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

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Answer 1 · 2016-09-11T16:02:12

$\frac{1}{x ln x}$ $1/(xlnx)$

First derviative: $\frac{d}{d x} ln (ln (x))$ $d/dxln(ln(x))$

Chain rule: $d/dxf(g(x))=f'(g(x))*g'(x)$

Let g=ln(x)

$f(g)=ln(g), g(x)=ln(x)$

$f'(g)=1/g, g'(x)=1/x$

$f'(x)=1/ln(a)$

$f'(g(x))*g'(x)=(1/ln(x)*1/x)$

$1/(xlnx)$

Second Derivative: $d/dx(1/ln(x)*1/x)$

Product rule: $f(x)g'(x)+f'(x)g(x)$

$f(x)=1/lnx, g(x)=1/x$

$1/lnxd/dx(x^-1)+d/dx(lnx)^-1(1/x)$

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

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