First, let's find the derivative of #arcsec(x)# by applying the Chain Rule to the equation #sec(arcsec(x))=x# (and using the fact that #d/dx(sec(x))=sec(x)tan(x)#)
#d/dx(sec(arcsec(x)))=d/dx(x)=1\Rightarrow sec(arcsec(x))tan(arcsec(x))d/dx(arcsec(x))=1#
#\Rightarrow d/dx(arcsec(x))=1/(x*sqrt{x^2-1})# (draw a right triangle with one non-right angle labeled #arcsec(x)# and use the Pythagorean Theorem to help you do this last simplification).
Now the Chain Rule implies:
#d/dx(arcsec(1/x))=d/dx(arcsec(x^{-1}))=(-1x^{-2})/(x^{-1}*sqrt{(x^{-1})^2-1})#
This simplifies, after multiplying the top and bottom of the last fraction by #x^2#, to:
#d/dx(arcsec(1/x))=-1/sqrt{1-x^2}#
Evidently this implies that #arcsec(1/x)=arccos(x)+C# for all allowed #x# and some #C#. In fact, it can be shown that #C=0# and #arcsec(1/x)=arccos(x)# for all #-1\leq x\leq 1# and #x!=0#.
Try graphing them both to check this!
