How do you find the derivative of the function: #arccos ([2x + 1]/2)#?
1 Answer
Apr 17, 2017
# d/dx arccos ([2x + 1]/2) = -1/sqrt(1-(x+1/2)^2)#
Explanation:
We want the derivative of:
# arccos ([2x + 1]/2) = arccos (x+1/2)#
We can use the known result:
#d/dx arccos x = -1/sqrt(1-x^2)#
Along with the chain rule to get:
# d/dx arccos ([2x + 1]/2) = -1/sqrt(1-(x+1/2)^2) * d/dx (x+1/2)#
# " " = -1/sqrt(1-(x+1/2)^2)#
