Question

Find `dy/dx if x^2y^2 - tan^-1(sqrt(x^2 + y^2)) = cot^-1(sqrt(x^2 + y^2))`

Answer 1

`x^2y^2 - tan^-1(sqrt(x^2 + y^2)) = cot^-1(sqrt(x^2 + y^2))`

∴ `x^2y^2 = tan^-1(sqrt(x^2 + y^2)) + cot^-1(sqrt(x^2 + y^2))`

∴ `x^2y^2 = π/2         ...[∵ tan^-1 x + cot^-1 x = π/2]`

Differentiating both sides w.r.t. x, we get,

`x^2.d/dx(y^2) + y^2.d/dx(x^2) = 0`

∴ `x^2 × 2y dy/dx + y^2 × 2x = 0`

∴ `2x^2y dy/dx = – 2xy^2`

∴ `x dy/dx = – y`

∴ `dy/dx = -y/x`.