作者LuisSantos (^______^)
看板trans_math
標題Re: [微分] 93中興科管
時間Wed Dec 5 18:46:18 2007
※ 引述《bitchdog (PEaCE)》之銘言：
: f(x)=sin[cos(sin2x)], find f'(x)
f(x) = sin[cos(sin2x)]
d
f'(x) = (cos[cos(sin2x)])(----(cos(sin2x)))
dx
d
= (cos[cos(sin2x)])(-sin(sin2x))(----(sin2x))
dx
d
= -(cos[cos(sin2x)])(sin(sin2x))(cos2x)(----(2x))
dx
= -(cos[cos(sin2x)])(sin(sin2x))(cos2x)(2)
= (-2)(cos[cos(sin2x)])(sin(sin2x))(cos2x)
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