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

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

ChrisV Group TitleBest ResponseYou've already chosen the best response.0
1/2x^1/2 + 1/2sin(2x) * cos(2x) * 2
 2 years ago

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

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

ChrisV Group TitleBest ResponseYou've already chosen the best response.0
well the book says
 2 years ago

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

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

ChrisV Group TitleBest ResponseYou've already chosen the best response.0
i mean yes the 1/2
 2 years ago

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

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

KingGeorge Group TitleBest ResponseYou've already chosen the best response.1
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.
 2 years ago

ChrisV Group TitleBest ResponseYou've already chosen the best response.0
k i was wondering myself
 2 years ago
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