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Did I expand it wrong? y= (1+cos^2 (x))^6 y' = 6 ( 1+cos^2 (x) )^5 (2) (cosx) (-sinx)
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Hi cmkc109 :) No I don't think so. It seems right buddy..
But then the answer given is -6 sin 2x (1+cos^2 (x))^5
sin(2x)?? double angle?
OK let me try..
it doesnt have the bracket..but i assume it is the same -6 sin 2x (1+cos^2 x)^5
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since 2sinxcosx = sin2x
That's all you do to simply further!
ohh..that rule, thanks both of you :)
\[\large y=(1+\cos^2x)^6\]\[\large y'=6(1+\cos^2x)^5\frac{d}{dx}(1+\cos^2x)\]\[\large y'=6(1+cos^2x)^5(2cosx)(-sinx)\]\[\large y'=-6(1+cos^2x)^5(2sinxcosx)\]\[\large y'=-6(1+cos^2x)^5(sin2x)\]
thanks both!
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