## Problem Statement

Based on the definition of the derivative, find $f^{\prime}(0)$:

$$
f(x)=\left\{\begin{array}{c}
\operatorname{arctg}\left(x \cos \left(\frac{1}{5 x}\right)\right), x \neq 0 \\
0, x=0
\end{array}\right.
$$
## Solution

By definition, the derivative at the point $x=0$:

$f^{\prime}(0)=\lim _{\Delta x \rightarrow 0} \frac{f(0+\Delta x)-f(0)}{\Delta x}$

Based on the definition, we find:

$$
\begin{aligned}
& f^{\prime}(0)=\lim _{\Delta x \rightarrow 0} \frac{f(0+\Delta x)-f(0)}{\Delta x}=\lim _{\Delta x \rightarrow 0} \frac{\operatorname{arctg}\left(\Delta x \cdot \cos \left(\frac{1}{5 \Delta x}\right)\right)-0}{\Delta x}= \\
& =\lim _{\Delta x \rightarrow 0} \frac{\operatorname{arctg}\left(\Delta x \cdot \cos \left(\frac{1}{5 \Delta x}\right)\right)}{\Delta x}=
\end{aligned}
$$

Using the substitution of equivalent infinitesimals:

$$
\begin{aligned}
& \operatorname{arctg}\left(\Delta x \cdot \cos \left(\frac{1}{5 \Delta x}\right)\right) \sim \Delta x \cdot \cos \left(\frac{1}{5 \Delta x}\right), \text { as } \\
& \Delta x \rightarrow 0\left(\Delta x \cdot \cos \left(\frac{1}{5 \Delta x}\right) \rightarrow 0\right)
\end{aligned}
$$

We get:

$$
=\lim _{\Delta x \rightarrow 0} \frac{\Delta x \cdot \cos \left(\frac{1}{5 \Delta x}\right)}{\Delta x}=\lim _{\Delta x \rightarrow 0} \cos \left(\frac{1}{5 \Delta x}\right)_{- \text { the limit does not exist. }}
$$

Thus, $f^{\prime}(0)$ - does not exist.

## Problem Kuznetsov Differentiation 2-3