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

1 Answer
ErmWhat
Sep 15, 2017

#-(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.

Related questions
See all questions in Chain Rule
Impact of this question
1808 views around the world
You can reuse this answer
Creative Commons License