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    I've been killing myself trying to do this one:-

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

    I've tried the quotient rule :-

    dy/dx = [((x-1)^(1/2)).2 - (1/2(x-1)^(-1/2)).2x] / (x-1)

    upon cancelling down i get [x(x-1) - 2(x-1)(x-1)] / [(x-1)^(3/2)]

    the books answer is (x-2) / (x-1)^(3/2)

    Can anyone help?
    Thanks in advance
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    howdy

    after differentiating, factorise in terms of (x-1)^-1/2
    then you get that * by 2(x-1)-x on the top
    and if you take the factor to the bottom, multiply through you get the x-1)^3/2 at the bottom and simplifiying the top gives x-2

    dy/dx = \frac{2(x-1)^(1/2)-\frac{1}{2}2x(x-1)^(-1/2)}{x-1} = (x-1)^(-1/2)\frac{2(x-1)-x}{x-1} = \frac{x-2}{(x-1)^(3/2)}
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    (Original post by robfinlay)
    I've been killing myself trying to do this one:-

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

    I've tried the quotient rule :-

    dy/dx = [((x-1)^(1/2)).2 - (1/2(x-1)^(-1/2)).2x] / (x-1)

    upon cancelling down i get [x(x-1) - 2(x-1)(x-1)] / [(x-1)^(3/2)]

    the books answer is (x-2) / (x-1)^(3/2)

    Can anyone help?
    Thanks in advance
    This is right. now split it up into two separate terms to get:

    [2(x-1)^.5]/(x-1) - x/[(x-1)^1.5]

    Multiply LH term by [(x-1)^.5]/[(x-1)^.5] to get them both over the denominator (x-1)^1.5. Then it should all work out.
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    hmm, still dont understand. my arithmetic is messy
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    ahhh thanks alot.
 
 
 

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