Derivação Implícita

[latex]\\\mathrm{arctan(x^2y)=x+xy^2\ \therefore\ \frac{d}{dx}\left [ arctan(x^2y) \right ]=\frac{d}{dx}(x+xy^2)}\\\\ \mathrm{\frac{d}{dx}[arctan(f(x))]=\frac{\frac{d}{dx}(f(x))}{[f(x)]^2+1}\ \therefore\ \frac{d}{dx}\left [ arctan(x^2y) \right ]=\frac{\frac{d}{dx}(x^2y)}{x^4y^2+1}}\\\\ \mathrm{\frac{d}{dx}\left [ arctan(x^2y) \right ]=\frac{y\frac{d}{dx}(x^2)+x^2\frac{d}{dx}(y)}{x^4y^2+1}=\frac{2xy+x^2y'}{x^4y^2+1}}\\\\ \mathrm{Por\ sua\ vez:\frac{d}{dx}(x+xy^2)=2xyy'+y^2+1.\ Assim:}\\\\ \mathrm{\frac{2xy+x^2y'}{x^4y^2+1}=2xyy'+y^2+1\ \therefore\ y'=-\frac{x^4y^4+(x^4+1)y^2-2xy+1}{2x^5y^3+2xy-x^2}}\\\\ \mathrm{A\ igualdade\ anterior\ \acute{e}\ o\ mesmo\ que:y'=-\frac{x^4y^4+(x^4+1)y^2-2xy+1}{x^2-2xy-2x^5y^3}}[/latex]