## Problem Statement

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

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

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

Using the substitution of equivalent infinitesimals:

$$
\begin{aligned}
& \ln \left(1-\sin \left(\Delta x^{3} \sin \frac{1}{\Delta x}\right)\right) \sim-\sin \left(\Delta x^{3} \sin \frac{1}{\Delta x}\right), \text { as } \\
& \Delta x \rightarrow 0\left(-\sin \left(\Delta x^{3} \sin \frac{1}{\Delta x}\right) \rightarrow 0\right)
\end{aligned}
$$

We get:

$=\lim _{\Delta x \rightarrow 0} \frac{-\sin \left(\Delta x^{3} \sin \frac{1}{\Delta x}\right)}{\Delta x}=$

Using the substitution of equivalent infinitesimals:

![](https://cdn.mathpix.com/cropped/2024_05_22_2ae1538796d8794fcb74g-01.jpg?height=155&width=1451&top_left_y=2313&top_left_x=158)

We get:

$$
=\lim _{\Delta x \rightarrow 0} \frac{-\Delta x^{3} \sin \frac{1}{\Delta x}}{\Delta x}=\lim _{\Delta x \rightarrow 0}-\Delta x^{2} \sin \frac{1}{\Delta x}=
$$

Since $\sin \frac{1}{\Delta x}$ is bounded, then

$\Delta x^{2} \sin \frac{1}{\Delta x} \rightarrow{ }_{, \text {as }} \Delta x \rightarrow 0\left(\Delta^{2} x \rightarrow 0\right)$

Thus,

$=-0=0$

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

## Problem Kuznetsov Differentiation 2-4