## Problem Statement

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

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

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

Using the substitution of equivalent infinitesimals:

$$
\ln \left(1+2 \Delta x^{2}+\Delta x^{3}\right) \sim 2 \Delta x^{2}+\Delta x^{3}, \text { as } \Delta x \rightarrow 0\left(2 \Delta x^{2}+\Delta x^{3} \rightarrow 0\right)
$$

We get:

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

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

## Problem Kuznetsov Differentiation $2-29$