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## Homework Statement

[tex]y= \frac{sin^2 x}{1+ ctgx} +\frac{cos^2 x}{1+ ctgx}[/tex]

Find: [tex]y'[/tex]

## The Attempt at a Solution

Answer is[tex]-cos2x[/tex] but again I can't find it.

Where am I making a mistake?

[tex] 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}\ + \ \frac{(cos^2 x)'(1+ tgx) - cos^2 x (1+ tgx)'}{(1+ tgx)^2} =[/tex]

[tex]

= \frac{sin2x( 1+ctgx) +1 }{(1+ ctgx)^2} +\frac{-sin2x(1+tgx)-1}{(1+ tgx)^2}

[/tex]