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Math Help - Differentiating natural logs

  1. #1
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    Differentiating natural logs

    I'm having a little trouble differentiating these two functions, if anyone could help I would greatlyyy appreciate it.

    g(x) = ln (x * sqrt(x^(2)-1))

    and

    f(u) = ln u / (1+ln(2u))

    Whatever I do that makes sense to me, I end up with (1/x) + ((x) / (x^(2) - 1)) for g(x) and for the second, I used the quotient rule and can't end up with anything similar to the answers for either. Thank you very much in advance!
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  2. #2
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    Quote Originally Posted by Celestiax View Post
    I'm having a little trouble differentiating these two functions, if anyone could help I would greatlyyy appreciate it.

    g(x) = ln (x * sqrt(x^(2)-1))

    and

    f(u) = ln u / (1+ln(2u))

    Whatever I do that makes sense to me, I end up with (1/x) + ((x) / (x^(2) - 1)) for g(x) and for the second, I used the quotient rule and can't end up with anything similar to the answers for either. Thank you very much in advance!
    Dear Celestiax,

    For the first one you should use the chain rule and the product rule of differentiation.

    That is, \frac{dy}{dx}=\frac{dy}{du}.\frac{du}{dx}~where~u~  and~y~are~fuctions~of~x.

    \frac{d}{dx}(v.u)=v.\frac{du}{dx}+u.\frac{dv}{dx}

    Also you should know that, \frac{d}{dx}\ln{x}=\frac{1}{x}

    Hope this will help you.
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  3. #3
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    I'll do the first step on each one:

    g'(x)=\frac{1}{x\sqrt{x^{2}-1}}\,\frac{d}{dx}(x\sqrt{x^{2}-1}), and

    f'(u)=\frac{(1+\ln(2u))\frac{d}{du}\,\ln(u)-\ln(u)\frac{d}{du}(1+\ln(2u))}{(1+\ln(2u))^{2}}. Can you continue?
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  4. #4
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    Quote Originally Posted by Celestiax View Post
    Whatever I do that makes sense to me, I end up with (1/x) + ((x) / (x^(2) - 1))
    Correct.
    ... and for the second, I used the quotient rule and can't end up with anything similar to the answers for either
    Post your answer/working. The chances are it's probably correct as well but it just isn't on the form in which it's in the book.
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  5. #5
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    Thank you to both of you, that is what I did for f(u), but I don't understand how you simplify the numerator to only (1 + ln 2), and how the 2u in the denominator cancels out? And for g(x), I cannot make it g(x) = lnx + (1/2) ln (x^(2) - 1)?
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  6. #6
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    The answers in the book are: g(x) = (2x^(2) - 1) / (x (x^(2) -1) and f(u) = (1 + ln 2) / ( (u) (1 + ln(2u))^(2) I guess I just didn't combine the terms for g(x)?
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  7. #7
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    I must confess, I don't see how an ln(2u) can become an ln(2). Are you sure there's no u there? Be careful with parentheses!
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  8. #8
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    Quote Originally Posted by Celestiax View Post
    The answers in the book are: g(x) = (2x^(2) - 1) / (x (x^(2) -1) I guess I just didn't combine the terms for g(x)?
    Yup!

    [LaTeX ERROR: Convert failed]
    Differentiate, either by writing it as [LaTeX ERROR: Convert failed] and using the chain rule, or using the quotient rule (as laid out earlier in the thread), and we will just work out how what we get is equivalent to the book's answer... don't worry about the simplification for now.
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  9. #9
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    Ah, thank you very much! I worked it all out now and I believe I have it correct...I didn't know ln(2u) - ln(u) could just be ln(2) for f(u). Although I'm not sure where the 2u in the denominator goes...but I guess I'll figure it out haha. For f(u) I got the same as in the book, but with a 2u in the denominator as well.
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  10. #10
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    Quote Originally Posted by Celestiax View Post
    Ah, thank you very much! I worked it all out now and I believe I have it correct...I didn't know ln(2u) - ln(u) could just be ln(2) for f(u). Although I'm not sure where the 2u in the denominator goes...but I guess I'll figure it out haha. For f(u) I got the same as in the book, but with a 2u in the denominator as well.
    Isn't the book's answer [LaTeX ERROR: Convert failed] If so, then you were obviously meant to have the [LaTeX ERROR: Convert failed] in the denominator. If it's [LaTeX ERROR: Convert failed] , then it's because [LaTeX ERROR: Convert failed] . It isn't important to get the answer in the exact form as in the book anyway... it's alright as far as it's correct and sufficiently simplified.
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  11. #11
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    Thank you to everyone that helped! I think I have it all worked out now.
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