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!

Homework Help: Finding the time derivative of a trigonometric function

  1. Jan 6, 2012 #1
    1. The problem statement, all variables and given/known data

    Finding the time derivative of [itex]sin^2( \alpha )[/itex], knowing that [itex] \dot \alpha ≠ 0[/itex]

    2. Relevant equations

    i know that [itex] \frac {d}{dt} sin ( \alpha ) = \dot \alpha cos ( \alpha) [/itex]

    3. The attempt at a solution

    That should give
    [itex] \dot \alpha ^2 cos^2( \alpha ) [/itex]

    But it's not and i'm not sure why.
  2. jcsd
  3. Jan 6, 2012 #2
    The best way to evaluate the derivate would be to use the chain rule. Are you familar with the chain rule?
  4. Jan 6, 2012 #3


    User Avatar

    Staff: Mentor

    You have one of the relevant equations; another relevant equation is the equation for the derivative of something squared.
  5. Jan 7, 2012 #4
    Another relevant equation [itex] \frac {d}{dx}x^2 = 2dx [/itex]

    [itex] \frac {d} {dt} sin^2 \alpha = 2 \dot \alpha cos \alpha [/itex]

    Is this it?
  6. Jan 7, 2012 #5
    d(x^2)/dx = 2x, not 2dx
  7. Jan 7, 2012 #6


    User Avatar

    Staff: Mentor

    as LawrenceC has pointed out, you don't have that expression quite right...

    But even after you correct it, that's a special case (u=x) of the more general expression

    [itex] \frac {d} {dx} u^2 = 2u \frac {du} {dx}[/itex]

    which you'll probably find more useful. Pay some attention to 1337's comment about the chain rule too.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook