## Problem Statement

Find the derivative $y_{x}^{\prime}$.

$$
\left\{\begin{array}{l}
x=\sqrt{2 t-t^{2}} \\
y=\frac{1}{\sqrt[3]{(1-t)^{2}}}
\end{array}\right.
$$
## Solution

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

$y_{t}^{\prime}=\left(\frac{1}{\sqrt[3]{(1-t)^{2}}}\right)^{\prime}=\left((1-t)^{-\frac{2}{3}}\right)^{\prime}=-\frac{2}{3} \cdot(1-t)^{-\frac{5}{3}} \cdot(-1)=\frac{2}{3 \sqrt[3]{(1-t)^{5}}}$

We obtain:

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

## Problem Kuznetsov Differentiation 16-3