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Math Help - Stuff on product rule problem

  1. #1
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    Stuff on product rule problem

    Question is:

    y = (x + 1) \sqrt[x - 1]

    So I convert to

    y = (x + 1) (x - 1)^(1/2)

    Then

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

    y= (x - 1)^{1/2} + \frac{(x + 1}{2(x - 1)^{\frac{1}{2}}}

    Is this correct?

    Book says numerator is 3x - 1!

    What have I done wrong?

    Angus
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  2. #2
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    Re: Stuff on product rule problem

    You haven't done anything wrong, you just need to combine the terms of your answer.
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  3. #3
    MHF Contributor Siron's Avatar
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    Re: Stuff on product rule problem

    You have given:
    y=(x+1)\cdot \sqrt{x-1}
    which is equal to:
    y=(x+1)\cdot (x-1)^{\frac{1}{2}}
    Using the product rule gives us:
    y'=\frac{d}{dx}(x+1)\cdot (x-1)^{\frac{1}{2}}+(x+1)\cdot\frac{d}{dx}\left[(x-1)^{\frac{1}{2}}\right]
    y'=(x-1)^{\frac{1}{2}}+\frac{(x+1)\cdot (x-1)^{\frac{-1}{2}}}{2}
    y'=\sqrt{x-1}+\frac{x+1}{2\sqrt{x-1}}

    Now make a common denominator and add the fractions.
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  4. #4
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    Re: Stuff on product rule problem

    Rationalize the denominator.
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  5. #5
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    Re: Stuff on product rule problem

    Quote Originally Posted by SammyS View Post
    Rationalize the denominator.
    I did struggle with the rationalising. I was thinking

    \sqrt{x-1} * \sqrt{x-1}

    would be multiplied like (x - 1)(x - 1) - ie x * x, x * -1 etc, but of course it is not. you just add the powers. ie (x - 1)^(1/2) (x - 1)^(1/2) = (x - 1)

    Sorry I will learn how to do latex some time. I put around [tex] tags but it doesn't work for me.
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  6. #6
    MHF Contributor Siron's Avatar
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    Re: Stuff on product rule problem

    Indeed: \sqrt{x-1}\cdot \sqrt{x-1}=x-1 therefore you get:
    y'=\sqrt{x-1}+\frac{x+1}{2\sqrt{x-1}}=\frac{2(x-1)}{2\sqrt{x-1}}+\frac{x+1}{2\sqrt{x-1}}=\frac{2(x-1)+x+1}{2\sqrt{x-1}}=\frac{3x-1}{2\sqrt{x-1}}

    Note:
    If you want to use LaTEX then use [ TEX ] ... [ /TEX ] in stead of [ MATH ] ... [ /MATH ]
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