The question is asking, "what is the equation for the derivative of #f#, when its input is #g(x)#?
So we need to do two things: (1) Find the general derivative of #f#, and (2) Plug in #g(x)# as the input to this derivative.
Step (1):
If
#f(x)=-sqrt(2x-1)=-(2x-1)^(1//2)#,
then
#f'(x)=-1/2(2x-1)^(-1//2)(2)#
#=>f'(x)=-1/(sqrt(2x-1))#
by use of both the power rule and the chain rule.
This is the general derivative of #f#, with respect to some input #x#. If we plug in different inputs, we get different outputs.
Step (2):
Now, we plug in the input #g(x)# (instead of just #x#) to the derivative #f':#
#f'(color(red)x)=-1/(sqrt(2color(red)x-1))#
#=>f'[color(blue)g(x)]=-1/(sqrt(2color(blue)((2-1/x)^2)-1))#
#=>f'[g(x)]=-1/(sqrt(2(2-1/x)^2-1))#.
Depending on your teacher, this is likely as far as you'll need to go.
Bonus:
Please note: The domain of #f'(x)=-1/(sqrt(2x-1))# is #{x|2x-1>0}#, which is #x>1/2#.
(Slightly more limiting than the domain of just #f(x)#, which is #x>=1/2#)
The domain of #g(x)=(2-1/x)^2# is #{x|x!=0}#.
That means the domain of this composite function #f'[g(x)]# has to satisfy both of these restrictions. Luckily, that's pretty easy, considering the domain of #f'(x)# is a subset of the domain of #g(x)#. So the domain of #f'[g(x)]# is
#{x|x>1/2nnx!=0}#, or just #{x|x>1/2}#.
