How do I find the derivative of #y = arccos((x-3)^2)#?

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Answer 1 · 2016-02-20T06:25:00

#dy/dx=d/dx(cos^-1 (x-3)^2)=(-2x+6)/sqrt(1-(x-3)^4)#

The formula to find the derivative of #cos^-1 u# is

#d/dx(cos^-1 u)=-1/sqrt(1-u^2)*d/dx(u)#

So from the given #y=cos^-1 (x-3)^2#
Let #u=(x-3)^2#

from the formula

#dy/dx=d/dx(cos^-1 u)=-1/sqrt(1-u^2)*d/dx(u)#

#dy/dx=d/dx(cos^-1 (x-3)^2)=-1/sqrt(1-((x-3)^2)^2)*d/dx((x-3)^2)#

#dy/dx=d/dx(cos^-1 (x-3)^2)=#

#-1/sqrt(1-(x-3)^4)*2(x-3)*d/dx(x-3)#

#dy/dx=d/dx(cos^-1 (x-3)^2)=(-2(x-3))/sqrt(1-(x-3)^4)*(1)#

#dy/dx=d/dx(cos^-1 (x-3)^2)=(-2x+6)/sqrt(1-(x-3)^4)#

have a nice day!