implicit differentiation solving for $y\prime$ I'm supposed to implicitly differentiate the following and give the answer in terms of $y\prime$. $$\tan(x-y)={y \over 1+x^2}$$
$${ (1+x^2)y\prime - 2xy \over (1+x^2)^2 }$$
How do I solve for $y \prime$?
Edit:
After a ludicrous amount of algebra, i finally ended up at
$$ { sec^2(x-y)(1+x^2)^2+2xy \over (1+x^2)(1+sec^2(x-y)(1+x^2)) } $$
which apparently is correct. 

Differentiating both sides from
$$\tan (x - y(x)) = \frac{{y(x)}}{{{x^2} + 1}}$$
gives:
$$\frac{1}{{{{\cos }^2}(x - y(x))}} - \frac{{y'(x)}}{{{{\cos }^2}(x - y(x))}} = \frac{{y'(x)}}{{{x^2} + 1}} - \frac{{2xy(x)}}{{{{\left( {{x^2} + 1} \right)}^2}}}$$
which can be solved and simplified for ${y'(x)}$ like so:
$$y'(x) = \frac{{2xy(x)}}{{{x^2} + 1}} + \frac{{2\left( {{x^2} - 2xy(x) + 1} \right)}}{{2{x^2} + \cos (2(x - y(x))) + 3}}$$
(I don't like sec-terms)
