Question

Differential the following functions w.r.t. $x$:
$\frac{\sec x+\tan x}{\sec x-\tan x}$

Answer 1

$y=\frac{secx+tanx}{secx-tanx}=\frac{1+sinx}{1-sinx}$

$\frac{dy}{dx}=\frac{(1-sinx)\left(cotx\right)-(1+sinx)(-cosx)}{(1-sinx{)}^2}$

$\frac{dy}{dx}=\frac{cosx-cosxsinx+cosx+cosxsinx}{(1-sinx{)}^2}$

$\frac{dy}{dx}=\frac{2cosx}{(1-sinx{)}^2}$