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