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Math Help - Calculus - Chain Rule Derivative

  1. #1
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    Calculus - Chain Rule Derivative

    The question is find dy/dx of \frac{1}{4}\sin^2(2x).

    which I rewrote as \frac{1}{4}(\sin2x)^2

    then, using the chain rule, \frac{1}{2}\sin2x\cos2x.

    But the book gives the answer \frac{1}{2}\sin4x.

    Is my answer wrong? Or do I just need to simplify it further?
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  2. #2
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    Hello, Happy!

    Find \frac{dy}{dx} of: . \frac{1}{4}\sin^2(2x).

    which I rewrote as: \frac{1}{4}(\sin2x)^2 . . . . good!

    then, using the chain rule, \frac{1}{2}\sin2x\cos2x . . . . You dropped a "2"
    \frac{dy}{dx}\:=\;\frac{1}{4}\cdot2\sin2x\cos2x\cd  ot2 \:=\:\sin2x\cos2x

    Then they used the identity: . \sin2\theta \:=\:2\sin\theta\cos\theta

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  3. #3
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    Oh yes! Thank you, I forgot to "chain" the derivative of the 2x to the rest of the chained derivative.
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