If #f(x) =-sqrt(3x-1) # and #g(x) = (2-1/x)^2 #, what is #f'(g(x)) #?

1 Answer
Sep 15, 2017

Answer:

#-(3x^2)/(22x^2-12x+6)#

Explanation:

First of all, figure out what #f'(x)# is.
#d/dx(-sqrt(3x-1))# = #-d/dx(sqrt(3x-1))#.
Using the chain rule, we get:
#-1/(2(3x-1))*3# = #-3/(6x-2)#.
Now, we just plug #g(x)# in, and we get the answer.
#-3/(6(2-1/x)^2-2)#
Expand the above equation...
#-3/(6(4-2/x+1/x^2)-2)#
Expand and simplify.
#-3/(24-12/x+6/x^2-2)# = #-3/(22-12/x+6/x^2)#
Move all the fractions in the denominator into one fraction.
#-3/((22x^2-12x+6)/x^2)#
Move the denominator of this new fraction into the numerator of the bigger fraction.
#-(3x^2)/(22x^2-12x+6)#
This bigger fraction cannot be simplified further.