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Math Help - Difference between In and Log in differentiating

  1. #1
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    Question Difference between In and Log in differentiating

    y=x^\frac{1}{x}
    \frac{dy}{dx} = x^ \frac{1}{x}(\frac{1}{x^2} - \frac{log(x)}{x^2})

    When you graph x^(1/x), the maximum point (gradient zero) should be e (2.71828...) for x. But it doesn't.



    \frac{dy}{dx} = x^\frac{1}{x} (\frac{1}{x^2} - \frac{In(x)}{x^2})
    By replacing the log with in, I get the correct answer. When dy/dx = 0, x = e.

    Can anyone explain the difference? And which one should I use the next time I use the chain rule to differentiate powers, In or Log?
    Last edited by DavidM; August 7th 2010 at 12:54 AM.
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  2. #2
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    Quote Originally Posted by DavidM View Post
    y=x^\frac{1}{x}
    \frac{dy}{dx} = x^ \frac{1}{x}(\frac{1}{x^2} - \frac{log(x)}{x^2})

    When you graph x^(1/x), the maximum point (gradient zero) should be e (2.71828...) for x. But it doesn't.



    \frac{dy}{dx} = x^\frac{1}{x} (\frac{1}{x^2} - \frac{In(x)}{x^2})
    By replacing the log with in, I get the correct answer. When dy/dx = 0, x = e.

    Can anyone explain the difference? And which one should I use the next time I use the chain rule to differentiate powers, In or Log?
    Another way to differentiate this is by differentiating implicitly which is easier compared to the chain rule.

    You can only differentiate a log to base e (ln). You will need to do a logarithmic conversion to base e if you are asked to differentiate log to other bases.
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  3. #3
    Senior Member eumyang's Avatar
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    Minor nitpick: "ln" starts with a lower case "L", not a capital "I".
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  4. #4
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    Oh thank-you I didn't realise you can only differentiate with the log to base e.

    I never knew "ln" was used with an "L". Always thought it was "In" for some reason...
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  5. #5
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    You can differentiate logs to other bases. It's just that the formula is a little different.

    If y= log_a(x), then x= a^y so that ln(x)= ln(a^y)= yln(a) and so y= log_a(x)= \frac{ln(x)}{ln(a)}. The derivative of that is \frac{dy}{dx}= \frac{1}{ln(a)}\frac{1}{x}= \frac{1}{x ln(a)}.
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  6. #6
    Senior Member eumyang's Avatar
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    Quote Originally Posted by DavidM View Post
    I never knew "ln" was used with an "L". Always thought it was "In" for some reason...
    According to the Talk page for the Natural Logarithm article in Wikipedia, someone learned that it was from the Latin "logarithmus naturalis." I never knew that so that was an interesting tidbit.
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  7. #7
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    My first calculus teacher kept calling sinh, "sinus hyperbolicus"
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  8. #8
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    Quote Originally Posted by DavidM View Post
    I never knew "ln" was used with an "L". Always thought it was "In" for some reason...
    Probably because on the calculator, you can't see much difference between I in caps and L in lower case
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