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Math Help - Logarithmic Differentiation

  1. #1
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    Logarithmic Differentiation

    Problem:
    $y=ln(x\sqrt{x^{2}-1}$)

    I can get this far:
    $y=ln(x)+ln(x^{2}-1)^{1/2}$

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

    But I get confused on where to go from here. Thanks in advance for any help.
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  2. #2
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    Quote Originally Posted by kwikness View Post
    Problem:
    $y=ln(x\sqrt{x^{2}-1}$)

    I can get this far:
    $y=ln(x)+ln(x^{2}-1)^{1/2}$

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

    But I get confused on where to go from here. Thanks in advance for any help.

    Do you have to differentiate that function? What's the problem? For the first summand

    apply directly the derivative of ln x, and for the second one use the chain

    rule or \ln (x^2-1)=\ln (x-1)+\ln (x+1) .

    Tonio
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  3. #3
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    Sorry, should have mentioned it was a differentiation problem. I get this far:
    $y=ln(x\sqrt{x^{2}-1}$)
       $y=ln(x)+ln(x^{2}-1)^{1/2}$
       $=ln(x)+(1/2)ln(x^{2}-1)$
       $y'=1/x+((1/2)/(x^{2}-1))$

    Is this right so far?
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  4. #4
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    Quote Originally Posted by kwikness View Post
    Sorry, should have mentioned it was a differentiation problem. I get this far:
    $y=ln(x\sqrt{x^{2}-1}$)
       $y=ln(x)+ln(x^{2}-1)^{1/2}$
       $=ln(x)+(1/2)ln(x^{2}-1)$
       $y'=1/x+((1/2)/(x^{2}-1))$
    The answer should be \displaystyle \frac{1}{x}+\frac{x}{x^2+1}.
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  5. #5
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    Quote Originally Posted by Plato View Post
    The answer should be \displaystyle \frac{1}{x}+\frac{x}{x^2+1}.
    Can you explain how you get there from my 3rd step? I'm trying to use the product rule for:
    (1/2)ln(x^2-1)

    Since (1/2) is a constant (0.5) it cancels right? And I'm left with just the
    (1/2)/(1/(x^2-1))
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  6. #6
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    You need to use the chain rule on \dfrac{1}{2}\ln(x^2-1).

    In other words multiply by the derivative of x^2-1 and see what cancels
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  7. #7
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    Sorry, i'm really struggling with this. It's been a while.

    Multiply what by the derivative of x^2-1 (2x)

    Can you show me step by step from where I went wrong?
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  8. #8
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    The derivative of \ln(\sqrt{x^2+1}) is \dfrac{x}{x^2+1}.

    Now you tell us what about that you do not understand.
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  9. #9
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    Quote Originally Posted by Plato View Post
    The derivative of \ln(\sqrt{x^2+1}) is \dfrac{x}{x^2+1}.

    Now you tell us what about that you do not understand.
    My problem is getting that derivative. I've only found the derivative of simple natural logs like ln(x) where I can just differentiate to 1/x.
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  10. #10
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    Quote Originally Posted by kwikness View Post
    My problem is getting that derivative. I've only found the derivative of simple natural logs like ln(x) where I can just differentiate to 1/x.
    It could be that you are simply not ready to do this question.
    In that case you need to have a sit down one to one with an tutor.
    We here do not provide that service.
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  11. #11
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    I'm not asking for tutoring. If I was ready to do this question I wouldn't have had to post it.

    I figured it out eventually though. Thanks for the help guys.

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

    y'=(0)(ln(x^{2}-1)+(\frac{1}{2})(\frac{2x}{x^{2}-2})

    =\frac{2x}{2x^{2}-2}

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

    =\frac{x}{x^{2}-1}
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  12. #12
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    The purpose of logarithmic differentiation is to make use of logarithm rules to simplify the function, so that difficult functions can be differentiated more easily.

    Make use of the fact that \displaystyle \ln{(x^2 - 1)} = \ln{[(x - 1)(x + 1)]} = \ln{(x- 1)} + \ln{(x + 1)}.

    I'm sure you can differentiate the last term...
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