#### Answer

The Derivative is:
$y'=\frac{-\csc^2 x}{(1+\cot x)^2}$
OR
$y'=-\frac{1}{1+\sin(2x)}$

#### Work Step by Step

$y=\frac{\cot x}{1+\cot x}$
$y'=\frac{(-\csc^2 x)(1+\cot x)-(\cot x)(-\csc^2 x)}{(1+\cot x)^2}$
$y'=\frac{-\csc^2 x-\frac{\cos x}{\sin^3 x}+\frac{\cos x}{\sin^3 x}}{(1+\cot x)^2}$
$y'=\frac{-\csc^2 x}{(1+\cot x)^2}$
$y'=\frac{-\frac{1}{\sin^2 x}}{1+2\frac{\cos x}{\sin x}+\frac{\cos^2 x}{\sin^2 x}}$
$y'=\frac{-\frac{1}{\sin^2 x}}{\frac{\sin^2 x+2\sin x\cos x+\cos^2 x}{\sin^2 x}}$
$y'=-\frac{1}{\sin^2 x}\frac{\sin^2 x}{\sin^2 x+\cos^2 x+2\sin x\cos x}$
$y'=-\frac{1}{\sin^2 x+\cos^2 x+2\sin x\cos x}$
Applying trigonometric Identities
$2\sin x\cos x =\sin(2x)$ and $\sin^2 x+\cos^2 x=1$
$y'=-\frac{1}{1+\sin(2x)}$