Derivação Implícita

[latex]\\\mathrm{\frac{x^2}{x+y}=y^2+1\ \therefore\ \frac{d}{dx}\left ( \frac{x^2}{x+y} \right )=\frac{d}{dx}(y^2+1)}\\\\ \mathrm{\frac{d}{dx}\left [ \frac{P(x)}{Q(x)} \right ]=\frac{Q(x)\frac{d}{dx}[P(x)]-P(x)\frac{d}{dx}[Q(x)]}{[Q(x)]^2}.\ Assim:}\\\\ \mathrm{\frac{d}{dx}\left ( \frac{x^2}{x+y} \right )=\frac{(x+y)(x^2)'-x^2(x+y)'}{(x+y)^2}=\frac{2x(x+y)-x^2(y'+1)}{(x+y)^2}}\\\\ \mathrm{\frac{d}{dx}\left ( \frac{x^2}{x+y} \right )=\frac{2x}{x+y}-\frac{x^2(y'+1)}{(x+y)^2}.\ Ademais,\frac{d}{dx}(y^2+1)=2y'y}\\\\ \mathrm{Deste\ modo:\frac{2x}{x+y}-\frac{x^2(y'+1)}{(x+y)^2}=2y'y\ \therefore\ y'=\frac{x(x+2y)}{2x^2y+4xy^2+2y^3+x^2}}[/latex]