## Task Condition

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

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

$x_{t}^{\prime}=(\operatorname{sh} t)^{\prime}=\operatorname{ch} t$

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

We obtain:

$$
\begin{aligned}
& y_{x}^{\prime}=\frac{y_{t}^{\prime}}{x_{t}^{\prime}}=\left(\frac{2 \operatorname{sh} t}{\operatorname{ch}^{3} t}\right) /(\operatorname{ch} t)=\frac{2 \operatorname{sh} t}{\operatorname{ch}^{4} t} \\
& \left(y_{x}^{\prime}\right)_{t}^{\prime}=\left(\frac{2 \operatorname{sh} t}{\operatorname{ch}^{4} t}\right)^{\prime}=2 \cdot \frac{\operatorname{ch} t \cdot \operatorname{ch}^{4} t-\operatorname{sh} t \cdot 4 \operatorname{ch}^{3} t \cdot \operatorname{sh} t}{\operatorname{ch}^{8} t}= \\
& =2 \cdot \frac{\operatorname{ch}^{2} t-4 \operatorname{sh}^{2} t}{\operatorname{ch}^{5} t}=\frac{2-6 \operatorname{sh}^{2} t}{\operatorname{ch}^{5} t}
\end{aligned}
$$

Then:

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

## Problem Kuznetsov Differentiation 20-14