## Task Condition

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

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

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

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

We obtain:

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

$=\frac{2-t}{1-t} \cdot \sqrt{\frac{t-1}{1-t}}=\frac{(2-t) \cdot \sqrt{-1}}{1-t}$

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

$=\sqrt{-1} \cdot \frac{-1+t+2-t}{(1-t)^{2}}=\frac{\sqrt{-1}}{(1-t)^{2}}$

Then:

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

## Problem Kuznetsov Differentiation 20-10