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Math Help - Derivative of Natural Log

  1. #1
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    Derivative of Natural Log

    Hey, I was just wondering if anyone could walk me through how to find the derivative of y^2=2x  ln(3x)

    I think that the beginning of the solution should look something like this (2y)y'= 2x(\frac{1}{3x})+2 ln(3x) , but I am not sure.

    Thank you for your help.
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  2. #2
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    Quote Originally Posted by Chris22 View Post
    Hey, I was just wondering if anyone could walk me through how to find the derivative of y^2=2x  ln(3x)

    I think that the beginning of the solution should look something like this (2y)y'= 2x(\frac{1}{3x})+2 ln(3x) , but I am not sure.

    Thank you for your help.
    Recall that \frac{d}{dx}(ln[f(x)]) = \frac{f'(x)}{f(x)}
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  3. #3
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    Quote Originally Posted by Chris22 View Post
    Hey, I was just wondering if anyone could walk me through how to find the derivative of y^2=2x  ln(3x)

    I think that the beginning of the solution should look something like this (2y)y'= 2x(\frac{1}{3x})+2 ln(3x) , but I am not sure.

    Thank you for your help.
    All that's left now is to simplify (if you are required to do so).

    Just move 2y to the other side and substitute y with what it equals.

    In your case y=\sqrt{2xln(2x)}

    Edit: Yeah, above post is correct. You forgot to take the derivative of the inside of ln. Otherwise it's good.
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  4. #4
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    Thanks for the help, but I just looked at the answers my teacher gave me and the answer it says there is \frac{1+ln(3x)}{y}.
    I don't know how he got that answer. Do either of you know what I am doing wrong?

    Thanks again for your guys' help.
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