1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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!

    Attached Files:

    • 2.JPG
      File size:
      32.7 KB
  2. jcsd
  3. Aug 18, 2011 #2


    User Avatar
    Gold Member

    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

    User Avatar
    Science Advisor
    Homework Helper

    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


    User Avatar
    Gold Member

    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)))
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook