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Math Help - differentiation question!

  1. #1
    Super Member
    Joined
    Oct 2007
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    Santiago
    Posts
    517

    differentiation question!

    <br /> <br />
f(x) = x\sqrt{x^2+1}<br /> <br />

    what i have so far is, leaving the x on the left aside for abit.

    <br /> <br />
f(x) = \sqrt{x^2+1} = (x^2 + 1)^\frac{1}{2}<br /> <br />

    then

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

    then

    <br /> <br />
f(x) = \frac{1}{2\sqrt{x^2 + 1}} \cdot (2x) = \frac{2x}{2\sqrt{x^2 + 1}}<br />

    now what do i do with the x? (thats if my calculations are correct)
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  2. #2
    Newbie
    Joined
    Sep 2007
    Posts
    6
    Product rule:
    <br />
\frac{d}{dx}\ (u\cdot v) = u \cdot \frac{dv}{dx} + v \cdot \frac{du}{dx}<br />
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  3. #3
    Junior Member
    Joined
    May 2008
    Posts
    50
    To continue from what dammitpoo said

    the product rule is

     \frac{d}{dx}[f(x)g(x)] = f(x) * \frac{d}{dx}[g(x)] + g(x)* \frac{d}{dx}[f(x)]

    where  f(x) = x and  g(x) = \sqrt{x^2 + 1}

    So  \frac{d}{dx}(f(x)) = 1 and  \frac{d}{dx}g(x) = \frac{x}{\sqrt{x^2 + 1}} ............which is what you calculated but you forgot to cancel the 2's.

    You should be able to finish it from here....just plug it in to the product rule above.
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