## Problem Statement

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

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

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

Using the substitution of equivalent infinitesimals:

$e^{\Delta x \sin 5 \Delta x}-1 \sim \Delta x \sin 5 \Delta x$, as $\Delta x \rightarrow 0(\Delta x \sin 5 \Delta x \rightarrow 0)$

We get:

$=\lim _{\Delta x \rightarrow 0} \frac{\Delta x \sin 5 \Delta x}{\Delta x}=\lim _{\Delta x \rightarrow 0} \sin 5 \Delta x=\sin 0=0$

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

## Problem Kuznetsov Differentiation 2-20