First let's apply u substitution making #u# =#sqrt(cos(x))# Making #d/dxarccos(u)#.
We must note #d/dx arccos(x) = (-1)/(1-x^2)#. Applying this to #arccos(sqrt(cos(x)))# we receive #d/dx arccos(x) = (-1)/(1-(u)^2)#. Applying the chain rule we then multiple the function by #u'# leaving #d/dx arccos(x) = (-u')/sqrt((1-(u)^2))#.
u' = #(-sin(x))/(2(sqrt(cos(x))))#
Now me must substitute #u# into the function #d/dx arccos(x) = (-(-sin(x))/(2(sqrt(cos(x)))))/sqrt((1-(sqrt(cos(x)))^2)#
We must further simplify this function #(1-(sqrt(cos(x)))^2)# = #1-cos(x)#
#(-sin(x))/(2(sqrt(cos(x))))# rationalizing the denominator we recieve
#(sqrt(cos(x)))(-sin(x))/(2(cos(x))# or since #sin(x)/(cos(x)) = tan(x)# then #(-tan(x))(sqrt(cos(x)))/2#
furthermore we remove the negative in front of tan(x) from the #-u# leaving #(tan(x)sqrt(cos(x)))/(2(sqrt(1-cos(x)))#
