Find the derivative of the function: $\frac{x}{1+\tan x}$

Let $f(x) = \frac{x}{1+\tan x}$.
Using the quotient rule,$\frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}$.
Here,$u = x$ and $v = 1 + \tan x$.
$\frac{du}{dx} = 1$ and $\frac{dv}{dx} = \sec^2 x$.
Substituting these into the quotient rule formula:
$f'(x) = \frac{(1 + \tan x)(1) - (x)(\sec^2 x)}{(1 + \tan x)^2}$.
Thus,$f'(x) = \frac{1 + \tan x - x \sec^2 x}{(1 + \tan x)^2}$.