## Task Condition

Find the derivative.

$$
y=\ln (\sqrt{x}+\sqrt{x+1})
$$
## Solution

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

## Problem Kuznetsov Differentiation 8-5