# Derivative Of A Natural Logarithm

• April 24th 2012, 12:37 PM
Bashyboy
Derivative Of A Natural Logarithm
I am given the function f(x) = ln(lnx^2) to differentiate; but I am not entirely certain how to do this. Could someone possibly help me?

Thank you
• April 24th 2012, 01:24 PM
Plato
Re: Derivative Of A Natural Logarithm
Quote:

Originally Posted by Bashyboy
I am given the function f(x) = ln(lnx^2) to differentiate; but I am not entirely certain how to do this. Could someone possibly help me?

Note that $f(x)=\ln(2)+\ln(\ln(x))$ so $f'(x)=\frac{1}{x\ln(x)}~.$
• April 24th 2012, 02:27 PM
HallsofIvy
Re: Derivative Of A Natural Logarithm
You need to know this: the derivative of f(x)= ln(u(x)) is [tex]\frac{1}{u}\frac{du}{dx}[/itex].

As Plato said, $ln(ln(x^2)= ln(2ln(x))= ln(ln(x))+ ln(2)$. "ln(2)" is a constant and its derivative is 0.

so you have ln(u(x)) with u(x)= ln(x).