Differentiate sinsinsin2x

Question

cwrw238 · Answer

i'm confused with this one
i guess you use the chain rule?

anonymous · Answer

You do, it's horrifying! :(

anonymous · Answer

d(sin(sin(sin(2x))))/dx=dsinu/du*du/dx
u=sin(sin(2x)) and dsinu/du=cosu

anonymous · Answer

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!

anonymous · Answer

*twice more

anonymous · Answer

Using the chain rule three times.

cwrw238 · Answer

ugh - yes!
ill try that

.sam. · Answer

you should be getting $$2 \cos (2 x) \cos (\sin (2 x)) \cos (\sin (\sin (2 x)))$$

cwrw238 · Answer

i got
2cos(sinsin2x)*cos(sin2x)*2cos2x

cwrw238 · Answer

not quite i've got one extra '2'

anonymous · Answer

Oh right, missed that one, sorry

cwrw238 · Answer

i think sam is right though

cwrw238 · Answer

yea the first 2 is wrong

cwrw238 · Answer

thnx guys

cwrw238 · Answer

i did it in steps from left to right

anonymous · Answer

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 :)

anonymous · Answer

http://www.wolframalpha.com/input/?i=Differentiate+sin%28sin%28sin%282x%29%29%29