## Problem Statement

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

$f(x)=\left\{\begin{array}{c}x^{2} \cos \left(\frac{4}{3 x}\right)+\frac{x^{2}}{2}, x \neq 0 \\ 0, x=0\end{array}\right.$
## 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:

$$
\begin{aligned}
& f^{\prime}(0)=\lim _{\Delta x \rightarrow 0} \frac{f(0+\Delta x)-f(0)}{\Delta x}=\lim _{\Delta x \rightarrow 0} \frac{x^{2} \cos \left(\frac{4}{3 x}\right)+\frac{x^{2}}{2}-0}{\Delta x}= \\
& =\lim _{\Delta x \rightarrow 0} \frac{x^{2} \cos \left(\frac{4}{3 x}\right)+\frac{x^{2}}{2}}{\Delta x}=\lim _{\Delta x \rightarrow 0}\left(x \cos \left(\frac{4}{3 x}\right)+\frac{x}{2}\right)=
\end{aligned}
$$

Since $\cos \left(\frac{4}{3 x}\right)$ is bounded, then

$\Delta x \cdot \cos \left(\frac{4}{3 x}\right) \rightarrow 0 \quad$, as $\Delta x \rightarrow 0$

Then:

$=0+\frac{0}{2}=0$

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

## Problem Kuznetsov Differentiation $2-8$