## Task Condition

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

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

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

We obtain:

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

Then:

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

## Problem Kuznetsov Differentiation 20-7