Find `(dy)/(dx)`, when:
`logsqrt(x^(2)+y^(2))=tan^(-1).(y)/(x)`

Answer

`(x+y)/(x-y)`

Explanation

`(1)/(2)log(x^(2)+y^(2))=tan^(-1)((y)/(x))`
`rArrlog(x^(2)+y^(2))=2tan^(-1).(y)/(x)rArr(x^(2)+y^(2))=e^(2tan^(-1)y//x)." ...(i)"`
On differentiating (i) w.r.t. x, we get
`2x+2y.(dy)/(dx)=e^(2tan^(-1)y//x).(2)/((1+(y^(2))/(x^(2)))).((x(dy)/(dx)-y))/(x^(2))`
`rArrx+y(dy)/(dx)=(x^(2)+y^(2)).(x^(2))/((x^(2)+y^(2))).((x(dy)/(dx)-y))/((x^(2)))." [using (i)]"`
`rArr x+y(dy)/(dx)=x(dy)/(dx)-y`
`rArr(x-y)(dy)/(dx)=(x+y)rArr(dy)/(dx)=((x+y))/((x-y)).`