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)-(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#
