## problem statement

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

$$
f(x)=\left\{\begin{array}{c}
\sqrt[3]{1-2 x^{3} \sin \frac{5}{x}}-1+x, x \neq 0 \\
0, x=0
\end{array}\right.
$$
## 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:

$$
\begin{aligned}
& f^{\prime}(0)=\lim _{\Delta x \rightarrow 0} \frac{f(0+\Delta x)-f(0)}{\Delta x}=\lim _{\Delta x \rightarrow 0} \frac{\sqrt[3]{1-2 \Delta x^{3} \sin \frac{5}{\Delta x}}-1+\Delta x-0}{\Delta x}= \\
& =\lim _{\Delta x \rightarrow 0} \frac{\sqrt[3]{1-2 \Delta x^{3} \sin \frac{5}{\Delta x}}-1}{\Delta x}+\lim _{\Delta x \rightarrow 0} \frac{\Delta x}{\Delta x}= \\
& =\lim _{\Delta x \rightarrow 0} \frac{\left(\sqrt[3]{1-2 \Delta x^{3} \sin \frac{5}{\Delta x}}-1\right)\left(\sqrt[3]{\left(1-2 \Delta x^{3} \sin \frac{5}{\Delta x}\right)^{2}}+\sqrt[3]{1-2 \Delta x^{3} \sin \frac{5}{\Delta x}}+1\right)}{\Delta x\left(\sqrt[3]{\left(1-2 \Delta x^{3} \sin \frac{5}{\Delta x}\right)^{2}}+\sqrt[3]{1-2 \Delta x^{3} \sin \frac{5}{\Delta x}}+1\right)}+\lim _{\Delta x \rightarrow 0} 1= \\
& =\lim _{\Delta x \rightarrow 0} \frac{1-2 \Delta x^{3} \sin \frac{5}{\Delta x}-1}{\Delta x\left(\sqrt[3]{\left(1-2 \Delta x^{3} \sin \frac{5}{\Delta x}\right)^{2}}+\sqrt[3]{1-2 \Delta x^{3} \sin \frac{5}{\Delta x}}+1\right)}+1= \\
& =\lim _{\Delta x \rightarrow 0} \frac{-2 \Delta x^{3} \sin \frac{5}{\Delta x}}{\Delta x\left(\sqrt[3]{\left(1-2 \Delta x^{3} \sin \frac{5}{\Delta x}\right)^{2}}+\sqrt[3]{1-2 \Delta x^{3} \sin \frac{5}{\Delta x}}+1\right)}+1= \\
& =\lim _{\Delta x \rightarrow 0} \frac{-2 \Delta x^{2} \sin \frac{5}{\Delta x}}{\sqrt[3]{\left(1-2 \Delta x^{3} \sin \frac{5}{\Delta x}\right)^{2}}+\sqrt[3]{1-2 \Delta x^{3} \sin \frac{5}{\Delta x}}+1}+1= \\
& \text { Since } \sin \frac{5}{\Delta x} \text { is bounded, then }
\end{aligned}
$$

$\Delta x \cdot \sin \frac{5}{\Delta x} \rightarrow 0$, as $\Delta x \rightarrow 0$

Then:

$=\frac{-2 \cdot 0}{\sqrt[3]{(1-2 \cdot 0)^{2}}+\sqrt[3]{1-2 \cdot 0}+1}+1=\frac{0}{\sqrt[3]{1}+\sqrt[3]{1}+1}+1=1$

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

## Problem Kuznetsov Differentiation $2-27$