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derivative of cos (chain 2x)

sin2x*2

Are you sure, Starica is getting ready to correct you

awww i thought we r to find dy/dx :S

o_o... but wasn't she right?

no i missed the negative sign

isn't it finding the derivative of cos(2x)?

lol, it's alright :)

i guessed it so

you were close =P

thanks:p

np ^_^, do you get it now sandip?

no its finding the derivative of cos^2x

OH!
so the question is:
\[\frac{dy}{dx} \cos^2(x)\]?

ya

but star ...dont u think it must b y= cos^2x n find dy/dx??

yeah lol

wasn't I right? o_o

she/he doesn't get the chain rule >_< lol

in a much simpler way do it this way:
cos(x) . cos(x)
u'v + uv' ^_^ and solve

I hope this way helps though :)

alright gtg, later ^_^

What can I say. Star likes manual labor.

lol, I'm just trying to get it clear for him/her ^_^?