Answer
$f'(x) = \dfrac{-\csc^2 x - \csc^3 x + \csc x.\cot^2 x}{(1+\csc x)^2}$
Work Step by Step
In order to derivate this function, you have to apply the quotient rule
$\dfrac{d}{dx}(\dfrac{a}{b})= \dfrac{a'b-ab'}{b^2}$
So, let's identify a and b and derivate them
$a=\cot x$
$a'=-\csc^2 x$
$b= 1+\csc x$
$b'=-\csc x.\cot x$
Then, substitute in the formula:
$f'(x)= \dfrac{ (-\csc^2 x) (1+\csc x) - (\cot x) (-\csc x.\cot x) }{(1+\csc x)^2}$
Simplify and get the answer:
$f'(x) = \dfrac{-\csc^2 x - \csc^3 x + \csc x.\cot ^2 x}{(1+\csc x)^2}$