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Math Help - Quotient Derivative

  1. #1
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    Quotient Derivative

    I need some help on a quotient derivative:

    \frac{a+\sqrt{x}}{a-\sqrt{x}}

    Here is what I have so far:

    1. \frac{(a-x^{1/2})(\frac{1}{2}x^{-1/2}) - (a+x^{1/2})(\frac{-1}{2}x^{-1/2})}{(a-\sqrt{x})^2}

    2. \frac{\frac{1}{2} ax^\frac{-1}{2} - \frac{1}{2}x^0 +\frac{1}{2} ax^\frac{-1}{2}- \frac{1}{2}x^0}{(a-\sqrt{x})^2}

    I know my algebra must be wrong because I'm coming up with a totally different answer than the book. Could someone please help give me a play by play?

    Thanks
    -db
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  2. #2
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    Quote Originally Posted by dbakeg00 View Post
    I need some help on a quotient derivative:

    \frac{a+\sqrt{x}}{a-\sqrt{x}}

    Here is what I have so far:

    1. \frac{(a-x^{1/2})(\frac{1}{2}x^{-1/2}) - (a+x^{1/2})(\frac{-1}{2}x^{-1/2})}{(a-\sqrt{x})^2}

    2. \frac{\frac{1}{2} ax^\frac{-1}{2} - \frac{1}{2}x^0 +\frac{1}{2} ax^\frac{-1}{2}- \frac{1}{2}x^0}{(a-\sqrt{x})^2}

    I know my algebra must be wrong because I'm coming up with a totally different answer than the book. Could someone please help give me a play by play?

    Thanks
    -db
    Multiply by \frac{a+\sqrt{x}}{a+ \sqrt{x}}

    \frac{a+\sqrt{x}}{a-\sqrt{x}} \times \frac{a+\sqrt{x}}{a+ \sqrt{x}} = \frac{(a+\sqrt{x})^2}{a^2-x}

    u = (a + \sqrt{x})^2 \: \rightarrow \: u' = a+\sqrt{x}

    v = a^2-x \: \rightarrow \: v' = -1



    y' = \frac{u'v-v'u}{v^2} = \frac{(a^2-x)(a+\sqrt{x}) + (a+\sqrt{x})^2}{(a^2-x)^2} = \frac{(a+\sqrt{x})(a^2-x+a+\sqrt{x})}{(a^2-x)^2}

    And that's the form I'd leave it in.
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  3. #3
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    Thank you. 2 questions though:

    1. How did you know to multiply by <br /> <br />
\frac{a+\sqrt{x}}{a+ \sqrt{x}}<br />

    2. the book is giving this answer:
    \frac{a}{(a-\sqrt{x})^2*\sqrt{x}}
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  4. #4
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    Quote Originally Posted by dbakeg00 View Post
    Thank you. 2 questions though:

    1. How did you know to multiply by <br /> <br />
\frac{a+\sqrt{x}}{a+ \sqrt{x}}<br />

    2. the book is giving this answer:
    \frac{a}{(a-\sqrt{x})^2*\sqrt{x}}
    I didn't, that was just a guess based on rationalising the denominator. Turned out to be wrong.

    Note that x^{-1/2} = \frac{1}{\sqrt{x}}

    u = a+\sqrt{x} \: \: \rightarrow \:\: u' = \frac{1}{2\sqrt{x}}

    v = a - \sqrt{x} \: \: \rightarrow \: \: v' = -\frac{1}{2\sqrt{x}}


    y' = \frac{\frac{1}{2\sqrt{x}}(a-\sqrt{x}) + \frac{1}{2\sqrt{x}}(a+\sqrt{x})}{(a - \sqrt{x})^2}

    Note that in the numerator there is a factor of \frac{1}{\sqrt{x}}

    =\frac{1}{2\sqrt{x}} \times \frac{(a-\sqrt{x})+(a+\sqrt{x})}{(a-\sqrt{x})^2}

    Because \frac{a}{b} \times \frac{c}{d} = \frac{ac}{bd} we can put that 2\sqrt{x} on the denominator. Also \sqrt{x} cancels in the numerator

    \frac{2a}{2\sqrt{x} \times (a-\sqrt{x})^2}

    Lastly 2 cancels to give the book's answer of \frac{a}{\sqrt{x} \cdot (a-\sqrt{x})^2}
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  5. #5
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    Quote Originally Posted by dbakeg00 View Post
    Thank you. 2 questions though:

    1. How did you know to multiply by <br /> <br />
\frac{a+\sqrt{x}}{a+ \sqrt{x}}<br />

    2. the book is giving this answer:
    \frac{a}{(a-\sqrt{x})^2*\sqrt{x}}
    If you simplify your answer in #1, you will get the answer given in the book

    <br />
\frac{(a-x^{1/2})(\frac{1}{2}x^{-1/2}) - (a+x^{1/2})(\frac{-1}{2}x^{-1/2})}{(a-\sqrt{x})^2}=\frac{\frac{1}{2} ax^\frac{-1}{2} - \frac{1}{2}x^0 +\frac{1}{2} ax^\frac{-1}{2}+ \frac{1}{2}x^0}{(a-\sqrt{x})^2}

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

    =\frac{a}{(a-\sqrt{x})^2*\sqrt{x}}
    Last edited by ione; March 8th 2010 at 01:57 PM.
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  6. #6
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    Quote Originally Posted by ione View Post
    If you simplify you original answer, you will get the answer given in the book

    <br />
\frac{\frac{1}{2} ax^\frac{-1}{2} - \frac{1}{2}x^0 +\frac{1}{2} ax^\frac{-1}{2}- \frac{1}{2}x^0}{(a-\sqrt{x})^2}=\frac{ax^\frac{-1}{2}}{(a-\sqrt{x})^2}

    =\frac{a}{(a-\sqrt{x})^2*\sqrt{x}}
    In the numerator, I see where you are getting the "-1" but I don't understand where the "+1" is coming from??
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  7. #7
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    Quote Originally Posted by dbakeg00 View Post
    In the numerator, I see where you are getting the "-1" but I don't understand where the "+1" is coming from??
    Sorry, I made a typo

    Check my edit
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  8. #8
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    Thank you everyone for all the help!
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