## Calculus: Early Transcendentals 8th Edition

$y'=\frac{2x}{(x^{2}+y^{2}-2y)}$
First simplify the given function using the properties of logarithms and then apply the chain rule of differentiation. $y'=\frac{d}{dx}[ln(x^{2}+y^{2})]$ $=\frac{1}{(x^{2}+y^{2})}(2x+2y\frac{dy}{dx})$ $=\frac{2x}{(x^{2}+y^{2})}+\frac{2yy'}{(x^{2}+y^{2})}$ $y'-\frac{2yy'}{(x^{2}+y^{2})}=\frac{2x}{(x^{2}+y^{2})}$ $y'(x^{2}+y^{2}-2y)=2x$ Hence, $y'=\frac{2x}{(x^{2}+y^{2}-2y)}$