How to take the derivative of ln(lnx)?

**Barthayn** · Nov 10th 2011, 02:28 PM

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.

**skeeter** · Nov 10th 2011, 02:32 PM

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

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

**Barthayn** · Nov 10th 2011, 02:37 PM

thanks I got it.

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

**Quacky** · Nov 10th 2011, 02:37 PM

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.

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

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

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

And you can see how this simplifies.

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