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Math Help - Differential problem!

  1. #1
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    Differential problem!

    ok, i was doing my hw and i just do not have a clue how to solve this.
    the problem is:
    (2x^3-x)cos(1-x^2)

    and this is what i got or attempted so far. i'm not even sure if this is the right way. can somebody tell me where i went wrong?

    [d/dx(2x^3-x)cos)1-x^2) x (2x^3-x)[d/dx(cos(1-x^2))]
    =(6x^2-1)cos(1-x^2) x 2x(2x^3-x)sin(1-x^2)

    thanks much!
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  2. #2
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    Quote Originally Posted by katieeej View Post
    ok, i was doing my hw and i just do not have a clue how to solve this.
    the problem is:
    (2x^3-x)cos(1-x^2)

    and this is what i got or attempted so far. i'm not even sure if this is the right way. can somebody tell me where i went wrong?

    [d/dx(2x^3-x)cos)1-x^2) x (2x^3-x)[d/dx(cos(1-x^2))]
    =(6x^2-1)cos(1-x^2) x 2x(2x^3-x)sin(1-x^2)

    thanks much!
    Replace the times-sign by -
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  3. #3
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    Quote Originally Posted by katieeej View Post
    ok, i was doing my hw and i just do not have a clue how to solve this.
    the problem is:
    (2x^3-x)cos(1-x^2)

    and this is what i got or attempted so far. i'm not even sure if this is the right way. can somebody tell me where i went wrong?

    [d/dx(2x^3-x)cos)1-x^2) x (2x^3-x)[d/dx(cos(1-x^2))]
    =(6x^2-1)cos(1-x^2) x 2x(2x^3-x)sin(1-x^2)

    thanks much!
    .
    No! The product rule: The derivative of fg is NOT f'g' nor is it f'- g'. It is (fg)'= f'g+ fg'.
    Last edited by HallsofIvy; February 13th 2009 at 06:56 AM.
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