## Problem Statement

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

$$
\left\{\begin{array}{l}
x=\operatorname{ch} t \\
y=\sqrt[3]{\operatorname{sh}^{2} t}
\end{array}\right.
$$
## Solution

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

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

We obtain:

$y_{x}^{\prime}=\frac{y_{t}^{\prime}}{x_{t}^{\prime}}=\left(\frac{2 \operatorname{ch} t}{3 \sqrt[3]{\operatorname{sh} t}}\right) / \operatorname{sh} t=\frac{2 \operatorname{cth} t}{3 \sqrt[3]{\operatorname{sh} t}}$

$\left(y_{x}^{\prime}\right)_{t}^{\prime}=\left(\frac{2 \operatorname{cth} t}{3 \sqrt[3]{\operatorname{sh} t}}\right)^{\prime}=\frac{2}{3} \cdot \frac{-\frac{1}{\operatorname{sh}^{2} t} \cdot \sqrt[3]{\operatorname{sh} t}-\operatorname{cth} t \cdot \frac{1}{3} \cdot \frac{1}{\sqrt[3]{\mathrm{sh}^{2} t}} \cdot \operatorname{ch} t}{\sqrt[3]{\operatorname{sh}^{2} t}}=$

$=\frac{2}{3} \cdot \frac{-\frac{1}{\operatorname{sh}^{2} t} \cdot \operatorname{sh} t-\frac{\operatorname{ch} t}{\operatorname{sh} t} \cdot \frac{1}{3} \cdot \operatorname{ch} t}{\sqrt[3]{\operatorname{sh}^{6} t}}=-\frac{2}{3} \cdot \frac{\frac{1}{\operatorname{sh}^{6} t}+\frac{\operatorname{ch}^{2} t}{3 \operatorname{sh} t}}{\operatorname{sh}^{2} t}=$

$=-\frac{2}{9} \cdot \frac{3+\operatorname{ch}^{2} t}{\operatorname{sh}^{3} t}$

Then:

$y_{x x}^{\prime \prime}=\frac{\left(y_{x}^{\prime}\right)_{t}^{\prime}}{x_{t}^{\prime}}=\left(-\frac{2}{9} \cdot \frac{3+\mathrm{ch}^{2} t}{\operatorname{sh}^{3} t}\right) / \operatorname{sh} t=-\frac{2\left(3+\operatorname{ch}^{2} t\right)}{9 \operatorname{sh}^{4} t}$

## Problem Kuznetsov Differentiation 20-25