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Math Help - Use logarithmic differentiation to find the derivative of y= (x√(x+1))/(2 + x^4)

  1. #1
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    Use logarithmic differentiation to find the derivative of y= (x√(x+1))/(2 + x^4)

     y= \frac{x\sqrt{x+1}}{(2 + x^4)}
    lny= ln \frac{x\sqrt{x+1}}{(2 + x^4)}
    lny= ln({x\sqrt{x+1}) - ln{(2 + x^4)}

    Is that correct so far? And any advice for what to do after...I've never done a question like this and it's confusing me!
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  2. #2
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    Write it as \ln (y) = \ln (x) + \frac{1}{2}\ln (x + 1) - \ln (2 + x^4 )
    Last edited by Plato; November 23rd 2010 at 11:12 AM.
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  3. #3
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    Thanks..I'm trying to work it out.

     \frac{y'}{y}= \frac{1}{x} + \frac{1}{2x+2}-\frac{1}{2+x^4}

    Am I on the right track or way off base?
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  4. #4
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    You missed a 4x^3 is the last term.
    Now multiply all by y.
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  5. #5
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    I don't understand your last term. Given that \dfrac{d}{dx} \ln[f(x)] = \dfrac{f'(x)}{f(x)}

    In your case \dfrac{d}{dx} \ln(2+x^2) = \dfrac{4x^3}{2+x^4}

    Rest of it is fine though, remember that you made a y sub when multiplying by y.
    Last edited by e^(i*pi); November 23rd 2010 at 11:13 AM. Reason: I copied the typo in post 2
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  6. #6
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    Thanks so much for the help
    I think I have the solution figured out now.
    y'= -\frac{(5x^5 + 6x^4 - 6x - 4)y}{2x(x + 1)(x^4 + 2)}
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  7. #7
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    Quite frankly, I prefer the answer form \displaystyle y' = \frac{{x\sqrt {x + 1} }}{{2 + x^4 }}\left\{ {\frac{1}{x} + \frac{1}<br />
{{2\left( {x + 1} \right)}} - \frac{{4x^3 }}<br />
{{2 + x^4 }}} \right\}

    But you must follow your instructions.
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