## Problem Statement

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

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

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

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

Then:

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

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

$$
\begin{aligned}
& y-y_{0}=y_{0}^{\prime}\left(x-x_{0}\right), \text { where } \\
& y_{0}^{\prime}=0 \\
& y_{0}=y\left(x_{0}\right)=\frac{2 \cdot 1}{1^{2}+1}=\frac{2}{2}=1
\end{aligned}
$$

We get:

$y-1=0 \cdot(x-(-2))$

$y=1$

That is, the equation of the tangent line is:

$y=1$

## Problem Kuznetsov Differentiation 3-23