If #f(x)= sqrt(x^2-1 # and #g(x) = 1/x #, what is #f'(g(x)) #?

1 Answer
Aug 24, 2016

Answer:

#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))#