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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

    Nugatory

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    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

    Nugatory

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    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.
     
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