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

1 Answer
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.