Question

Question #f3dc0

Answer 1

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

Explanation:

`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

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

Explanation:

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)`