Logarithmic Differentiation

**john doe** · October 13th 2008, 12:27 AM

How would i differentiate this

$<br /> ln(lnx^2)<br />$

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)

**Opalg** · October 13th 2008, 12:54 AM

Your answer is correct.

You could have made life a bit easier for yourself by using properties of logarithms. In fact, $\ln(x^2)=2\ln x$ , so $\ln(\ln(x^2)) = \ln(2\ln x) = \ln2 + \ln(\ln x)$ . Then when you differentiate, the first term is constant, so it disappears and you only have to deal with the second term.

**Jasonium** · October 13th 2008, 02:18 AM

Isn't it 1/(xlnx) ?

**mr fantastic** · October 13th 2008, 02:53 AM

Originally Posted by Jasonium

Isn't it 1/(xlnx) ?

No, it's $\frac{1}{x \ln|x|}$ which is a simplified form of the answer found by the OP.

**mr fantastic** · October 13th 2008, 02:56 AM

Originally Posted by Opalg

Your answer is correct.

You could have made life a bit easier for yourself by using properties of logarithms. In fact, $\ln(x^2)=2\ln x$ , so $\ln(\ln(x^2)) = \ln(2\ln x) = \ln2 + \ln(\ln x)$ . 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 $\ln(\ln(x^2)) = \ln(2\ln x) = \ln2 + \ln(\ln {\color{red}|}x{\color{red}|})$ .....

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