Answer
$y'=\frac{2x}{(x^{2}+y^{2}-2y)}$
Work Step by Step
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, $y'=\frac{2x}{(x^{2}+y^{2}-2y)}$
