Example. According to the definition, find the derivative of the function

$$
f(x)=\left[\begin{array}{ll}
1-\cos \left(x \sin \frac{1}{x}\right), & x \neq 0 \\
0, & x=0
\end{array}\right.
$$

at the point $x=0$.
Solution.

1. By definition

$$
f^{\prime}(0)=\lim _{x \rightarrow 0} \frac{f(x)-f(0)}{x}=\lim _{x \rightarrow 0} \frac{1-\cos (x \sin (1 / x))-0}{x}
$$

2. Since $\sin (1 / x)$ is a bounded function and $x$ is an infinitesimal function as $x \rightarrow 0$, by the theorem on the product of an infinitesimal function and a bounded function, $x \sin (1 / x) \rightarrow 0$ as $x \rightarrow 0$. Replacing the infinitesimal function in the numerator with an equivalent one and again using the mentioned theorem, we get

$$
\lim _{x \rightarrow 0} \frac{1-\cos (x \sin (1 / x))-0}{x} \lim _{x \rightarrow 0} \frac{x^{2} \sin ^{2}(1 / x)}{2 x}=0
$$

3. Thus, the limit exists and is equal to zero. Therefore, $f^{\prime}(0)=0$.

Answer. $f^{\prime}(0)=0$.

![](https://cdn.mathpix.com/cropped/2024_05_22_5878ce417e2ca061000bg-099.jpg?height=46&width=1062&top_left_y=1264&top_left_x=118)

1. $f(x)=\left[\begin{array}{l}\sin \left(x^{3}+x^{2} \sin \frac{2}{x}\right), x \neq 0, \\ 0, x=0 .\end{array}\right.$
2. $f(x)=\left[\begin{array}{l}\operatorname{tg}\left(x^{2} \cos \frac{1}{9 x}\right)+2 x, x \neq 0, \\ 0, x=0 .\end{array}\right.$
3. $f(x)=\left[\begin{array}{l}\arcsin \left(x \cos \frac{1}{5 x}\right) \\ 0, x=0 .\end{array}\right.$
4. $f(x)=\left[\begin{array}{l}\ln \left(1-\operatorname{tg}\left(x^{2} \sin \frac{1}{x}\right)\right), x \neq 0, \\ 0, x=0 .\end{array}\right.$
5. $f(x)=\left[\begin{array}{l}\operatorname{tg}\left(x \sin \frac{3}{x}\right), x \neq 0, \\ 0, x=0 .\end{array}\right.$
6. $f(x)=\left[\begin{array}{l}\sqrt{1+\ln \left(1+x^{2} \sin \frac{1}{x}\right)}-1, x \neq 0, \\ 0, x=0 .\end{array}\right.$
7. $f(x)=\left[\begin{array}{l}\sin \left(e^{x^{2} \sin (5 / x)}-1\right), x \neq 0, \\ 0, \quad x=0 .\end{array}\right.$
8. $f(x)=\left[\begin{array}{l}x^{3} \cos \frac{4}{3 x}+3 x, \quad x \neq 0, \\ 0, \quad x=0 .\end{array}\right.$
9. $f(x)=\left[\begin{array}{l}\operatorname{arctg}\left(x-x^{2} \sin \frac{1}{3 x}\right), x \neq 0, \\ 0, x=0 .\end{array}\right.$
10. $f(x)=\left[\begin{array}{l}\sin ^{2} x \cos \frac{5}{x}+2 x, x \neq 0, \\ 0, x=0 .\end{array}\right.$

Answers. 1. 0. 2. 2. 3. Does not exist. 4. 0. 5. Does not exist. 6. 0. 7. 0. 8. 3. 9. 1. 10. 2.

## 4.2. Calculation of Derivatives

Problem Statement. Find the derivative of the function $y=f(x)$.

Solution Plan. The problem is solved in several stages. At each stage, it is necessary to recognize the type of function and apply the corresponding differentiation rule.

The following types of functions are possible.

- The function has the form $C_{1} u_{1}+C_{2} u_{2}+\ldots+C_{n} u_{n}$, where $u_{1}(x)$, $u_{2}(x), \ldots, u_{n}(x)$ are some functions and $C_{1}, C_{2}, \ldots, C_{n}$ are some constants. Use the formula for the derivative of a linear combination

$$
\left(C_{1} u_{1}+C_{2} u_{2}+\ldots+C_{n} u_{n}\right)^{\prime}=C_{1} u_{1}^{\prime}+C_{2} u_{2}^{\prime}+\ldots+C_{n} u_{n}^{\prime}
$$

- The function has the form $u \cdot v$. Use the formula for the derivative of a product

$$
(u \cdot v)^{\prime}=u^{\prime} \cdot v+u \cdot v^{\prime}
$$

- The function has the form $\frac{u}{v}$. Use the formula for the derivative of a quotient:

$$
\left(\frac{u}{v}\right)^{\prime}=\frac{u^{\prime} \cdot v-u \cdot v^{\prime}}{v^{2}}
$$

- The function has the form $u(v(x))$. Use the formula for the derivative of a composite function

$$
u(v(x))^{\prime}=u^{\prime}(v) \cdot v^{\prime}(x)
$$

- The function has the form $u(x)^{v(x)}$. The derivative of such a function is calculated using the formula

$$
u(x)^{v(x)}=e^{v \ln u}
$$

The transition from one stage to the next is made until each sign of the derivative contains a tabular function.