What is the derivative of #f(x)=(cos^-1(x))/x# ?

1 Answer
Jul 28, 2014

#f'(x)=-1/(xsqrt(1-x^2))-(cos^-1x)/x^2#

Using Quotient Rule, which is

#y=f(x)/g(x)#, then #y'=(f'(x)g(x)−f(x)g'(x))/(g(x))^2#

Applying this for given problem, which is #f(x)=(cos^-1x)/x#

#f'(x)=((cos^-1x)'(x)-(cos^-1x)(x)')/x^2#

#f'(x)=(-1/sqrt(1-x^2)*x-cos^-1x)/x^2#

#f'(x)=-1/(xsqrt(1-x^2))-(cos^-1x)/x^2#, where #-1#<#x#<#1#