If `y=btan^(-1)(x/a+tan^(-1)y/x),fin d(dy)/(dx)dot`

Answer

`((1)/(a)-(y)/(x^(2)+y^(2)))/((1)/(b)sec^(2)((y)/(b))-(x)/(x^(2)+y^(2)))`

Explanation

We have
`y=b tan^(-1)((x)/(a)+tan^(-1)""(y)/(x))`
`"or "tan""(y)/(b)=(x)/(a)+tan^(-1)""(y)/(x)`
Differentiating both sides w.r.t. x, we get
`(1)/(b)sec^(2)((y)/(b))(dy)/(dx)=(1)/(a)+(1)/(1+((y)/(x))^(2))xx(x(dy)/(dx)-y)/(x^(2))`
`"or "(1)/(b)sec^(2)((y)/(b))(dy)/(dx)=(1)/(a)+(x(dy)/(dx)-y)/(x^(2)+y^(2))`
`"or "(dy)/(dx){(1)/(b)sec^(2)((y)/(b))-(x)/(x^(2)+y^(2))}=(1)/(a)-(y)/(x^(2)+y^(2))`
`"or "(dy)/(dx)=((1)/(a)-(y)/(x^(2)+y^(2)))/((1)/(b)sec^(2)((y)/(b))-(x)/(x^(2)+y^(2)))`