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    Hiya,

    I've been doing a differentiation question and when I got the final answer, i've checked my answer many times but it doesn't match the book's answer at all, so i'm just wondering if you guys can help me see if i have done a mistake or the book is wrong.

    dy/dx = 3x/(2√(3x-2)) + √(3x-2)

    i simplified it to 9x-4/(2(3x-2))

    but the book says ((9x-4)√3x-2)/(2(3x-2)

    i don't know how they still have √3x-2 on top of the fraction..... =S

    sorry it's a bit messy..
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    Can you post your working? As far as I can see, the book's answer is correct.
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    (Original post by strwbry_short_cake)
    Hiya,

    I've been doing a differentiation question and when I got the final answer, i've checked my answer many times but it doesn't match the book's answer at all, so i'm just wondering if you guys can help me see if i have done a mistake or the book is wrong.

    dy/dx = 3x/(2√(3x-2)) + √(3x-2)

    i simplified it to 9x-4/(2(3x-2))

    but the book says ((9x-4)√3x-2)/(2(3x-2)

    i don't know how they still have √3x-2 on top of the fraction..... =S

    sorry it's a bit messy..
    The book is right. When you ADD fractions, you don't times the denominator, they become the original. Thats where you went wrong

    Ie:

     \frac{1}{x} + \frac {1}{x} = \frac {2}{x}
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    alright, here's my working......(i've probably done something really stupid.... ><
    sorry it's so messy...i don't know how to input fractions here...

    1. 3x/(2√3x-2) + √3x-2
    2. 3x/(2√3x-2) + 2(3x-2)/(2√3x-2).........(times by 2√3x-2 to the second fraction)
    i got 2(3x-2) because doesn't a √ x √ cancels........(√3x-2)(√3x-2) = 3x-2?
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    To be honest, I got (9x-4)/2Root(3x-2) which is same as book afaik

    just get denominators the same and add.
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    (Original post by strwbry_short_cake)
    [OP]
    ...
    i simplified it to 9x-4/(2(3x-2))
    ...
    Did you mean this to be \frac{9x-4}{2\sqrt{3x-2}} ?
    If so, then multiply top and bottom by \sqrt{3x-2} and compare with the book.
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    (Original post by strwbry_short_cake)
    alright, here's my working......(i've probably done something really stupid.... ><
    sorry it's so messy...i don't know how to input fractions here...

    1. 3x/(2√3x-2) + √3x-2
    2. 3x/(2√3x-2) + 2(3x-2)/(2√3x-2).........(times by 2√3x-2 to the second fraction)
    i got 2(3x-2) because doesn't a √ x √ cancels........(√3x-2)(√3x-2) = 3x-2?
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    ah i see what mistake i have made now...thanks guys ><

    so is it ok to leave it as 9x-4/(2√3x-2)? cos the answer in the book is ((9x-4)√3x-2)/2(3x-2)..
    is it not as tidy to leave a square root in the denominator?
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    (Original post by strwbry_short_cake)
    ah i see what mistake i have made now...thanks guys ><

    so is it ok to leave it as 9x-4/(2√3x-2)? cos the answer in the book is ((9x-4)√3x-2)/2(3x-2)..
    is it not as tidy to leave a square root in the denominator?
    Some books/people/syllabi don't like roots in the denominator. I can't really see the problem with them.
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    1/2 + 1/2 = 1, not 1/2
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    :confused:
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    *remembers why he stopped doing maths at GCSE
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    (Original post by Barton1)
    *remembers why he stopped doing maths at GCSE
    This is towards the low end of AS/high end of GCSE! Maths gets a lot more fun than this! :yes::nutcase:
 
 
 
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