## Problem Statement

Find the second-order derivative $y_{x x}^{\prime \prime}$ of the function given parametrically.

$\left\{\begin{array}{l}x=\sqrt{t-1} \\ y=\frac{1}{\sqrt{t}}\end{array}\right.$
## Solution

$x_{t}^{\prime}=(\sqrt{t-1})^{\prime}=\frac{1}{2 \sqrt{t-1}}$

$y_{t}^{\prime}=\left(\frac{1}{\sqrt{t}}\right)^{\prime}=\left(t^{-\frac{1}{2}}\right)^{\prime}=-\frac{1}{2} \cdot t^{-\frac{3}{2}}=-\frac{1}{2 t \sqrt{t}}$

We obtain:

$$
\begin{aligned}
& y_{x}^{\prime}=\frac{y_{t}^{\prime}}{x_{t}^{\prime}}=\left(-\frac{1}{2 t \sqrt{t}}\right) /\left(\frac{1}{2 \sqrt{t-1}}\right)=-\frac{\sqrt{t-1}}{t \sqrt{t}}=-\sqrt{\frac{t-1}{t^{3}}} \\
& \left(y_{x}^{\prime}\right)_{t}^{\prime}=\left(-\sqrt{\frac{t-1}{t^{3}}}\right)^{\prime}=-\frac{1}{2} \cdot \sqrt{\frac{t^{3}}{t-1}} \cdot \frac{1 \cdot t^{3}-(t-1) \cdot 3 t^{2}}{t^{6}}= \\
& =\frac{1}{2} \cdot \sqrt{\frac{t}{t-1}} \cdot \frac{-t+3(t-1)}{t^{3}}=\frac{(2 t-3) \sqrt{t}}{2 t^{3} \sqrt{t-1}}
\end{aligned}
$$

Then:

$y_{x x}^{\prime \prime}=\frac{\left(y_{x}^{\prime}\right)_{t}^{\prime}}{x_{t}^{\prime}}=\left(\frac{(2 t-3) \sqrt{t}}{2 t^{3} \sqrt{t-1}}\right) /\left(\frac{1}{2 \sqrt{t-1}}\right)=\frac{(2 t-3) \sqrt{t}}{t^{3}}$