## Condition of the problem

To derive the equation of the tangent line to the given curve at the point with abscissa $x_{0}$.

$$
y=\frac{3 x-2 x^{3}}{3}, x_{0}=1
$$
## Solution

Let's find $y^{\prime}:$

$y^{\prime}=\left(\frac{3 x-2 x^{3}}{3}\right)^{\prime}=\frac{3-6 x^{2}}{3}=1-2 x^{2}$

Then:

$y_{0}^{\prime}=y^{\prime}\left(x_{0}\right)=1-2 \cdot 1^{2}=-1$

Since the function $y^{\prime}$ at the point $x_{0}$ has a finite derivative, the equation of the tangent line is:

$y-y_{0}=y_{0}^{\prime}\left(x-x_{0}\right)$, where

$y_{0}^{\prime}=-1$

$y_{0}=y\left(x_{0}\right)=\frac{3 \cdot 1-2 \cdot 1^{3}}{3}=\frac{3-2}{3}=\frac{1}{3}$

We get:

$$
\begin{aligned}
& y-\frac{1}{3}=-1 \cdot(x-1) \\
& y=-x+1+\frac{1}{3} \\
& y=-x+1 \frac{1}{3}
\end{aligned}
$$

That is, the equation of the tangent line is:

$y=-x+1 \frac{1}{3}$

## Problem Kuznetsov Differentiation 3-28