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ChrisV

  • 2 years ago

Find the derivative of the function. sqrt(x) + 1/4(sin(2x)^2)

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  1. ChrisV
    • 2 years ago
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    \[\sqrt{x}+1/4\sin(2x)^{2}\]

  2. ChrisV
    • 2 years ago
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    1/2x^-1/2 + 1/2sin(2x) * cos(2x) * 2

  3. ChrisV
    • 2 years ago
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    \[1/\sqrt{x} + \sin(2x)\cos(2x)\]

  4. KingGeorge
    • 2 years ago
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    \[{d\over dx} \;\;\; \sqrt x+{1\over4}\sin(2x)^2 = {1\over{2\sqrt x}}+ {2\over2}\sin(2x)\cos(2x)\]

  5. ChrisV
    • 2 years ago
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    well the book says

  6. KingGeorge
    • 2 years ago
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    Remember you're also multiplying the \[{1\over \sqrt{x}}\]by \(1\over2\)

  7. ChrisV
    • 2 years ago
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    \[1/\sqrt{x} + 2xcos(2x)^{2}\]

  8. ChrisV
    • 2 years ago
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    i mean yes the 1/2

  9. ChrisV
    • 2 years ago
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    is there an identity that makes it 2xcos(2x)^2

  10. KingGeorge
    • 2 years ago
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    Not that I know of. Let me see if I can get wolfram to change it.

  11. KingGeorge
    • 2 years ago
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    Well, when I use wolfram to integrate \({1\over \sqrt{x}}+2x \cos^2 (2x)\) I don't get the original function back, so I'm thinking the book is wrong.

  12. ChrisV
    • 2 years ago
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    k i was wondering myself

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