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cwrw238 Group Title

Differentiate sinsinsin2x

  • 2 years ago
  • 2 years ago

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  1. cwrw238 Group Title
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    i'm confused with this one i guess you use the chain rule?

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

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

    • 2 years ago
  4. Noliec Group Title
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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!

    • 2 years ago
  5. Noliec Group Title
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    *twice more

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

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

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

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

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

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

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

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

    • 2 years ago
  14. cwrw238 Group Title
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    thnx guys

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

    • 2 years ago
  16. Chlorophyll Group Title
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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 :)

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

    • 2 years ago
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