There is a pattern amongst the inverse trig derivatives:

#d/(dx)[arcsinu] = 1/sqrt(1-u^2)*((du)/(dx))#

#d/(dx)[arccosu] = -1/sqrt(1-u^2)*((du)/(dx))#

#d/(dx)[arc secu] = 1/(|u|sqrt(u^2-1))*((du)/(dx))#

#d/(dx)[arc cscu] = -1/(|u|sqrt(u^2-1))*((du)/(dx))#

#d/(dx)[arctanu] = 1/(1+u^2)*((du)/(dx))#

#d/(dx)[arc cotu] = -1/(1+u^2)*((du)/(dx))#

Notice though that Wolfram Alpha would say that the derivative of #arc cscu# is #-1/(sqrt(1-1/(u^2))*u^2#, but if you use its answer and manipulate it by calling the outer #u^2# equal to #|u||u| = |u|*sqrt(u^2)# and distributing it into the square root, it will be equivalent. We can use both, though.

#d/(dx)[arc csc(x^2+1)] = -1/(|x^2+1|sqrt((x^2+1)^2-1))*(2x)#

#= -(2x)/(|x^2+1|sqrt((x^2+1)^2-1))#

or...

#d/(dx)[arc csc(x^2+1)] = -1/((x^2+1)^2sqrt(1-1/(x^2+1)^2))*(2x)#

#= -(2x)/((x^2+1)^2sqrt(1-1/(x^2+1)^2))#