## Task Condition

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

$$
y=\frac{x^{2}}{10}+3, x_{0}=2
$$
## Solution

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

$y^{\prime}=\left(\frac{x^{2}}{10}+3\right)^{\prime}=\frac{2 x}{10}=\frac{x}{5}$

Then:

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

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

$y_{0}^{\prime}=\frac{2}{5}$

$y_{0}=y\left(x_{0}\right)=\frac{2^{2}}{10}+3=\frac{2}{5}+3=\frac{17}{5}$

We get:

$y-\frac{17}{5}=\frac{2}{5} \cdot(x-2)$

$y=\frac{2}{5} \cdot x-\frac{4}{5}+\frac{17}{5}$

$y=\frac{2}{5} \cdot x+\frac{13}{5}$

Thus, the equation of the tangent line is:

$y=\frac{2}{5} \cdot x+\frac{13}{5}$

## Problem Kuznetsov Differentiation 3-29