## Problem Statement

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

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

$x_{t}^{\prime}=\left(\operatorname{sh}^{2} t\right)^{\prime}=2 \operatorname{sh} t \cdot \operatorname{ch} t$

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

We obtain:

$$
\begin{aligned}
& y_{x}^{\prime}=\frac{y_{t}^{\prime}}{x_{t}^{\prime}}=\left(\frac{-2}{\operatorname{ch}^{3} t} \cdot \operatorname{sh} t\right) /(2 \operatorname{sh} t \cdot \operatorname{ch} t)=-\frac{1}{\operatorname{ch}^{4} t} \\
& \left(y_{x}^{\prime}\right)_{t}^{\prime}=\left(-\frac{1}{\operatorname{ch}^{4} t}\right)^{\prime}=-\frac{-4}{\operatorname{ch}^{5} t} \cdot \operatorname{sh} t=\frac{4 \operatorname{sh} t}{\operatorname{ch}^{5} t}
\end{aligned}
$$

Then:

$y_{x x}^{\prime \prime}=\frac{\left(y_{x}^{\prime}\right)_{t}^{\prime}}{x_{t}^{\prime}}=\left(\frac{4 \operatorname{sh} t}{\operatorname{ch}^{5} t}\right) /(2 \operatorname{sh} t \cdot \operatorname{ch} t)=\frac{2}{\operatorname{ch}^{6} t}$

## Problem Kuznetsov Differentiation 20-4