anonymous
  • anonymous
Find the derivative of the function. sqrt(x) + 1/4(sin(2x)^2)
Mathematics
katieb
  • katieb
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anonymous
  • anonymous
\[\sqrt{x}+1/4\sin(2x)^{2}\]
anonymous
  • anonymous
1/2x^-1/2 + 1/2sin(2x) * cos(2x) * 2
anonymous
  • anonymous
\[1/\sqrt{x} + \sin(2x)\cos(2x)\]

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KingGeorge
  • KingGeorge
\[{d\over dx} \;\;\; \sqrt x+{1\over4}\sin(2x)^2 = {1\over{2\sqrt x}}+ {2\over2}\sin(2x)\cos(2x)\]
anonymous
  • anonymous
well the book says
KingGeorge
  • KingGeorge
Remember you're also multiplying the \[{1\over \sqrt{x}}\]by \(1\over2\)
anonymous
  • anonymous
\[1/\sqrt{x} + 2xcos(2x)^{2}\]
anonymous
  • anonymous
i mean yes the 1/2
anonymous
  • anonymous
is there an identity that makes it 2xcos(2x)^2
KingGeorge
  • KingGeorge
Not that I know of. Let me see if I can get wolfram to change it.
KingGeorge
  • KingGeorge
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.
anonymous
  • anonymous
k i was wondering myself

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