Question

What is the derivative of `tan^(-1)(x^2 y^5)`?

Answer 1

let `u = tan^-1 (x^2y^5)`

`=> tanu = x^2y^5`

By differentiating implicitly with respect to `x` we have,

`=> (du)/(dx)sec^2u = (2x)y^2 + x^2(5y^4)(dy)/(dx)`

`=> (du)/(dx) = (2xy^2 + 5y^4x^2(dy)/(dx))/(sec^2u) = (2xy^2 + 5y^4x^2(dy)/(dx))/(tan^2u + 1)`

but `u = tan^-1 (x^2y^5)`

`(du)/(dx) = (2xy^2 + 5y^4x^2(dy)/(dx))/([tan(tan^-1(x^2y^5))]^2 + 1) = (2xy^2 + 5y^4x^2(dy)/(dx))/((x^2y^5)^2 + 1) = (2xy^2 + 5y^4x^2(dy)/(dx))/(x^4y^10 + 1)`