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Derivative of sin^2

  1. Jul 21, 2011 #1
    1. The problem statement, all variables and given/known data
    Derivative of Sin22x



    2. Relevant equations
    dy/dx = dy/du * du/dx

    y=U2


    3. The attempt at a solution
    Just want to make sure I am doing this right*.

    Do I let U = Sin2x or U = 2x?

    Let's say U = Sin2x
    y=U2

    then y` = 2Sin2x * cos2x?

    Or if U = 2x.

    y = SinU2

    y` = 2cos2x * 2
    y` = 4cos2x

    am i on the right track with either of these? any help is appreciated! thanks!
     
    Last edited by a moderator: Feb 14, 2013
  2. jcsd
  3. Jul 21, 2011 #2

    ehild

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    Re: Derivative of Sin^2(2x)?

    You are halfway on the right track :smile:.

    Sin^2(2x)=F(U(V)): V=2x, U=sin(V), F=U^2.

    dF/dx=dF/dU*dU/dV*dV/dx.

    ehild
     
  4. Jul 21, 2011 #3

    HallsofIvy

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    Re: Derivative of Sin^2(2x)?

    You first let U= sin 2x so that [itex]y= U^2[/itex], [itex]y'= 2U U'[/itex].

    Then, to find U', let V= 2x so U= sin V. U'= cos(V)(V') and, of course, V'= 2.
    Put those together.

    No, because the derivative of sin2X is not cos2X. Use the chain rule again.

    No, because the derivative of [itex]sin^2(x)[/itex] is not [itex]cos^2(x)[/itex]

     
  5. Jul 22, 2011 #4
    Re: Derivative of Sin^2(2x)?

    Essentially, what this entire question boils down to:
    We need two applications of the chain rule.
    The first one started well.

    Instead of
    y` = 2Sin2x * cos2x
    I recommend beginning Calculus students write.
    y` = 2Sin2x * ( Sin2x )'
    The left factor, 2 Sin(2x) is finished.
    Then to evaluate the derivative of Sin 2x, apply chain rule a second time, with v =2x



    General hint, way to think of chain rule:
    Take deriv. of the outside, leave the inside alone , then multiply by deriv. of inside.
     
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