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Math Help - Sin Derivatives

  1. #1
    Junior Member
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    Sin Derivatives

    Well, this should be simpler than my previous posts =P

    I need to find the derivative of this: Sin SqreRoot 2x

    I use the chain rule, first to get this: Sin ((2x^1/2)) = Cos ((2x^1/2))

    Then, I took the inside.

    2x^1/2 = 1/2(2x)^/1/2 = X^/1/2

    So, I ended up with Cos(SqreRoot2x) x 1/SqreRootX. I thought it was right, but the book is telling me the denominator is SqreRoot2x. I don't know where I went wrong.

    Any help is appreciated!
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  2. #2
    MHF Contributor red_dog's Avatar
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    (\sin\sqrt{2x})'=\cos\sqrt{2x}\cdot(\sqrt{2x})'=\c  os\sqrt{2x}\cdot\frac{2}{2\sqrt{2x}}=\cos\sqrt{2x}  \cdot\frac{1}{\sqrt{2x}}
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  3. #3
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    I don't see how the derivative of SqreRoot2x is 2/2SqreRoot2x
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  4. #4
    Math Engineering Student
    Krizalid's Avatar
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    (\sqrt u)'=\frac1{2\sqrt u}\cdot u'

    It's the chain rule.
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