## Problem Statement

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

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

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

Using the substitution of equivalent infinitesimals:

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

We get:

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

$=\lim _{\Delta x \rightarrow 0} \frac{3}{2}-\lim _{\Delta x \rightarrow 0} \Delta x \sin \frac{1}{\Delta x}=$

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

$\Delta x \sin \frac{1}{\Delta x} \rightarrow 0$, as $\Delta x \rightarrow 0$

## Then:

$=\frac{3}{2}-0=\frac{3}{2}$

Thus,

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

## Problem Kuznetsov Differentiation $2-25$