# Logarithmic Differentiation

• Oct 13th 2008, 12:27 AM
john doe
Logarithmic Differentiation
How would i differentiate this

$\displaystyle ln(lnx^2)$

im trying to do it right now but im getting alot
of fractions

i think i got it does it look something like this

(1/lnx^2) (2/x)
• Oct 13th 2008, 12:54 AM
Opalg

You could have made life a bit easier for yourself by using properties of logarithms. In fact, $\displaystyle \ln(x^2)=2\ln x$, so $\displaystyle \ln(\ln(x^2)) = \ln(2\ln x) = \ln2 + \ln(\ln x)$. Then when you differentiate, the first term is constant, so it disappears and you only have to deal with the second term.
• Oct 13th 2008, 02:18 AM
Jasonium
Isn't it 1/(xlnx) ?
• Oct 13th 2008, 02:53 AM
mr fantastic
Quote:

Originally Posted by Jasonium
Isn't it 1/(xlnx) ?

No, it's $\displaystyle \frac{1}{x \ln|x|}$ which is a simplified form of the answer found by the OP.
• Oct 13th 2008, 02:56 AM
mr fantastic
Quote:

Originally Posted by Opalg
You could have made life a bit easier for yourself by using properties of logarithms. In fact, $\displaystyle \ln(x^2)=2\ln x$, so $\displaystyle \ln(\ln(x^2)) = \ln(2\ln x) = \ln2 + \ln(\ln x)$. Then when you differentiate, the first term is constant, so it disappears and you only have to deal with the second term.
I think it might be $\displaystyle \ln(\ln(x^2)) = \ln(2\ln x) = \ln2 + \ln(\ln {\color{red}|}x{\color{red}|})$ .....