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Question about derivative

  1. Aug 18, 2011 #1
    Hello everybody!

    I am currently reading Feinman course of physics and there is one question about derivative -

    I attached the formula. How did they got this derivative? I cannot understand this transformation, please help) Maybe smb could give a link where it is shown a rule how to get derivative in such case- A pity that I am not smart enough to solve this problem myself, it makes no fun when there are places in the book where I am not 100% sure what did the author meant.

    thanks a lot in advance!

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  3. Aug 18, 2011 #2


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    It's just the chain rule. Remember that [itex]v = v(t)[/itex] so it's just like taking the derivative [itex]{{d}\over{dt}}\left(f^2(t)) \right) = 2f(t) f'(t)[/itex]
  4. Aug 18, 2011 #3

    Andrew Mason

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    This is just an application of the chain rule. See: http://en.wikipedia.org/wiki/Chain_rule

  5. Aug 18, 2011 #4

    thanks a lot for your answers:

    sorry ( I am not really smart) but I would write just = 2f(t)

    my problem is I cannot get why I have to multiply also with f '(t)

    I cannot find exact explanation that passes to this case in wiki(
  6. Aug 18, 2011 #5


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    It should be in your calculus text, it's the chain rule.

    Pretend for a second you already knew the form of the velocity to be something trivial (and totally non-physical so don't assume you'll ever ever see the velocity written like this) as [itex]v(t) = t^2[/itex]. So your kinetic energy is

    [itex] T = mv(t)^2/2[/itex]

    So let's plug in what we know about v(t) and we find [itex]T = mt^4/2[/itex] is your kinetic energy. Do the time derivative and you get that it is [itex]2mt^3[/itex]. So that's the answer we KNOW is correct.

    Now, using the v(t) we have, you would only get [itex] T = mt^2[/itex] with your way of thinking, which is not what we know is true. You have to multiply by [itex]v'(t) = 2t[/itex] to get the correct answer.
  7. Aug 19, 2011 #6
    Thank you very much for support Pengwuino! You explanation on example is perfect- now I got it.

    I also found the rule:
    d(ab)dx = a db/dx + b da/dx

    I imagined it like d(v v)dx = v dv/dx + v dv/dx = 2v dv/dx

    thank you and good weekend)))
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