## Problem Statement

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

$$
f(x)=\left\{\begin{array}{c}
x+\arcsin \left(x^{2} \sin \frac{6}{x}\right), x \neq 0 \\
0, x=0
\end{array}\right.
$$
## 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:

$$
\begin{aligned}
& f^{\prime}(0)=\lim _{\Delta x \rightarrow 0} \frac{f(0+\Delta x)-f(0)}{\Delta x}=\lim _{\Delta x \rightarrow 0} \frac{\Delta x+\arcsin \left(\Delta x^{2} \sin \frac{6}{\Delta x}\right)-0}{\Delta x}= \\
& =\lim _{\Delta x \rightarrow 0} \frac{\Delta x}{\Delta x}+\lim _{\Delta x \rightarrow 0} \frac{\arcsin \left(\Delta x^{2} \sin \frac{6}{\Delta x}\right)}{\Delta x}=
\end{aligned}
$$

Using the substitution of equivalent infinitesimals:

$\arcsin \left(\Delta x^{2} \sin \frac{6}{\Delta x}\right) \sim \Delta x^{2} \sin \frac{6}{\Delta x} \text{ as } \Delta x \rightarrow 0 \left(\Delta x^{2} \sin \frac{6}{\Delta x} \rightarrow 0\right)$

We get:

$=\lim _{\Delta x \rightarrow 0} 1+\lim _{\Delta x \rightarrow 0} \frac{\Delta x^{2} \sin \frac{6}{\Delta x}}{\Delta x}=1+\lim _{\Delta x \rightarrow 0} \Delta x \sin \frac{6}{\Delta x}=$

Since $\sin \frac{6}{\Delta x}$ is bounded, then

$\Delta x \sin \frac{6}{\Delta x} \rightarrow 0$, as $\Delta x \rightarrow 0$

Thus:

$=1+0=1$

Therefore, $f^{\prime}(0)=1$

## Problem Kuznetsov Differentiation 2-11