## Problem Statement

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

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

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

Using the substitution of equivalent infinitesimals:

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

We get:

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

Since $\sin \left(\frac{1}{\Delta x}\right)_{\text { is bounded, then }}$

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

Then:

$=\frac{0}{\sqrt{1+\ln (1+0)}+1}=\frac{0}{\sqrt{1}+1}=0$

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

## Problem Kuznetsov Differentiation 2-6