## Problem Statement

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

$f(x)=\left\{\begin{array}{c}\tan\left(2^{x^{2} \cos (1 /(8 x))}-1+x\right), x \neq 0 ; \\ 0, x=0\end{array}\right.$
## 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:

$$
\begin{aligned}
& f^{\prime}(0)=\lim _{\Delta x \rightarrow 0} \frac{f(0+\Delta x)-f(0)}{\Delta x}=\lim _{\Delta x \rightarrow 0} \frac{\tan\left(2^{\Delta x^{2} \cos (1 /(8 \Delta x))}-1+\Delta x\right)-0}{\Delta x}= \\
& =\lim _{\Delta x \rightarrow 0} \frac{\tan\left(2^{\Delta x^{2} \cos (1 /(8 \Delta x))}-1+\Delta x\right)}{\Delta x}=
\end{aligned}
$$

We use the substitution of equivalent infinitesimals:

$\tan\left(2^{\Delta x^{2} \cos (1 /(8 \Delta x))}-1+\Delta x\right) \sim 2^{\Delta x^{2} \cos (1 /(8 \Delta x))}-1+\Delta x$, as

$\Delta x \rightarrow 0\left(2^{\Delta x^{2} \cos (1 /(8 \Delta x))}-1+\Delta x \rightarrow 0\right)$

We get:

$$
\begin{aligned}
& =\lim _{\Delta x \rightarrow 0} \frac{2^{\Delta x^{2} \cos (1 /(8 \Delta x))}-1+\Delta x}{\Delta x}=\lim _{\Delta x \rightarrow 0} \frac{2^{\Delta x^{2} \cos (1 /(8 \Delta x))}-1}{\Delta x}+\lim _{\Delta x \rightarrow 0} \frac{\Delta x}{\Delta x}= \\
& =\lim _{\Delta x \rightarrow 0} \frac{\left(e^{\ln 2}\right)^{\Delta x^{2} \cos (1 /(8 \Delta x))}-1}{\Delta x}+\lim _{\Delta x \rightarrow 0} 1=1+\lim _{\Delta x \rightarrow 0} \frac{e^{\ln 2 \cdot \Delta x^{2} \cos (1 /(8 \Delta x))}-1}{\Delta x}=
\end{aligned}
$$

We use the substitution of equivalent infinitesimals:

$e^{\ln 2 \cdot \Delta x^{2} \cos (1 /(8 \Delta x))}-1 \sim \ln 2 \cdot \Delta x^{2} \cos (1 /(8 \Delta x))$, as

$\Delta x \rightarrow 0\left(\ln 2 \cdot \Delta x^{2} \cos (1 /(8 \Delta x)) \rightarrow 0\right)$

We get:

$$
=1+\lim _{\Delta x \rightarrow 0} \frac{\ln 2 \cdot \Delta x^{2} \cos (1 /(8 \Delta x))}{\Delta x}=1+\lim _{\Delta x \rightarrow 0} \ln 2 \cdot \Delta x \cos (1 /(8 \Delta x))=
$$

Since $\cos (1 /(8 \Delta x))$ is bounded, then

$\Delta x \cos (1 /(8 \Delta x)) \rightarrow 0$, as $\Delta x \rightarrow 0$

Then:

$=1+\ln 2 \cdot 0=1$

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

## Problem Kuznetsov Differentiation 2-12