Thread: How to take the derivative of ln(lnx)?

How do you take the derivative of ln(lnx)?

This is where I go with it then I get confuse:

y=ln(lnx)
e^y=lnx.

From here I get lost. Everything I try does not work.

$\displaystyle \frac{d}{dx} \ln(u) = \frac{1}{u} \cdot \frac{du}{dx}$

now, let $\displaystyle u = \ln{x}$

thanks I got it. It is (x ln(x))^-1. However, I do not understand why 1 over u?

Originally Posted by Barthayn

How do you take the derivative of ln(lnx)?

This is where I go with it then I get confuse:

y=ln(lnx)
e^y=lnx.

From here I get lost. Everything I try does not work.

Alternatively, you can just do implicit differentiation from there.

$\displaystyle e^y\frac{dy}{dx}=\frac{1}{x}$

$\displaystyle \frac{dy}{dx}=\frac{1}{e^yx}$

$\displaystyle \frac{dy}{dx}=\frac{1}{e^{ln(ln(x))}x}$

And you can see how this simplifies.