## Task Condition

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

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

$x_{t}^{\prime}=(\sqrt{t})^{\prime}=\frac{1}{2 \sqrt{t}}$

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

We obtain:

$$
\begin{aligned}
& y_{x}^{\prime}=\frac{y_{t}^{\prime}}{x_{t}^{\prime}}=\left(\frac{1}{3 \sqrt[3]{(t-1)^{2}}}\right) /\left(\frac{1}{2 \sqrt{t}}\right)=\frac{2 \sqrt{t}}{3 \sqrt[3]{(t-1)^{2}}} \\
& \left(y_{x}^{\prime}\right)_{t}^{\prime}=\left(\frac{2 \sqrt{t}}{3 \sqrt[3]{(t-1)^{2}}}\right)^{\prime}=\frac{2}{3} \cdot \frac{\frac{1}{2 \sqrt{t}} \cdot \sqrt[3]{(t-1)^{2}}-\sqrt{t} \cdot \frac{2}{3} \cdot \frac{1}{\sqrt[3]{t-1}}}{\sqrt[3]{(t-1)^{4}}}= \\
& =\frac{2}{3} \cdot \frac{3 \cdot(t-1)-t \cdot 4}{6 \sqrt{t} \cdot \sqrt[3]{(t-1)^{5}}}=\frac{-t-3}{9 \sqrt{t} \cdot \sqrt[3]{(t-1)^{5}}}
\end{aligned}
$$

Then:

$$
y_{x x}^{\prime \prime}=\frac{\left(y_{x}^{\prime}\right)_{t}^{\prime}}{x_{t}^{\prime}}=\left(\frac{-t-3}{9 \sqrt{t} \cdot \sqrt[3]{(t-1)^{5}}}\right) /\left(\frac{1}{2 \sqrt{t}}\right)=-\frac{2(t+3)}{9 \sqrt[3]{(t-1)^{5}}}
$$

## Problem Kuznetsov Differentiation 20-11