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Math Help - My 2 worst enemies combined...

  1. #1
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    Post My 2 worst enemies combined...

    Fractions and Square Roots...

    Differentiate and simplify.

    G(x) = square root of (3x+1/2x-1)

    First, do I deal with this as a whole or break it down as two seperate functions: (3x+1) and (2x-1^1/2)
    Second, I am not sure if I have written it to the correct power: 3x+1/2x-1^-1/2

    -1/2(3x+1/2x-1)^-3/2
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  2. #2
    Member Glaysher's Avatar
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    Use Quotient rule with

    u = \sqrt{3x +1} = (3x + 1)^{\frac{1}{2}}

    v = \sqrt{2x - 1} = (2x - 1)^{\frac{1}{2}}

    EDIT

    My approach is slightly different from topsquark. You will need to use the chain rule twice with mine but they will both work in exactly the same way
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  3. #3
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by becky
    Fractions and Square Roots...

    Differentiate and simplify.

    G(x) = square root of (3x+1/2x-1)

    First, do I deal with this as a whole or break it down as two seperate functions: (3x+1) and (2x-1^1/2)
    Second, I am not sure if I have written it to the correct power: 3x+1/2x-1^-1/2

    -1/2(3x+1/2x-1)^-3/2
    You need the chain rule and quotient rule.

    G(u) = \sqrt{u} and u(x) = \frac{3x+1}{2x-1}.

    \frac{dG}{dx} = \frac{dG}{du} \cdot \frac{du}{dx}.

    So
    \frac{dG}{du} = \frac{d}{du}u^{1/2} = \frac{1}{2}u^{-1/2} = \frac{1}{2\sqrt{u}} = \frac{ \sqrt{2x-1}}{2 \sqrt{3x+1}}

    \frac{du}{dx} = \frac{(3)(2x-1) - (3x+1)(2)}{(2x-1)^2} = \frac{-5}{(2x-1)^2}

    Finally: \frac{dG}{dx} = \frac{ \sqrt{2x-1}}{2 \sqrt{3x+1}}\frac{-5}{(2x-1)^2}

    \frac{dG}{dx} = \frac{-5}{2 \sqrt{3x+1}(2x-1)^{3/2}}

    -Dan
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