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Math Help - Help with logarithmic differention

  1. #1
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    Help with logarithmic differention

    Hey I need help with this problem:

    Solve using logarithmic differentiation:
    y=((x^2+1)/(x^2-1))^(1/4)

    Thanks!
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  2. #2
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    Hello

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

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

    differentiating w.r.t.x

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

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

    therefore

     \frac{dy}{dx} = \frac{1}{2} \left( \frac{x^2 +1}{x^2 -1} \right)^{\frac{1}{4}} \left( \frac{x}{x^2 +1} - \frac{x}{x^2 -1} \right)
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  3. #3
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by luckeyduck View Post
    Hey I need help with this problem:

    Solve using logarithmic differentiation:
    y=((x^2+1)/(x^2-1))^(1/4)

    Thanks!
    first log both sides:

    \Rightarrow \ln y = \ln \left[ \frac {x^2 + 1}{(x + 1)(x - 1)} \right]^{1/4}

    \Rightarrow \ln y = \frac 14 \ln (x^2 + 1) - \frac 14 \ln (x + 1) - \frac 14 \ln (x - 1)

    now differentiate implicitly
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