Find the derivative of f(x) = 1+tan(x) / 1-tan(x)

We need to know
the following trigonometric function here.

`cos2x = cos^x-sin^2x`

`sin2x = 2sinxcosx`

`sin^2x+cos^2x = 1`

 

`(1+tanx)/(1-tanx)`

`=
(1+tanx)/(1-tanx)xx(1+tanx)/(1+tanx)`

`= (1+tanx)^2/(1-tan^2x)`

`= (1+sinx/cosx)^2/(1-(sin^2(x))/(cos^2(x)))`

`=
((cosx+sinx)^2/(cos^2x))/((cos^2x-sin^2x)/(cos^2x))`

`=
(cosx+sinx)^2/(cos^2x-sin^2x)`

`=
(cos^2x+sin^2x+2sinxcosx)/(cos2x)`

`= (1+sin2x)/(cos2x)`

 

`f(x) = (1+tanx)/(1-tanx) = (1+sin2x)/(cos2x)`

 

`f'(x)`

`=
(cos2x*2cos2x-(1+sin2x)(-2sin2x))/(cos^2(2x))`

`=
(2(cos^2(2x)+sin^2(2x)+sin2x))/(cos^2(2x))`

`= (2(1+sin2x))/(cos^2(2x))`

`= (2(1+sin2x))/(1-sin^2(2x))`

`=
(2(1+sin2x))/((1+sin2x)(1-sin2x))`

`= 2/(1-sin2x)`

 

So the derivative of f(x)
is `f'(x)= 2/(1-sin2x)`