## Problem Statement

Find the differential $d y$.

$y=\sqrt{1+2 x}-\ln |x+\sqrt{1+2 x}|$
## Solution

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

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

## Problem Kuznetsov Differentiation 4-3