## Condition of the problem

Find the differential $d y$.

$$
y=\ln \left(x+\sqrt{1+x^{2}}\right)-\sqrt{1+x^{2}} \operatorname{arctg} x
$$
## Solution

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

## Problem Kuznetsov Differentiation 4-10