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Math Help - quotent rule for natural logs

  1. #1
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    Post quotent rule for natural logs

    f(x) = (lnx)x
    f'(x) = x * (lnx)' - (lnx) * x'/(x)^2
    f'(x) = x*(1/x) - (lnx)*1/(x)^2
    f'(x) = 1 - lnx/(x)^2

    I hope this is correct
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by becky View Post
    f(x) = (lnx)x
    f'(x) = x * (lnx)' - (lnx) * x'/(x)^2
    f'(x) = x*(1/x) - (lnx)*1/(x)^2
    f'(x) = 1 - lnx/(x)^2

    I hope this is correct
    it's not correct i'm afraid. you would use the product rule here, since you don't have a quotient (or did you mean to type a slash after the lnx but forgot?)

    y = (ln)x
    => y' = lnx + x(1/x) = lnx + 1
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  3. #3
    is up to his old tricks again! Jhevon's Avatar
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    Was this what you meant? If so, you were correct (though you need to have parenthesis in your answer to show that this is the same thing)
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  4. #4
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    Inx/x

    Yes, that's what the problem...
    Glad you caught my mistake in writing the problem, I probably wouldn't have and would really be confused right now...
    Yeah, I got it right!
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  5. #5
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by becky View Post
    Yes, that's what the problem...
    Glad you caught my mistake in writing the problem, I probably wouldn't have and would really be confused right now...
    Yeah, I got it right!
    Good Job

    so you understand it well now?

    why not try a few more problems and post your solutions
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  6. #6
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    Re:

    Jhevon's is correct, but I think he will agree that unless you are told to do so using the Quotient rule isn't the easiest way to solve this problem...
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  7. #7
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    Re:

    Re:
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