## Calculus: Early Transcendentals 8th Edition

$y'=\dfrac{x^{2}+2xy}{x^{2}+2y(x+y)^{2}}$
$\dfrac{x^{2}}{x+y}=y^{2}+1$ Differentiate the whole equation: $\Big(\dfrac{x^{2}}{x+y}\Big)'=(y^{2}+1)'$ Apply the quotient rule to differentiate the left side: $\dfrac{(x+y)(x^{2})'-(x^{2})(x+y)'}{(x+y)^{2}}=(2y)(y')$ $\dfrac{(x+y)(2x)-(x^{2})(1+y')}{(x+y)^{2}}=2yy'$ Solve for $y'$: $2x^{2}+2xy-x^{2}-x^{2}y'=2yy'(x+y)^{2}$ $-x^{2}y'-2yy'(x+y)^{2}=-x^{2}-2xy$ $y'[-x^{2}-2y(x+y)^{2}]=-x^{2}-2xy$ $y'=\dfrac{-x^{2}-2xy}{-x^{2}-2y(x+y)^{2}}$ $y'=\dfrac{x^{2}+2xy}{x^{2}+2y(x+y)^{2}}$