## Task Condition

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

$$
\left\{\begin{array}{l}
x=\sqrt{t^{3}-1} \\
y=\ln t
\end{array}\right.
$$
## Solution

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

$y_{t}^{\prime}=(\ln t)^{\prime}=\frac{1}{t}$

We obtain:

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

Then:

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

## Problem Kuznetsov Differentiation 20-13