If #f(x)= sqrt(x^2-1 # and #g(x) = 1/x #, what is #f'(g(x)) #? Calculus Basic Differentiation Rules Chain Rule 1 Answer Alan N. Aug 24, 2016 #f'(g(x)) = 1/sqrt(1-x^2)# Explanation: #f(x) = sqrt(x^2-1) = (x^2-1)^(1/2)# #f'(x) = 1/2(x^2-1)^(-1/2)*2x# (Power rule and Chain rule) #= x/sqrt(x^2-1)# Given #g(x) = 1/x# #f'(g(x)) = (1/x)/sqrt((1/x)^2-1)# #= 1/(x(sqrt((1-x^2)/x^2))# #=1/(x/x*sqrt((1-x^2))# #= 1/(sqrt((1-x^2))# Answer link Related questions What is the Chain Rule for derivatives? How do you find the derivative of #y= 6cos(x^2)# ? How do you find the derivative of #y=6 cos(x^3+3)# ? How do you find the derivative of #y=e^(x^2)# ? How do you find the derivative of #y=ln(sin(x))# ? How do you find the derivative of #y=ln(e^x+3)# ? How do you find the derivative of #y=tan(5x)# ? How do you find the derivative of #y= (4x-x^2)^10# ? How do you find the derivative of #y= (x^2+3x+5)^(1/4)# ? How do you find the derivative of #y= ((1+x)/(1-x))^3# ? See all questions in Chain Rule Impact of this question 19464 views around the world You can reuse this answer Creative Commons License