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Math Help - Help finding derivative of logs

  1. #1
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    Help finding derivative of logs

    I have a problem that states: Find the derivative of y with respect to x

    y = ln (1/x(sqroot x + 1)

    The answer in the back of the book has (3x + 2) / 2x(x+1) as the answer and I don't understand how they came to this.
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  2. #2
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    Hello, KarlosK!

    Exactly where is your difficulty?
    . . Log rules?
    . . The derivative of logs?
    . . You really suck at algebra?


    Differentiate: . y \:=\:\ln\left(\frac{1}{x\sqrt{x + 1}}\right)

    The answer in the back of the book has: \frac{3x + 2}{2x(x+1)}
    . . This should have a "minus" in front.
    Simplify first . . .

    y \;=\;\ln\left(\frac{1}{x\sqrt{x+1}}\right) \;=\;\underbrace{\ln(1)}_{\text{This is 0}} - \ln\left(x\sqrt{x+1}\right) \;=\;-\ln\left(x\sqrt{x+1}\right)

    . =\;-\ln\left[x(x+1)^{\frac{1}{2}}\right] \;=\; -\bigg[\ln x + \ln(x+1)^{\frac{1}{2}}\bigg]


    Now differentiate:

    . . y' \;=\;-\bigg[\frac{1}{x} + \frac{1}{2}\!\cdot\!\frac{1}{x+1}\bigg] \;=\;-\bigg[\frac{2(x+1) + x}{2x(x+1)}\bigg] \;=\;-\frac{3x+2}{2x(x+1)}

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