?
anonymous
  • anonymous
yes, v is independent, m(v) is dependent
anonymous
  • 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

Looking for something else?

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