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

1. anonymous

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

2. anonymous

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

3. anonymous

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

4. KingGeorge

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

5. anonymous

well the book says

6. KingGeorge

Remember you're also multiplying the ${1\over \sqrt{x}}$by $$1\over2$$

7. anonymous

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

8. anonymous

i mean yes the 1/2

9. anonymous

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

10. KingGeorge

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

11. 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.

12. anonymous

k i was wondering myself