## Problem Statement

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

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

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

Using the substitution of equivalent infinitesimals:

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

$\sin \left(\frac{3 \Delta x}{2}\right) \sim \frac{3 \Delta x}{2}$, as $\Delta x \rightarrow 0\left(\frac{3 \Delta x}{2} \rightarrow 0\right)$

We get:

$$
=\lim _{\Delta x \rightarrow 0} \frac{-2\left(\frac{\Delta x}{2}\right)^{2}}{\Delta x^{2}}+\lim _{\Delta x \rightarrow 0} \frac{2\left(\frac{3 \Delta x}{2}\right)^{2}}{\Delta x^{2}}=\lim _{\Delta x \rightarrow 0} \frac{-1}{2}+\lim _{\Delta x \rightarrow 0} \frac{9}{2}=-\frac{1}{2}+\frac{9}{2}=4
$$

Thus, $f^{\prime}(0)=4$

## Problem Kuznetsov Differentiation 2-30