If #f(x)= 1/x # and #g(x) = 1/x #, how do you differentiate #f'(g(x)) # using the chain rule?

1 Answer
Mar 25, 2016

Answer:

#(df(g(x)))/(dx)=color(red)(-x^2)*color(blue)((-1/x^2)= 1 #

Explanation:

Given: #f(x) = 1/x# and #g(x)=1/x#
Required: #(df(g(x)))/(dx)=f'(g(x))#
Definition and Principles - Chain Rule:
#(df(g(x)))/(dx)=color(red)((df(g))/(dg))*color(blue)((dg(x))/(dx)) #
#color(red)((df(g))/(dg)) = -1/g^2# but #g=1/x#, thus
#color(red)((df(g))/(dg)) = -1/(1/(x^2))=-x^2#
#color(blue)((dg(x))/(dx))=-1/x^2# thus
#(df(g(x)))/(dx)=color(red)(-x^2)*color(blue)((-1/x^2)= 1 #