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

1 Answer
Apr 17, 2015

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)#