Question

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

Answer 1

`-(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.