Question

LGBADERIVATIVE:

\[\huge y = \cos^2 x\]

y " = ?

Answer 1

\[y' = -2\sin x \cos x\]

Answer 2

\[y'' = 2\sin ^2 x - 2 \cos ^2 x?\]

Answer 3

\[y'' = 2(\cos (2x) )?\]

Answer 4

close

Answer 5

so it's wrong?

Answer 6

yes...just a little

Answer 7

\[cos^2 x = \frac{cos2x + 1}{2}\]
\[y' = \frac{d}{dx}cos^2 x = \frac{d}{dx}\frac{cos2x + 1}{2} = \frac{1}{2}(-sin2x)(2) = -sin2x\]
\[y'' = \frac{d}{dx} -sin2x = -2cos2x\]
I haven't done diff. for long ... Sorry :(

Answer 8

wait....i think it should be 2... @Zarkon ?

Answer 9

\[2\sin^2 x + 2\cos ^2 x\]
\[2(1)\]

Answer 10

no

Answer 11

callisto and i had same answer :/

Answer 12

nope

Answer 13

Not the same :|

Answer 14

close, but not the same

Answer 15

huh? what's the difference between \[-2 (\cos (2x) )\] and \[-2 \cos 2x\]
???

Answer 16

you didn't put a - in your answer above

Answer 17

You didn't type the negative sign.

Answer 18

oops lol

Answer 19

it shouldve been \[-2\sin^2 x + 2\cos^2 x?\]

Answer 20

what is a \(-\) between friends?

Answer 21

hmm wonder what