Answer
For $x=-1$, $$(fog)'=1$$
Work Step by Step
$f(u)=1-\frac{1}{u}$, $u=g(x)=\frac{1}{1-x}$, $x=-1$
Applying the Chain Rule: $$(fog)'=f'(g(x))g'(x)=f'(u)u'$$
We have
$f'(u)=(1-\frac{1}{u})'=0-\Big(\frac{-1(u)'}{u^2}\Big)=\frac{1}{u^2}$
$u'=\Big(\frac{1}{1-x}\Big)'=\frac{-1(1-x)'}{(1-x)^2}=\frac{-1(-1)}{(1-x)^2}=\frac{1}{(1-x)^2}$
Therefore,
$$(fog)'=\frac{1}{u^2}\times\frac{1}{(1-x)^2}=\frac{1}{\frac{1}{(1-x)^2}}\times\frac{1}{(1-x)^2}$$
$$=(1-x)^2\times\frac{1}{(1-x)^2}=1$$
For $x=-1$, $$(fog)'=1$$
