Actually it isn't too bad!
Step I
Let #theta = csc^-1x#
#:. x = csc theta#
I shall assume that there is no need to show the step by step differentiation of #csc#, if I may (it just requires the chain rule, with #x= (sin theta)^-1#, and skip straight to:
#dx/(d theta) = -csc theta cot theta#
We see that #csc theta# is of course just #-x#. Then use the Pythagorean Identity to get #cot theta# in terms of #x#.
#1 + cot^2x = csc^2x#
#:.cot theta = sqrt (x^2-1)# (assume the positive square root here).
#:. dx/(d theta) = -xsqrt(x^2-1)#
#:. (d theta)/dx = -1/(xsqrt(x^2-1)#
Step II
Let #alpha = cot^-1x#
#:. x = cot alpha#
Again, if I may, I shall skip right to:
#dx / (d alpha) = -csc^2alpha#
Again, using the same Pythagorean Identity, we have:
#1 + x^2 = csc^2 alpha#
#:. -(1 + x^2) = dx/(d alpha)#
#:. (d alpha)/dx = -1/(1 +x^2)#
Step III
Now combine the lot:
#dy/dx = -1/(|x|sqrt(x^2-1)) + 4/(1+x^2)#, #x in (-oo,-1)uu(1, oo)#
Note: remember to re-introduce the constant #4# which I left out whilst differentiating #4cot^-1x#
Second note: the absolute value bars rather mysteriously appeared on the denominator of that derivative. I believe this is because actually sketching the graph of #y=csc^-1x#, the function is strictly decreasing, so by placing absolute value bars around the stray #x#, ensures that the derivative function stays negative for all #x#.
Third note: the domain comes in because the #csc# function is defined across this domain. The #cot# function is defined for all #x#, so the combined function takes the domain of the former.
