## Problem Statement

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

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

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

Using the substitution of equivalent infinitesimals:

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

We get:

$$
=\lim _{\Delta x \rightarrow 0} \frac{3 \Delta x^{2} \cos \frac{2}{\Delta x}}{\Delta x\left(\sqrt{1+\ln \left(1+3 \Delta x^{2} \cos \frac{2}{\Delta x}\right)}+1\right)}=\lim _{\Delta x \rightarrow 0} \frac{3 \Delta x \cos \frac{2}{\Delta x}}{\sqrt{1+\ln \left(1+3 \Delta x^{2} \cos \frac{2}{\Delta x}\right)}+1}=
$$

Since $\cos \frac{2}{\Delta x}$ is bounded, then

$\Delta x \cdot \cos \frac{2}{\Delta x} \rightarrow 0$, as $\Delta x \rightarrow 0$

Then:

$=\frac{3 \cdot 0}{\sqrt{1+\ln (1+3 \cdot 0)}+1}=\frac{0}{\sqrt{1}+1}=0$

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

## Problem Kuznetsov Differentiation 2-22