Answer
$$ y'= (1-x)^{-2}(2-x)^{-1/2}+\frac{1}{2}(1-x)^{-1}(2-x)^{-3/2} .$$
Work Step by Step
Recall the product rule: $(uv)'=u'v+uv'$
Recall that $(x^n)'=nx^{n-1}$
Since $ y=\frac{1}{(1-x)\sqrt{2-x}}$, we rewrite $ y $ as follows
$$ y=\frac{1}{(1-x)\sqrt{2-x}}= (1-x)^{-1}(2-x)^{-1/2}.$$
Now, the derivative $ y'$, by using the product rule, is given by
$$ y'=(-1)(-1)(1-x)^{-2}(2-x)^{-1/2}-\frac{1}{2}(-1)(1-x)^{-1}(2-x)^{-3/2} = (1-x)^{-2}(2-x)^{-1/2}+\frac{1}{2}(1-x)^{-1}(2-x)^{-3/2} .$$
