How do you find the derivative of #arccos ([2x + 1]/2)#?

#d/{dx}arccos({2x+1}/2)={-1}/{\sqrt{3/4-x-x^2}}#

First I will prove a result needed to do the differentiation:

#d/{du}arccos(u)={-1}/{\sqrt{1-u^2}}#

Proof:
Let #y=arccos(u)#, take the cosine of each side.

#\impliescos(y)=u#, now differentiate each side with respect to #u#

#\implies -sin(y){dy}/{du}=1#

#\implies {dy}/{du}=d/{du}[arccosu]={-1}/{sin(y)}#

Draw a triangle that agrees with the earlier expression, #cos(y)=u# to find out what #sin(y)# is equal to. From the diagram below, #sin(y)=\sqrt{1-u^2}#

Substitute #sin(y)# to get the desired result:#d/{du}arccos(u)={-1}/{\sqrt{1-u^2}}#

Now use the chain rule to take the derivative of #y=arccos({2x+1}/2)#

Let #u={2x+1}/2# so that #y=arccos(u)#

Chain Rule
#d/{dx}y={dy}/{du}*{du}/{dx}=( {-1}/{ \sqrt{1-u^2} })(1)#

Substitute #u# in the above result to get

#d/{dx}arccos({2x+1}/2)={-1}/{\sqrt{3/4-x-x^2}}#