#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}}#
