## Condition of the problem

Find the differential $d y$.

$y=\sqrt{x}-(1+x) \operatorname{arctg} \sqrt{x}$
## Solution

$d y=y^{\prime} \cdot d x=(\sqrt{x}-(1+x) \operatorname{arctg} \sqrt{x})^{\prime} d x=$

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

## Problem Kuznetsov Differentiation 4-29