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Math Help - Chain Rule Derivative

  1. #1
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    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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    Never mind, I've got a new answer that is right, and I checked it by graphing both my new equation and the equation the books gives w/ my TI-89 and the graphs are identical.

    Now, the real derivative I got is \sin2x\cos2x.

    Now I just need to convert this into the equation the book wants, which is \frac{1}{2}\sin4x.

    I'm guessing its a random trig identity...but can't find which one
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  3. #3
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Happy View Post
    Never mind, I've got a new answer that is right, and I checked it by graphing both my new equation and the equation the books gives w/ my TI-89 and the graphs are identical.

    Now, the real derivative I got is \sin2x\cos2x.

    Now I just need to convert this into the equation the book wants, which is \frac{1}{2}\sin4x.

    I'm guessing its a random trig identity...but can't find which one
    recall that \sin 2u = 2 \sin u \cos u

    \Rightarrow \sin u \cos u = \frac 12 \sin 2u

    Here you have u = 2x
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  4. #4
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    Ah, I see. Thank you for your help.
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