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



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

ChrisV
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1/2x^1/2 + 1/2sin(2x) * cos(2x) * 2

ChrisV
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\[1/\sqrt{x} + \sin(2x)\cos(2x)\]

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

ChrisV
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well the book says

KingGeorge
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Remember you're also multiplying the \[{1\over \sqrt{x}}\]by \(1\over2\)

ChrisV
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\[1/\sqrt{x} + 2xcos(2x)^{2}\]

ChrisV
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i mean yes the 1/2

ChrisV
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is there an identity that makes it 2xcos(2x)^2

KingGeorge
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Not that I know of. Let me see if I can get wolfram to change it.

KingGeorge
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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.

ChrisV
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k i was wondering myself