What is the derivative of # arctan sqrt(x^2 -1)#?

1 Answer
Feb 15, 2017

Explanation:

#y = tan^(-1)(sqrt(x^2-1))#
#dy/(dx) = 1/(1+(sqrt(x^2-1))^2) *1/(2sqrt(x^2-1)) . 2x#
#= 1/(x.sqrt(x^2-1))#

Another way to solve this is to use different variables and then reduce the equation.
Let #u = sqrt(x^2-1)#, #v=x^2#, then
#y = tan^(-1)u#
#u = sqrt(v-1)#
#dy/(dx) = dy/(du).(du)/(dv).(dv)/(dx)#
#= 1/(1+u^2).(1/(2sqrt(v-1))).2x#
substituting the relationships back we get,#dy/(dx) = 1/(1+(sqrt(x^2-1))^2) *1/(2sqrt(x^2-1)) . 2x#
#= 1/(x.sqrt(x^2-1))#