# How do you find the first and second derivative of  ln(ln x)?

Sep 11, 2016

$\frac{1}{x \ln x}$

#### Explanation:

First derviative: $\frac{d}{\mathrm{dx}} \ln \left(\ln \left(x\right)\right)$

Chain rule: $\frac{d}{\mathrm{dx}} f \left(g \left(x\right)\right) = f ' \left(g \left(x\right)\right) \cdot g ' \left(x\right)$

Let g=ln(x)

$f \left(g\right) = \ln \left(g\right) , g \left(x\right) = \ln \left(x\right)$

$f ' \left(g\right) = \frac{1}{g} , g ' \left(x\right) = \frac{1}{x}$

$f ' \left(x\right) = \frac{1}{\ln} \left(a\right)$

$f ' \left(g \left(x\right)\right) \cdot g ' \left(x\right) = \left(\frac{1}{\ln} \left(x\right) \cdot \frac{1}{x}\right)$

$\frac{1}{x \ln x}$

Second Derivative: $\frac{d}{\mathrm{dx}} \left(\frac{1}{\ln} \left(x\right) \cdot \frac{1}{x}\right)$

Product rule: $f \left(x\right) g ' \left(x\right) + f ' \left(x\right) g \left(x\right)$

$f \left(x\right) = \frac{1}{\ln} x , g \left(x\right) = \frac{1}{x}$

$\frac{1}{\ln} x \frac{d}{\mathrm{dx}} \left({x}^{-} 1\right) + \frac{d}{\mathrm{dx}} {\left(\ln x\right)}^{-} 1 \left(\frac{1}{x}\right)$

-1/(x^2lnx)-1/(x^2ln^2(x)