text: 1. ## Derivative Calculation
  
  Given:
  
  $\displaystyle y= \frac{sin^2 x}{1+ ctgx} +$$\displaystyle \frac{cos^2 x}{1+ ctgx}$
  
  Find $\displaystyle y'$
  
  I know that answer is $\displaystyle -cos2x$, but can't calculate it.
  
  I got so far:
  
  $\displaystyle y'=(\frac{sin^2 x}{1+ ctgx} +\frac{cos^2 x}{1+ ctgx} )'= \frac{(sin^2 x)'(1+ ctgx) - sin^2 x (1+ ctgx)'}{(1+ ctgx)^2}$ + $\displaystyle \frac{(cos^2 x)'(1+ tgx) - cos^2 x (1+ tgx)'}{(1+ tgx)^2} =$
  
  $\displaystyle =\frac{2sinx(1+ctgx)+1}{(1+ ctgx)^2} +\frac{2cos x(1+tgx)-1}{(1+ tgx)^2}$
  
  2. I spot one crucial error. There might be more but you need to correct this one first.
  
  $\displaystyle (\cos^2(x))' = -2\cos(x)\sin(x)$ and $\displaystyle (\sin^2(x))' = 2\sin(x)\cos(x)$