## Condition of the problem

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

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

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

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

Then:

$y_{0}^{\prime}=y^{\prime}\left(x_{0}\right)=-\frac{3}{(3 \cdot 2+2)^{2}}=-\frac{3}{8^{2}}=-\frac{3}{64}$

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

$y_{0}^{\prime}=-\frac{3}{64}$

$y_{0}=y\left(x_{0}\right)=\frac{1}{3 \cdot 2+2}=\frac{1}{8}$

We get:

$y-\frac{1}{8}=-\frac{3}{64} \cdot(x-2)$

$y=-\frac{3}{64} \cdot x+\frac{3}{32}+\frac{1}{8}$

$y=-\frac{3}{64} \cdot x+\frac{7}{32}$

Thus, the equation of the tangent line is:

$y=-\frac{3}{64} \cdot x+\frac{7}{32}$

Problem Kuznetsov Differentiation 3-20