## Problem Statement

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

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

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

Using the substitution of equivalent infinitesimals:

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

We get:

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

Using the substitution of equivalent infinitesimals:

![](https://cdn.mathpix.com/cropped/2024_05_22_9f9f9ef347bdb90989d8g-01.jpg?height=143&width=867&top_left_y=2178&top_left_x=157)

We get:

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

Thus, $f^{\prime}(0)=-\frac{1}{2}$

## Problem Kuznetsov Differentiation 2-17