Question

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

Answer 1

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

Answer 2

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

Answer 3

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

Answer 4

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

Answer 5

well the book says

Answer 6

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

Answer 7

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

Answer 8

i mean yes the 1/2

Answer 9

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

Answer 10

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

Answer 11

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.

Answer 12

k i was wondering myself