## Problem Statement

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

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

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

Using the substitution of equivalent infinitesimals:

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

We get:

$$
\begin{aligned}
& =\lim _{\Delta x \rightarrow 0}\left(\ln 3 \cdot \Delta x^{2} \sin \frac{2}{\Delta x}\right) / \Delta x-\lim _{\Delta x \rightarrow 0} 2= \\
& =\lim _{\Delta x \rightarrow 0} \ln 3 \cdot \Delta x \sin \frac{2}{\Delta x}-2=
\end{aligned}
$$

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

$$
\Delta x \cdot \sin \frac{2}{\Delta x} \rightarrow 0, \text { as } \Delta x \rightarrow 0
$$

Thus

$=\ln 3 \cdot 0-2=-2$

Therefore, $f^{\prime}(0)=-2$

## Problem Kuznetsov Differentiation $2-21$