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cwrw238
Differentiate sinsinsin2x
i'm confused with this one i guess you use the chain rule?
You do, it's horrifying! :(
d(sin(sin(sin(2x))))/dx=dsinu/du*du/dx u=sin(sin(2x)) and dsinu/du=cosu
And then you apply it once more but that gets too messy for me to type - bound to make a mistake. Hehe, hope my answer helped to some extent atleast; good luck!
Using the chain rule three times.
ugh - yes! ill try that
you should be getting \[2 \cos (2 x) \cos (\sin (2 x)) \cos (\sin (\sin (2 x)))\]
i got 2cos(sinsin2x)*cos(sin2x)*2cos2x
not quite i've got one extra '2'
Oh right, missed that one, sorry
i think sam is right though
yea the first 2 is wrong
i did it in steps from left to right
Sam's correct! You should take derivative from RIGHT to left: ( sin2x )' = 2 cosx sin ( sin2x) ' = 2 cosx * cos ( sin2x ) sin [ sin ( sin2x)] ' = 2 cosx * cos ( sin2x ) * cos [ sin ( sin2x)] Just go step by step, you won't miss it :)
http://www.wolframalpha.com/input/?i=Differentiate+sin%28sin%28sin%282x%29%29%29