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ChrisV
 3 years ago
Best ResponseYou've already chosen the best response.0\[\sqrt{x}+1/4\sin(2x)^{2}\]

ChrisV
 3 years ago
Best ResponseYou've already chosen the best response.01/2x^1/2 + 1/2sin(2x) * cos(2x) * 2

ChrisV
 3 years ago
Best ResponseYou've already chosen the best response.0\[1/\sqrt{x} + \sin(2x)\cos(2x)\]

KingGeorge
 3 years ago
Best ResponseYou've already chosen the best response.1\[{d\over dx} \;\;\; \sqrt x+{1\over4}\sin(2x)^2 = {1\over{2\sqrt x}}+ {2\over2}\sin(2x)\cos(2x)\]

KingGeorge
 3 years ago
Best ResponseYou've already chosen the best response.1Remember you're also multiplying the \[{1\over \sqrt{x}}\]by \(1\over2\)

ChrisV
 3 years ago
Best ResponseYou've already chosen the best response.0\[1/\sqrt{x} + 2xcos(2x)^{2}\]

ChrisV
 3 years ago
Best ResponseYou've already chosen the best response.0is there an identity that makes it 2xcos(2x)^2

KingGeorge
 3 years ago
Best ResponseYou've already chosen the best response.1Not that I know of. Let me see if I can get wolfram to change it.

KingGeorge
 3 years ago
Best ResponseYou've already chosen the best response.1Well, 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
 3 years ago
Best ResponseYou've already chosen the best response.0k i was wondering myself
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