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Answer 1 · 2016-10-16T15:45:40

$1/(sin(x)cos(x)ln(tan(x))$

$y=ln(ln(tan(x)))$

We will use the chain rule as applied to the natural logarithm function:

$d/dxln(x)=1/x" "=>" "d/dxln(f(x))=1/f(x)*f'(x)$

In the given function we will need to use the chain rule multiple times. The first time will peel away the first $ln$ function:

$dy/dx=1/ln(tan(x))*[d/dxln(tan(x))]$

Reapplying the chain rule:

$dy/dx=1/ln(tan(x)) * [1/tan(x) * d/dxtan(x)]$

Since $d/dxtan(x)=sec^2(x)$ :

$dy/dx=1/ln(tan(x)) * 1/tan(x) * sec^2(x)$

Simplifying:

$dy/dx=1/ln(tan(x)) * 1/(sin(x)/cos(x)) * 1/cos^2(x)$

$dy/dx=1/ln(tan(x)) * cos(x)/sin(x) * 1/cos^2(x)$

$dy/dx=1/(sin(x)cos(x)ln(tan(x))$

Answer 2 · 2016-10-17T18:34:45

$1/ ln(tan x) 1/(sin x cos x)$

Simpler way for arriving at the result is to let y=ln (ln(tan x), so that it is $e^y = ln(tan x)$

now differentiate w.r.t x, $e^y dy/dx= 1/(tan x) sec^2 x = 1/(sinx cos x)$

This gives $dy/dx = 1/e^y 1/( sin x cos x)$

= $1/ ln(tan x) 1/(sin x cos x)$