## Condition of the problem

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

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

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

$$
\begin{aligned}
& y^{\prime}=\left(\frac{x}{x^{2}+1}\right)^{\prime}=\frac{x^{\prime}\left(x^{2}+1\right)-x\left(x^{2}+1\right)^{\prime}}{\left(x^{2}+1\right)^{2}}= \\
& =\frac{x^{2}+1-x \cdot 2 x}{\left(x^{2}+1\right)^{2}}=\frac{1-x^{2}}{\left(x^{2}+1\right)^{2}}
\end{aligned}
$$

Then:

$y_{0}^{\prime}=y^{\prime}\left(x_{0}\right)=\frac{1-(-2)^{2}}{\left((-2)^{2}+1\right)^{2}}=\frac{-3}{5^{2}}=-\frac{3}{25}$

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

$y_{0}^{\prime}=-\frac{3}{25}$

$y_{0}=y\left(x_{0}\right)=\frac{-2}{(-2)^{2}+1}=-\frac{2}{5}$

We get:

$y-\left(-\frac{2}{5}\right)=-\frac{3}{25} \cdot(x-(-2))$

$y=-\frac{3}{25} \cdot(x+2)-\frac{2}{5}$

$y=-\frac{3}{25} \cdot x-\frac{6}{25}-\frac{2}{5}$

$y=-\frac{3}{25} \cdot x-\frac{16}{25}$

Thus, the equation of the tangent line is:

$$
y=-\frac{3}{25} \cdot x-\frac{16}{25}
$$

## Problem Kuznetsov Differentiation 3-21