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Math Help - Chain rule problem?

  1. #1
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    Chain rule problem?

    Please help

    f(x)= 1/2x - 1/3x^2
    =(2x)^-1 -(3x)^-2
    f ' =(-2x)^-2 + (6x)^-3
    =-2/x^2 + 6/x^-3
    This is as far as I can get, the finished answer should be 4-3x/6x^3

    Can you show me how to finish it please? Thanks
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  2. #2
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    Quote Originally Posted by wolfhound View Post
    Please help

    f(x)= 1/2x - 1/3x^2
    =(2x)^-1 -(3x)^-2
    f ' =(-2x)^-2 + (6x)^-3
    =-2/x^2 + 6/x^-3
    This is as far as I can get, the finished answer should be 4-3x/6x^3

    Can you show me how to finish it please? Thanks
    For clarification: Do you mean

    f(x)= 1/2*x - 1/3*x^2

    or

    f(x)= 1/(2x) - 1/(3x^2)

    or

    f(x)= 1/(2x) - 1/((3x)^2)
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  3. #3
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    Quote Originally Posted by earboth View Post
    For clarification: Do you mean



    or

    f(x)= 1/(2x) - 1/((3x)^2)
    Hello, this one please help
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  4. #4
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    f(x) = (2x)^{-1} - (3x)^{-2}

    f'(x) = -(2x)^{-2} \cdot 2 + 2(3x)^{-3} \cdot 3

    f'(x) = -\frac{2}{(2x)^2} + \frac{6}{(3x)^3}

    you can finish the algebra grunt work
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  5. #5
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    Quote Originally Posted by skeeter View Post
    f(x) = (2x)^{-1} - (3x)^{-2}

    f'(x) = -(2x)^{-2} \cdot 2 + 2(3x)^{-3} \cdot 3

    f'(x) = -\frac{2}{(2x)^2} + \frac{6}{(3x)^3}

    you can finish the algebra grunt work
    I dont know how to finish it!
    someone help please
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  6. #6
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    Quote Originally Posted by wolfhound View Post
    I dont know how to finish it!
    someone help please
    Take skeeter's result:

    1. Expand the terms at the RHS. Cancel if possible.

    2. Determine a common denominator. Transform the fractions such that they have the same denominator. Collect like terms in the numerator.

    First step:

    <br />
f'(x) = -\frac{2}{(2x)^2} + \frac{6}{(3x)^3} = -\frac{1}{2x^2}+\frac{2}{9x^3}<br />

    Next steps for you.
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