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Math Help - derivatives - double check my work please

  1. #1
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    Post derivatives - double check my work please

    Please double check my work - just doesn't 'feel' right and I've always had a problem with what I call 'three stacked fractions'.
    Instructions: Compute the derivative of the given function .

    h(u) = 1/square root of u

    My solution: h'(u) lim h->0 1/square root x+h - 1/square root u/h=

    square root u/x+h - square root x+h/u /h(square root x+h)(square root u)=

    square root u(x+h) - square root x+h(x+h/h(square root u)(x+h)^2=

    -square root x.
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  2. #2
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    Quote Originally Posted by becky
    Please double check my work - just doesn't 'feel' right and I've always had a problem with what I call 'three stacked fractions'.
    Instructions: Compute the derivative of the given function .

    h(u) = 1/square root of u

    My solution: h'(u) lim h->0 1/square root x+h - 1/square root u/h=

    square root u/x+h - square root x+h/u /h(square root x+h)(square root u)=

    square root u(x+h) - square root x+h(x+h/h(square root u)(x+h)^2=

    -square root x.
    You should get for, \frac{1}{\sqrt{u}} the answer of,
    2\sqrt{u}
    Last edited by ThePerfectHacker; May 22nd 2006 at 03:03 PM.
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by ThePerfectHacker
    You should get for, \frac{1}{\sqrt{u}} the answer of,
    2\sqrt{u}
    <br />
\frac{d}{du} \left[\frac{1}{\sqrt{u}}\right]=-\frac{1}{2}\frac{1}{(\sqrt{u})^3}<br />

    RonL
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  4. #4
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    Quote Originally Posted by CaptainBlack
    <br />
\frac{d}{du} \left[\frac{1}{\sqrt{u}}\right]=-\frac{1}{2}\frac{1}{(\sqrt{u})^3}<br />

    RonL
    Sorry
    I took the integral
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