6.14. Find the derivative of the function

$$
f(x)=\left\{\begin{array}{l}
\operatorname{arctg}\left(x^{2} \cdot \sin (1 /(5 x))\right), \quad x \neq 0, \\
0, x=0
\end{array}\right.
$$

and compute its value at the point $x_{0}=0$.
Solution. Let's find the derivatives for $x \neq 0$. We have

$$
\begin{gathered}
f^{\prime}(x)=\frac{1}{1+\left(x^{2} \cdot \sin (1 / 5 x)\right)^{2}} \cdot\left(x^{2} \cdot \sin (1 /(5 x))\right)^{\prime}= \\
=\frac{1}{1+\left(x^{2} \cdot \sin (1 /(5 x))\right)^{2}} \cdot\left(\left(x^{2}\right)^{\prime} \cdot \sin (1 /(5 x))+x^{2} \cdot(\sin (1 /(5 x)))^{\prime}\right)= \\
=\frac{1}{1+\left(x^{2} \cdot \sin (1 /(5 x))\right)^{2}} \cdot\left(2 x \cdot \sin (1 /(5 x))+x^{2} \cdot \cos (1 /(5 x)) \cdot\left(-1 /\left(5 x^{2}\right)\right)\right)= \\
=\frac{1}{1+\left(x^{2} \cdot \sin (1 /(5 x))\right)^{2}} \cdot\left(2 x \cdot \sin (1 /(5 x))-\frac{\cos (1 /(5 x))}{5}\right), x \neq 0 .
\end{gathered}
$$

Since the one-sided limits

$$
\lim _{x \rightarrow 0 \pm 0} \sin (1 / 5 x)
$$

do not exist, the one-sided limits of $f^{\prime}(x)$ at $x_{0}=0$ do not exist either.

To find the derivative at the given point, we will use the definition of the derivative.

$$
f^{\prime}(0)=\lim _{\Delta x \rightarrow 0} \frac{\operatorname{arctg}\left((0+\Delta x)^{2} \cdot \sin \left(\frac{1}{5(0+\Delta x)}\right)\right)-0}{\Delta x}
$$

Since the limit of the product of the bounded function $\sin (1 /(5 \Delta x))$ and the infinitesimal $\Delta x$ is zero, we get that $y^{\prime}(0)=0$.

Note that the original function $f(x)$ is continuous at the point $x_{0}=0$. Indeed, the function is defined both at the point $x_{0}=0$ and in some neighborhood of it, and

$$
\lim _{x \rightarrow 0} \operatorname{arctg}(x \cdot \sin (1 /(5 x)))=0=f(0)
$$