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Math Help - Derivative, really tricky one!! Am I right???

  1. #1
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    Derivative, really tricky one!! Am I right???

    y = [x + (x + sin²x)³]^4

    I ended up getting



    4[x + x(x+sin²x)³]³ * 3x(sin²x)² * 2cosx



    Am I right???! If not, can someone seriously walk me through this? This was a really messy problem for me.
    Last edited by AntiSandman; October 12th 2008 at 07:27 PM.
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by AntiSandman View Post
    y = [x + x(sin²x)³]^4

    I ended up getting



    4[x + x(x+sin²x)³]³ * 3x(sin²x)² * 2cosx



    Am I right???! If not, can someone seriously walk me through this? This was a really messy problem for me.
    nope, not right.

    how about i give you a few hints and you try again and see where we get

    to find the derivative of x(\sin^2 x)^3 you need to use the product rule

    when you are differentiating the (\sin^2 x)^3 part, you need to use the chain rule. it would be easier to write (\sin^2 x)^3 as \sin^6 x by the way
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  3. #3
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    y = [x + (x + sin²x)³]^4

    I made a mistake, sorry, that's the real problem.

    Am I still right?

    (editted the above post too)
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by AntiSandman View Post
    y = [x + (x + sin²x)³]^4

    I made a mistake, sorry, that's the real problem.

    Am I still right?

    (editted the above post too)
    nope, still note right. you need the chain rule to differentiate (x + \sin^2 x)^3. plus, what about the derivative of the lone x?
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  5. #5
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    Is this it?

    4[x + x(x+sin²x)³]³ * [1 + 3(x + sin²x)² * (1 + 2cosxsinx)]
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  6. #6
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by AntiSandman View Post
    Is this it?

    4[x + x(x+sin²x)³]³ * [1 + 3(x + sin²x)² * (1 + 2cosxsinx)]
    yup, that's it

    you can write 2cos(x)sin(x) as sin(2x) though. maybe simplify even more if you are so inclined (i wouldn't be )

    wait, you have an extra x. the one in red shouldn't be there
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  7. #7
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    Quote Originally Posted by Jhevon View Post
    yup, that's it

    you can write 2cos(x)sin(x) as sin(2x) though. maybe simplify even more if you are so inclined (i wouldn't be )

    wait, you have an extra x. the one in red shouldn't be there
    Thanks so much, I seriously appreciate you helping me out so late at night.
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