## Problem Statement

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

$f(x)=\left\{\begin{array}{c}\operatorname{arctg}\left(x^{3}-x^{\frac{3}{2}} \sin \frac{1}{3 x}\right), x \neq 0 ; \\ 0, x=0\end{array}\right.$
## 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:

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

Using the substitution of equivalent infinitesimals:

$$
\begin{aligned}
& \operatorname{arctg}\left(\Delta x^{3}-\Delta x^{\frac{3}{2}} \sin \frac{1}{3 \Delta x}\right) \sim \Delta x^{3}-\Delta x^{\frac{3}{2}} \sin \frac{1}{3 \Delta x}, \text { as } \\
& \Delta x \rightarrow 0\left(\Delta x^{3}-\Delta x^{\frac{3}{2}} \sin \frac{1}{3 \Delta x} \rightarrow 0\right)
\end{aligned}
$$

We get:

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

Since $\sin \frac{1}{3 \Delta x}$ is bounded, then

$\Delta x^{\frac{1}{2}} \sin \frac{1}{3 \Delta x} \rightarrow 0$, as $\Delta x \rightarrow 0\left(\Delta x^{\frac{1}{2}} \rightarrow 0\right)$

Then:
$=0^{2}-0=0$

Thus, $f^{\prime}(0)=0$

## Problem Kuznetsov Differentiation 2-9