## Task Condition

Find the differential $d y$.

$$
y=\ln \left|\frac{x+\sqrt{x^{2}+1}}{2 x}\right|
$$
## Solution

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

## Problem Kuznetsov Differentiation 4-17