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Math Help - Help: Product Differentiation!

  1. #1
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    Question Help: Product Differentiation!

    Hi, Could someone give me somw much needed guidance on the following question:
    Differentiate: f(x) = 2cos(3x - 1)/3x ?
    I'm struggling to start with this one.
    Firstly does 2cos(3x - 1)/3x become: 2/3 xcos(1 - 3x), I'm trying to remove the division by 3x?

    Can someone please give me some help so I can get on with this?

    Thanks in advance.
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  2. #2
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    Quote Originally Posted by joemc22 View Post
    Hi, Could someone give me somw much needed guidance on the following question:
    Differentiate: f(x) = 2cos(3x - 1)/3x ?
    I'm struggling to start with this one.
    Firstly does 2cos(3x - 1)/3x become: 2/3 xcos(1 - 3x), I'm trying to remove the division by 3x?

    Can someone please give me some help so I can get on with this?

    Thanks in advance.
    Use the quotient rule \left(\frac{u}{v}\right)' = \frac{u'v - u v'}{v^2}
    Last edited by Jester; May 18th 2009 at 11:07 AM. Reason: Fixing what the Gremlins changed :-)
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  3. #3
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    Product differentiation

    OK, it's taking me a bit of time to get back into it!

    So I have:
    f(x)= 2cos(3x-1)/3x.
    So u= 2cos(3x-1) and v= 3x
    du/dx= 6sin(1-3x) and dv/dx= 3?
    Using quotent rule I get:
    (3x)6sin(1-3x) - 2cos(3x-1) x 3/3x^2, changing cos to sin and multiplying the last part by 3 I would then get the answer as 3x/3x^2

    Does this look right, if not could you give me some pointers where i've gone wrong?

    Thanks
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  4. #4
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    Quote Originally Posted by joemc22 View Post
    OK, it's taking me a bit of time to get back into it!

    So I have:
    f(x)= 2cos(3x-1)/3x.
    So u= 2cos(3x-1) and v= 3x
    du/dx= 6sin(1-3x) and dv/dx= 3?

    Using quotent rule I get:
    (3x)6sin(1-3x) - 2cos(3x-1) x 3/3x^2, (3x)^2 = 3^2 * x^2 = 9x^2. You must also square the three.
    changing cos to sin and multiplying the last part by 3 I would then get the answer as 3x/3x^2

    Does this look right, if not could you give me some pointers where i've gone wrong?

    Thanks
    The function inside the brackets does not change when using the chain rule:

    f(x) = \frac{2cos(3x-1)}{3x}

    u = 2cos(3x-1)
    \frac{du}{dx} = -6sin(3x-1)

    v = 3x
    \frac{dv}{dx} = 3

    f'(x) = \frac{vu'-uv'}{v^2} = \frac{-18xsin(3x-1)-6cos(3x-1)}{9x^2}
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  5. #5
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    Thanks,

    Wasn't sure what to change the 2sin(3-1x) into, the self help lessons i'm learning from sometimes don't have the examples i need, but you help has made it clearer.
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  6. #6
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    Quote Originally Posted by danny arrigo View Post
    Use the quotient rule \left(\frac{u}{v}\right)' = \frac{u'v + u v'}{v^2}
    Don't you mean \left(\frac{u}{v}\right)' = \frac{u'v - u v'}{v^2}
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  7. #7
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    Quote Originally Posted by Spec View Post
    Don't you mean \left(\frac{u}{v}\right)' = \frac{u'v - u v'}{v^2}
    You're so right - how'd that negative turn to a postive. Must be the MHF gremlins
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