## Condition of the problem

Find the differential $d y$.
$y=x \cdot \operatorname{arctg} x-\ln \sqrt{1+x^{2}}$
## Solution

$$
\begin{aligned}
& d y=y^{\prime} \cdot d x=\left(x \cdot \operatorname{arctg} x-\ln \sqrt{1+x^{2}}\right)^{\prime} d x= \\
& =\left(\operatorname{arctg} x+x \cdot \frac{1}{1+x^{2}}-\frac{1}{\sqrt{1+x^{2}}} \cdot \frac{1}{2 \sqrt{1+x^{2}}} \cdot 2 x\right) d x= \\
& =\left(\operatorname{arctg} x+\frac{x}{1+x^{2}}-\frac{x}{1+x^{2}}\right) d x=\operatorname{arctg} x \cdot d x
\end{aligned}
$$

Kuznetsov Differentiation 4-30