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Math Help - Logarithm Derivative (easy...except for me)

  1. #1
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    Logarithm Derivative (easy...except for me)

    Ok we have 2 problems here....one is just a slightly harder than the other


    y=ln x^2+1

    y=ln (x2+1)/x^2



    Please post steps!!! Thanks
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  2. #2
    Senior Member Pinkk's Avatar
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    Remember: \frac{d}{dx}ln(u)=\frac{1}{u}\cdot\frac{d}{dx}u

    So for the first problem, u=x^{2}+1,

    \frac{d}{dx}ln(x^{2}+1)=\frac{1}{x^{2}+1}\cdot\fra  c{d}{dx}(x^{2}+1)=\frac{2x}{x^{2}+1}

    You should be able to do the second problem now.
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  3. #3
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    Do I split the numerator and the denominator?


    So...  2x/(x+1) + 2/x
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  4. #4
    Senior Member Pinkk's Avatar
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    What you wrote is somewhat unclear. If the problem is:

    \frac{ln(x^{2}+1)}{x^{2}} You'll need to invoke log differentiation AND the quotient rule:

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

    And you can just simplify from there.
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  5. #5
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    Hmm my calculus professor has the answer as


     2x/(x^2+1)-2/x


    The only way I could figure out how to get that answer was to make the problem

     ln x^2+1- ln x^2
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  6. #6
    Super Member redsoxfan325's Avatar
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    Maybe the problem was \frac{d}{dx}\left[\ln\left(\frac{x^2+1}{x^2}\right)\right] instead of \frac{d}{dx}\left[\frac{\ln(x^2+1)}{x^2}\right].

    Because \frac{d}{dx}\left[\ln\left(\frac{x^2+1}{x^2}\right)\right] = \frac{d}{dx}[\ln(x^2+1)] - \frac{d}{dx}[\ln(x^2)] = \frac{2x}{x^2+1}-\frac{2}{x} like you said.
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  7. #7
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    Quote Originally Posted by redsoxfan325 View Post
    Maybe the problem was \frac{d}{dx}\left[\ln\left(\frac{x^2+1}{x^2}\right)\right] instead of \frac{d}{dx}\left[\frac{\ln(x^2+1)}{x^2}\right].

    Because \frac{d}{dx}\left[\ln\left(\frac{x^2+1}{x^2}\right)\right] = \frac{d}{dx}[\ln(x^2+1)] - \frac{d}{dx}[\ln(x^2)] = \frac{2x}{x^2+1}-\frac{2}{x} like you said.

    You are correct. When I was first doing the problem I was using the quotient rule and I was getting the wrong answer


    So I put two and two together (pun intended) and I figured I had to split the numerator/denominator


    +rep to everyone for helping
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