## Calculus (3rd Edition)

$$y'=\frac{ 1 -\frac{z}{2}}{(1-z)^{3/2}}.$$
Recall the quotient rule: $(\dfrac{u}{v})'=\dfrac{vu'-uv'}{v^2}$ Recall that $(x^n)'=nx^{n-1}$ Since $y= \frac{z}{\sqrt{1-z}}$, then by the quotient rule the derivative $y'$ is given by $$y'=\frac{\sqrt{1-z} +\frac{z}{2\sqrt{1-z}}}{1-z} =\frac{ 1-z +\frac{z}{2}}{(1-z)\sqrt{1-z}}=\frac{ 1 -\frac{z}{2}}{(1-z)^{3/2}}.$$