Calculus and Its Applications, Chapter 1, 1.8, Section 1.8, Problem 62

If $\displaystyle y = \frac{x}{\sqrt{x - 1}}$, find $y''$
We have $\displaystyle y = \frac{x}{(x - 1)^{\frac{1}{2}}}$, so by using Quotient Rule

$\begin{equation} \begin{aligned} y' &= \frac{(x - 1)^{\frac{1}{2}} \cdot \frac{d}{dx} (x) - (x) \cdot \frac{d}{dx} (x -1)^{\frac{1}{2}} }{\left[ (x - 1)^{\frac{1}{2}} \right]^2}\\ \\ y' &= \frac{(x - 1)^{\frac{1}{2}} (1) - x \cdot \frac{1}{2}(x - 1)^{\frac{1}{2}- 1} \cdot \frac{d}{dx} (x - 1) }{x - 1}\\ \\ y' &= \frac{(x - 1)^{\frac{1}{2}} - \frac{x}{2} (x - 1)^{-\frac{1}{2}}(1) }{x - 1} \\ \\ &= \frac{(x - 1)^{\frac{1}{2}} - \frac{x}{2(x - 1)^{\frac{1}{2}}} }{x - 1}\\ \\ &= \frac{2(x - 1) - x}{x - 1}\\ \\ &= \frac{2x - 2 - x}{x - 1}\\ \\ &= \frac{x - 2}{x - 1} \end{aligned} \end{equation}$


Again, by applying Quotient Rule

$\begin{equation} \begin{aligned} y'' &= \frac{(x - 1) \cdot \frac{d}{dx} (x -2 ) - (x - 2) \cdot \frac{d}{dx} (x - 1) }{(x - 1)^2}\\ \\ y'' &= \frac{(x - 1)(1) - (x - 2)(1)}{(x - 1)^2}\\ \\ &= \frac{x - 1 - x + 2}{(x - 1)^2}\\ \\ &= \frac{1}{(x - 2)^2} \end{aligned} \end{equation}$