How do you find the derivative of #g(x)=3 arccos (x/2)#?

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Answer 1 · 2015-07-08T22:16:03

Assuming you do not remember the derivative of #arccosu#, which is #-1/sqrt(1-u^2) ((du)/(dx))#:

#y = 3arccos(x/2)#

#cos(y/3) = x/2#

#-1/3sin(y/3)((dy)/(dx)) = 1/2#

#(dy)/(dx) = -3/(2sin(y/3))#

Since #sin^2u + cos^2u = 1#:

#= -3/(2sqrt(1-cos^2(y/3))#

#= -3/(2sqrt(1-(x/2)^2)#

Getting it even simpler:

#= -3/(2sqrt(1-x^2/4)#

#= -3/(2sqrt(1/4)sqrt(4-x^2)#

#= color(blue)(-3/sqrt(4-x^2))#