## Problem Statement

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

$$
f(x)=\left\{\begin{array}{c}
\arcsin \left(x^{2} \cos \left(\frac{1}{9 x}\right)\right)+\frac{2}{3} x, x \neq 0 \\
0, x=0
\end{array}\right.
$$
## 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:

$$
\begin{aligned}
& f^{\prime}(0)=\lim _{\Delta x \rightarrow 0} \frac{f(0+\Delta x)-f(0)}{\Delta x}=\lim _{\Delta x \rightarrow 0} \frac{\arcsin \left(\Delta x^{2} \cos \left(\frac{1}{9 \Delta x}\right)\right)+\frac{2}{3} \Delta x-0}{\Delta x}= \\
& =\lim _{\Delta x \rightarrow 0} \frac{\arcsin \left(\Delta x^{2} \cos \left(\frac{1}{9 \Delta x}\right)\right)}{\Delta x}+\lim _{\Delta x \rightarrow 0} \frac{\frac{2}{3} \Delta x}{\Delta x}=
\end{aligned}
$$

Using the substitution of equivalent infinitesimals:

$\arcsin \left(\Delta x^{2} \cos \left(\frac{1}{9 \Delta x}\right)\right) \sim \Delta x^{2} \cos \left(\frac{1}{9 \Delta x}\right)$, as

$\Delta x \rightarrow 0\left(\Delta x^{2} \cos \left(\frac{1}{9 \Delta x}\right) \rightarrow 0\right)$

We get:

$=\lim _{\Delta x \rightarrow 0} \frac{\Delta x^{2} \cos \left(\frac{1}{9 \Delta x}\right)}{\Delta x}+\lim _{\Delta x \rightarrow 0} \frac{2}{3}=\lim _{\Delta x \rightarrow 0} \Delta x \cos \left(\frac{1}{9 \Delta x}\right)+\frac{2}{3}=$

Since $\cos \left(\frac{1}{9 \Delta x}\right)$ is bounded, then

$\Delta x \cos \left(\frac{1}{9 \Delta x}\right) \rightarrow 0 \quad$, as $\Delta x \rightarrow 0$

Thus:
$=0+\frac{2}{3}=\frac{2}{3}$

Therefore,

$$
f^{\prime}(0)=\frac{2}{3}
$$

## Problem Kuznetsov Differentiation 2-2