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Math Help - If [math]y = In \left[ \frac{x(x^2 + 1)}{\sqrt{x^3 - 1}} \right] [/math]

  1. #1
    Junior Member bijosn's Avatar
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    If [math]y = In \left[ \frac{x(x^2 + 1)}{\sqrt{x^3 - 1}} \right] [/math]

    y = In \left[ \frac{x(x^2 + 1)}{\sqrt{x^3 - 1}} \right] find y'

    I began this by
    letting u=\frac{x(x^2+1)}{\sqrt{x^3 - 1}}

    y' = \frac{1}{u}.u'

    the answer I get is

    \frac{3x}{x^2 + 1} + \frac{1}{x(x^2+1)} -  \frac{3x^2}{2(x^3 - 1)}

    which is wrong
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  2. #2
    MHF Contributor
    Prove It's Avatar
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    What are you trying to do? Differentiate this function?

    \displaystyle y = \ln{\left[\frac{x(x^2 + 1)}{\sqrt{x^3 - 1}}\right]}

    \displaystyle e^y = (x^3 + x)(x^3 - 1)^{-\frac{1}{2}}

    \displaystyle \frac{d}{dx}\left(e^y\right) = \frac{d}{dx}\left[(x^3 + x)(x^3 - 1)^{-\frac{1}{2}}\right]

    \displaystyle \frac{d}{dy}\left(e^y\right)\,\frac{dy}{dx} = (x^3 + x)\,\frac{d}{dx}\left[(x^3 - 1)^{-\frac{1}{2}}\right] + (x^3 - 1)^{-\frac{1}{2}}\,\frac{d}{dx}(x^3 + x).


    Go from here...
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  3. #3
    Junior Member bijosn's Avatar
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    Quote Originally Posted by Prove It View Post
    What are you trying to do? Differentiate this function?

    \displaystyle y = \ln{\left[\frac{x(x^2 + 1)}{\sqrt{x^3 - 1}}\right]}

    \displaystyle e^y = (x^3 + x)(x^3 - 1)^{-\frac{1}{2}}

    \displaystyle \frac{d}{dx}\left(e^y\right) = \frac{d}{dx}\left[(x^3 + x)(x^3 - 1)^{-\frac{1}{2}}\right]

    \displaystyle \frac{d}{dy}\left(e^y\right)\,\frac{dy}{dx} = (x^3 + x)\,\frac{d}{dx}\left[(x^3 - 1)^{-\frac{1}{2}}\right] + (x^3 - 1)^{-\frac{1}{2}}\,\frac{d}{dx}(x^3 + x).




    Go from here...
    thank you, yes, I am trying to differentiate it
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  4. #4
    Junior Member bijosn's Avatar
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    \displaystyle \frac{d}{dy}\left(e^y\right)\,\frac{dy}{dx} = (x^3 + x)\,\frac{d}{dx}\left[(x^3 - 1)^{-\frac{1}{2}}\right] + (x^3 - 1)^{-\frac{1}{2}}\,\frac{d}{dx}(x^3 + x).


    I am a bit lost and rusty, I could do this a month back but now I've forgotten what to do.

    the right hand side looks like this
    \displaystyle = (x^3 + x)\,(-\frac{1}{2}(x^3 - 1)^{-\frac{3}{2}} 3x^2) + (x^3 - 1)^{-\frac{1}{2}}\,(3x^2 + 1)

    what do I do now
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