## Problem Statement

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

$f(x)=\left\{\begin{array}{c}\sin \left(e^{x^{2} \sin \frac{5}{x}}-1\right)+x, x \neq 0 \\ 0, x=0\end{array}\right.$
## 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:

$$
\begin{aligned}
& f^{\prime}(0)=\lim _{\Delta x \rightarrow 0} \frac{f(0+\Delta x)-f(0)}{\Delta x}=\lim _{\Delta x \rightarrow 0}\left(\sin \left(e^{\Delta x^{2} \sin \frac{s}{\Delta x}}-1\right)+\Delta x-0\right) / \Delta x= \\
& =\lim _{\Delta x \rightarrow 0}\left(\sin \left(e^{\Delta x^{2} \sin \frac{5}{\Delta x}}-1\right)\right) / \Delta x+\lim _{\Delta x \rightarrow 0} \frac{\Delta x}{\Delta x}=
\end{aligned}
$$

Using the substitution of equivalent infinitesimals:

$$
e^{\Delta x^{2} \sin \frac{5}{\Delta x}}-1 \sim \Delta x^{2} \sin \frac{5}{\Delta x}, \text { as } \Delta x \rightarrow 0\left(\Delta x^{2} \sin \frac{5}{\Delta x} \rightarrow 0\right)
$$

We get:

$$
=\lim _{\Delta x \rightarrow 0}\left(\sin \left(\Delta x^{2} \sin \frac{5}{\Delta x}\right)\right) / \Delta x+\lim _{\Delta x \rightarrow 0} 1=
$$

Using the substitution of equivalent infinitesimals:

![](https://cdn.mathpix.com/cropped/2024_05_22_6e5525776c432bba8201g-01.jpg?height=148&width=1451&top_left_y=2162&top_left_x=158)

We get:

$=\lim _{\Delta x \rightarrow 0} \frac{\Delta x^{2} \sin \frac{5}{\Delta x}}{\Delta x}+1=\lim _{\Delta x \rightarrow 0} \Delta x \sin \frac{5}{\Delta x}+1=$

Since $\sin \left(\frac{5}{\Delta x}\right)$ is bounded,

$\Delta x \cdot \sin \left(\frac{5}{\Delta x}\right) \rightarrow 0 \quad$, as $\Delta x \rightarrow 0$

Then:

$=0+1=1$

Thus, $f^{\prime}(0)=1$

## Problem Kuznetsov Differentiation $2-7$