## Condition of the problem

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

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

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

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

Then:

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

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

$$
\begin{aligned}
& y-y_{\overline{0}}=y_{\overline{0}}^{\prime}\left(x-x_{0}\right), \text { where } \\
& y_{\overline{0}}^{\prime}=1 \\
& y_{\overline{0}}=y\left(x_{\overline{0}}\right)=2 x_{\overline{0}}+\frac{1}{x_{\overline{0}}}=2 \cdot 1+\frac{1}{1}=2+1=3
\end{aligned}
$$

We get:

$$
\begin{aligned}
& y-3=1 \cdot(x-1) \\
& y=x-1+3 \\
& y=x+2
\end{aligned}
$$

Thus, the equation of the tangent line is: $y=x+2$

## Problem Kuznetsov Differentiation 3-15