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Math Help - Derivatives of Natural Logs question

  1. #1
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    Derivatives of Natural Logs question

    Hello, I'm having difficulty with the following question:

    Solve the equation f'(x)=0.

    f(x)= (x^2 + 1)^-1 ln (x^2 + 1)


    I've worked out the derivative to be ln(x+1) divided by (x^2 + 1)^2. However, I'm unsure if this is correct, and if it is, I don't know how to solve for the x values once I set the derivative equal to zero- the natural log in the numerator throws me off. Thank you for any help!
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  2. #2
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    f(x)=\frac{ln(x^{2}+1)}{x^{2}+1}

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

    Now, what makes the numerator = 0?.

    -2x(ln(x^{2}+1)-1)=0

    0 is obviously one solution.

    ln(x^{2}+1)-1=0

    ln(x^{2}+1)=1

    x^{2}+1=e

    x=\pm\sqrt{e-1}
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  3. #3
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    Quote Originally Posted by galactus View Post
    f(x)=\frac{ln(x^{2}+1)}{x^{2}+1}

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

    Now, what makes the numerator = 0?.

    -2x(ln(x^{2}+1)-1)=0

    0 is obviously one solution.

    ln(x^{2}+1)-1=0

    ln(x^{2}+1)=1

    x^{2}+1=e

    x=\pm\sqrt{e-1}
    Thanks very much- but how did you get the derivative? I can't seem to work out the derivative correctly.
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  4. #4
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    Actually, I ran it through my calculator, but we can do it the 'old-fashioned' way.

    We can use the quotient rule:

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

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

    Factor:

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

    Which is the same thing as I posted. My calculator just displayed it that way.
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