Condition of the problem

Find the derivative.

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

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

## Problem Kuznetsov Differentiation 8-1