This was part of an implicit differentiation question. However, the book showed the step -sin²(x+y) and then the next step -x²/x²+1 with both equal to y' just one being a simplification of the other, which is why I posted it on this Trigonometry thread.

This was part of an implicit differentiation question. However, the book showed the step -sin²(x+y) and then the next step -x²/x²+1 with both equal to y' just one being a simplification of the other, which is why I posted it on this Trigonometry thread.

To answer your original question, \(\displaystyle -sin^2(x+ y)\) does not simplify to \(\displaystyle -\frac{x^2}{x^2+ 1}\). For one thing, the first is a function of x and y, the second a function only of x.

The original equation being implicitly differentiated is tan(x+y)=x. The steps are as follows:
(1+y')sec²(x+y)=1
y'=1-sec²(x+y)/sec²(x+y)
=-tan²(x+y)/tan²(x+y)
=sin²(x+y)
=-x²/x²+1