SOLVED Difference between In and Log in differentiating

Jul 2010
16
1
Hastings
\(\displaystyle y=x^\frac{1}{x}\)
\(\displaystyle \frac{dy}{dx} = x^ \frac{1}{x}(\frac{1}{x^2} - \frac{log(x)}{x^2})\)

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



\(\displaystyle \frac{dy}{dx} = x^\frac{1}{x} (\frac{1}{x^2} - \frac{In(x)}{x^2})\)
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:
Sep 2008
1,261
539
West Malaysia
\(\displaystyle y=x^\frac{1}{x}\)
\(\displaystyle \frac{dy}{dx} = x^ \frac{1}{x}(\frac{1}{x^2} - \frac{log(x)}{x^2})\)

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



\(\displaystyle \frac{dy}{dx} = x^\frac{1}{x} (\frac{1}{x^2} - \frac{In(x)}{x^2})\)
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.
 
Jan 2010
278
138
Minor nitpick: "ln" starts with a lower case "L", not a capital "I".
 
Jul 2010
16
1
Hastings
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...
 

HallsofIvy

MHF Helper
Apr 2005
20,249
7,909
You can differentiate logs to other bases. It's just that the formula is a little different.

If \(\displaystyle y= log_a(x)\), then \(\displaystyle x= a^y\) so that \(\displaystyle ln(x)= ln(a^y)= yln(a)\) and so \(\displaystyle y= log_a(x)= \frac{ln(x)}{ln(a)}\). The derivative of that is \(\displaystyle \frac{dy}{dx}= \frac{1}{ln(a)}\frac{1}{x}= \frac{1}{x ln(a)}\).
 
Jan 2010
278
138
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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HallsofIvy

MHF Helper
Apr 2005
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My first calculus teacher kept calling sinh, "sinus hyperbolicus"
 
Aug 2010
20
1
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