## Calculus (3rd Edition)

$$y'= (1-x)^{-2}(2-x)^{-1/2}+\frac{1}{2}(1-x)^{-1}(2-x)^{-3/2} .$$
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 $$y'=(-1)(-1)(1-x)^{-2}(2-x)^{-1/2}-\frac{1}{2}(-1)(1-x)^{-1}(2-x)^{-3/2} = (1-x)^{-2}(2-x)^{-1/2}+\frac{1}{2}(1-x)^{-1}(2-x)^{-3/2} .$$