## Problem Statement

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

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

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

Using the substitution of equivalent infinitesimals:

$$
\begin{aligned}
& \tan\left(\Delta x^{3}+\Delta x^{2} \sin \left(\frac{2}{\Delta x}\right)\right) \sim \Delta x^{3}+\Delta x^{2} \sin \left(\frac{2}{\Delta x}\right), \text { as } \\
& \Delta x \rightarrow 0\left(\Delta x^{3}+\Delta x^{2} \sin \left(\frac{2}{\Delta x}\right) \rightarrow 0\right)
\end{aligned}
$$

We get:

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

Since $\sin \left(\frac{2}{\Delta x}\right)$ is bounded, then

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

Thus:
$=0^{2}+0=0$

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

## Problem Kuznetsov Differentiation 2-1