how to solve dy/dx: cos*2X

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how to solve dy/dx: cos*2X

Mathematics
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derivative of cos (chain 2x)
sin2x*2
Are you sure, Starica is getting ready to correct you

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awww i thought we r to find dy/dx :S
you just derive the outside and inside ^_^ outside = cos inside 2x so dy/dx of cos = -sin and dy/dx of 2x = 2 soooo, dy/dx cos(2x) = -2sin(2x)
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
alright so we have 2 steps, first outside then inside, our outside here is the power and the inside is cos(x) so dy/dx = - 2cos(x) sin(x) ^_^
but star ...dont u think it must b y= cos^2x n find dy/dx??
yeah lol
wasn't I right? o_o
... Go by the definition of the chain rule. \[Let\ (cos\ x)^2 = f(g(x))\] What is our f(x), what is our g(x)?
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 ^_^
he-she isn't going to be able to do any of the rest of their homework if they don't understand the chain rule. This is a good opportunity to learn the chain rule since you need it to solve this problem.
What can I say. Star likes manual labor.
lol, I'm just trying to get it clear for him/her ^_^?

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