If `log(x^2+y^2)=tan^(-1)(y/x),` show that `(dy)/(dx)=(x+y)/(x-y)`

Explanation

Differentiating both sides w.r.t. x, we get
`(d)/(dx){log(x^(2)+y^(2))}=2(d)/(dx){tan^(-1)((y)/(x))}`
`"or "(1)/(x^(2)+y^(2))xx(d)/(dx)(x^(2)+y^(2))=2(1)/(1+(y//x)^(2))xx(d)/(dx)((y)/(x))`
`"or "(1)/(x^(2)+y^(2)){(d)/(dx)(x^(2))+(d)/(dx)(y^(2))}=2xx(x^(2))/(x^(2)+y^(2)){(x(dy)/(dx)-yxx1)/(x^(2))}`
`"or "(1)/(x^(2)+y^(2)){2x+22y(dy)/(dx)}=(2)/(x^(2)+y^(2)){x(dy)/(dx)-y}`
`"or "{x+y(dy)/(dx)}=2{x(dy)/(dx)-y}`
`"or "x+y(dy)/(dx)=x(dy)/(dx)-y`
`"or "(dy)/(dx)(y-x)=-(x+y)`
`"or "(dy)/(dx)=(x+y)/(x-y)`