Question

What is the derivative of the inverse tan(y/x)?

Answer 1

The derivative would be `1/sqrt(x^2+y^2) (dy/dx -y/x)`

If u is `tan^-1(y/x)` then tan u =`y/x`. Differentiating w.r.t. x,

`sec^2u (du)/dx= 1/x^2 (xdy/dx -y)`

`(du)/dx= cos^2 u [1/x^2(x dy/dx -y)]`

= `x/sqrt(x^2+y^2)` `1/x^2 (xdy/dx -y)`

=`1/sqrt(x^2+y^2) (dy/dx -y/x)`