## Problem Statement

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

$$
\left\{\begin{array}{l}
x=\sqrt{1-t^{2}} \\
y=\frac{1}{t}
\end{array}\right.
$$
## Solution

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

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

We obtain:

$$
\begin{aligned}
& y_{x}^{\prime}=\frac{y_{t}^{\prime}}{x_{t}^{\prime}}=\left(-\frac{1}{t^{2}}\right) /\left(-\frac{t}{\sqrt{1-t^{2}}}\right)=\frac{\sqrt{1-t^{2}}}{t^{3}} \\
& \left(y_{x}^{\prime}\right)_{t}^{\prime}=\left(\frac{\sqrt{1-t^{2}}}{t^{3}}\right)^{\prime}=\frac{\frac{1}{2 \sqrt{1-t^{2}}} \cdot(-2 t) \cdot t^{3}-\sqrt{1-t^{2}} \cdot 3 t^{2}}{t^{6}}=\frac{-t^{2}-3\left(1-t^{2}\right)}{t^{4} \sqrt{1-t^{2}}}= \\
& =\frac{-3+2 t^{2}}{t^{4} \sqrt{1-t^{2}}}
\end{aligned}
$$

Then:

$y_{x x}^{\prime \prime}=\frac{\left(y_{x}^{\prime}\right)_{t}^{\prime}}{x_{t}^{\prime}}=\left(\frac{-3+2 t^{2}}{t^{4} \sqrt{1-t^{2}}}\right) /\left(-\frac{t}{\sqrt{1-t^{2}}}\right)=\frac{3-2 t^{2}}{t^{5}}$

## Problem Kuznetsov Differentiation 20-2