## Problem Statement

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

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

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

Using the substitution of equivalent infinitesimals:

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

We get:

$=\lim _{\Delta x \rightarrow 0} \frac{2 \cdot\left(\frac{1}{2} \cdot \Delta x \sin \frac{1}{\Delta x}\right)^{2}}{\Delta x}=\lim _{\Delta x \rightarrow 0} \frac{1}{2} \cdot \Delta x \cdot \sin ^{2} \frac{1}{\Delta x}=$

Since $\sin ^{2} \frac{1}{\Delta x}$ is bounded, then

$\Delta x \cdot \sin ^{2} \frac{1}{\Delta x} \rightarrow 0$, as $\Delta x \rightarrow 0$

Thus:

$=2 \cdot 0=0$

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

## Problem Kuznetsov Differentiation 2-31