## Problem Statement

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

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

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

Using the substitution of equivalent infinitesimals:

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

We get:

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

Using the substitution of equivalent infinitesimals:

![](https://cdn.mathpix.com/cropped/2024_05_22_41fcb48922b41c38bc2fg-01.jpg?height=146&width=1472&top_left_y=2388&top_left_x=153)

We get:
$=\lim _{\Delta x \rightarrow 0} \frac{\Delta x^{\frac{3}{2}} \sin \frac{2}{\Delta x}}{\Delta x}+0=\lim _{\Delta x \rightarrow 0} \Delta x^{\frac{1}{2}} \sin \frac{2}{\Delta x}=$

Since $\sin \frac{2}{\Delta x}$ is bounded, then

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

Thus,

$=0$

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

Problem Kuznetsov Differentiation $2-26$