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.
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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. And you can see how this simplifies.
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