## Problem Statement

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

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

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

Using the substitution of equivalent infinitesimals:

$$
\begin{aligned}
& e^{\Delta x^{2}}-1 \sim \Delta x^{2}, \text { as } \Delta x \rightarrow 0\left(\Delta x^{2} \rightarrow 0\right) \\
& \sin \frac{\Delta x}{2} \sim \frac{\Delta x}{2}, \text { as } \\
& \Delta x \rightarrow 0\left(\frac{\Delta x}{2} \rightarrow 0\right)
\end{aligned}
$$

We get:

$$
=\lim _{\Delta x \rightarrow 0} \frac{\Delta x^{2}}{\Delta x^{2}}+\lim _{\Delta x \rightarrow 0} \frac{2 \cdot\left(\frac{\Delta x}{2}\right)^{2}}{\Delta x^{2}}=\lim _{\Delta x \rightarrow 0} 1+\lim _{\Delta x \rightarrow 0} \frac{1}{2}=1+\frac{1}{2}=1.5
$$

Thus, $f^{\prime}(0)=1.5$

## Problem Kuznetsov Differentiation 2-19