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Thread: Beginner in finding derivative of a function...

  1. #1
    Junior Member
    Joined
    May 2007
    Posts
    36

    Beginner in finding derivative of a function...

    Find the derivative of:


    $\displaystyle f(x) = \sqrt {(x^2 + 1)(\sqrt x + 1)^3 }$

    My work so far:


    $\displaystyle f'(x) = \frac{1}
    {2}[(x^2 + 1)(\sqrt x + 1)^3 ]^{ - \frac{1}
    {2}} \bullet \frac{d}
    {{dx}}[(x^2 + 1)(\sqrt x + 1)^3 $


    $\displaystyle
    f'(x) = \frac{1}
    {2}[(x^2 + 1)(\sqrt x + 1)^3 ]^{ - \frac{1}
    {2}} \bullet (2x)(\sqrt x + 1)^3 + (x^2 + 1)3(x^{\frac{1}
    {2}} + 1)^2 $
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  2. #2
    Senior Member
    Joined
    Dec 2007
    From
    Anchorage, AK
    Posts
    276

    Close

    You're close, but just a little off on your last shown step:
    $\displaystyle \frac{d}{dx}(\sqrt{x}+1)^3=3(\sqrt{x}+1)^2\cdot\fr ac{d}{dx}\sqrt{x}$, via the chain rule, so the last line should be:
    $\displaystyle f'(x)=\frac{1}{2}[(x^2+1)(\sqrt{x}+1)^3]^{-\frac{1}{2}}\cdot\left[(2x)(\sqrt{x}+1)^3+(x^2+1)\cdot3(\sqrt{x}+1)^2\fra c{1}{2}x^{-\frac{1}{2}}\right]$
    Which can be simplified via algebra.

    --Kevin C.
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