## Problem Statement

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

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

$$
\begin{aligned}
& f^{\prime}(0)=\lim _{\Delta x \rightarrow 0} \frac{f(0+\Delta x)-f(0)}{\Delta x}=\lim _{\Delta x \rightarrow 0} \frac{\Delta x^{2} e^{|\Delta x|} \sin \frac{1}{\Delta x^{2}}-0}{\Delta x}= \\
& =\lim _{\Delta x \rightarrow 0} e^{|\Delta x|} \cdot \Delta x \cdot \sin \frac{1}{\Delta x^{2}}= \\
& \quad \sin \frac{1}{\Delta x^{2}} \text { - is bounded, so }
\end{aligned}
$$

$\Delta x \sin \frac{1}{\Delta x^{2}} \rightarrow 0$, as $\Delta x \rightarrow 0$

Then:

$=e^{|0|} \cdot 0=1 \cdot 0=0$

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

## Problem Kuznetsov Differentiation $2-28$