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cwrw238

  • 2 years ago

Differentiate sinsinsin2x

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  1. cwrw238
    • 2 years ago
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    i'm confused with this one i guess you use the chain rule?

  2. Noliec
    • 2 years ago
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    You do, it's horrifying! :(

  3. Noliec
    • 2 years ago
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    d(sin(sin(sin(2x))))/dx=dsinu/du*du/dx u=sin(sin(2x)) and dsinu/du=cosu

  4. Noliec
    • 2 years ago
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    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!

  5. Noliec
    • 2 years ago
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    *twice more

  6. Noliec
    • 2 years ago
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    Using the chain rule three times.

  7. cwrw238
    • 2 years ago
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    ugh - yes! ill try that

  8. .Sam.
    • 2 years ago
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    you should be getting \[2 \cos (2 x) \cos (\sin (2 x)) \cos (\sin (\sin (2 x)))\]

  9. cwrw238
    • 2 years ago
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    i got 2cos(sinsin2x)*cos(sin2x)*2cos2x

  10. cwrw238
    • 2 years ago
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    not quite i've got one extra '2'

  11. Noliec
    • 2 years ago
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    Oh right, missed that one, sorry

  12. cwrw238
    • 2 years ago
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    i think sam is right though

  13. cwrw238
    • 2 years ago
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    yea the first 2 is wrong

  14. cwrw238
    • 2 years ago
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    thnx guys

  15. cwrw238
    • 2 years ago
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    i did it in steps from left to right

  16. Chlorophyll
    • 2 years ago
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    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 :)

  17. Aron_West
    • 2 years ago
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    http://www.wolframalpha.com/input/?i=Differentiate+sin%28sin%28sin%282x%29%29%29

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