## Problem Statement

Find the derivative.

$$
y=\ln \frac{1+2 \sqrt{-x-x^{2}}}{2 x+1}+\frac{4}{2 x+1} \sqrt{-x-x^{2}}
$$
## Solution

$$
\begin{aligned}
& y^{\prime}=\left(\ln \frac{1+2 \sqrt{-x-x^{2}}}{2 x+1}+\frac{4}{2 x+1} \sqrt{-x-x^{2}}\right)^{\prime}= \\
& =\frac{2 x+1}{1+2 \sqrt{-x-x^{2}}} \cdot \frac{\frac{1}{\sqrt{-x-x^{2}}} \cdot(-1-2 x) \cdot(2 x+1)-\left(1+2 \sqrt{-x-x^{2}}\right) \cdot 2}{(2 x+1)^{2}}+ \\
& +\frac{4 \cdot \frac{1}{2 \sqrt{-x-x^{2}}} \cdot(-1-2 x) \cdot(2 x+1)-4 \sqrt{-x-x^{2}} \cdot 2}{(2 x+1)^{2}}= \\
& =\frac{2 x+1}{1+2 \sqrt{-x-x^{2}}} \cdot \frac{-(2 x+1)^{2}-2 \sqrt{-x-x^{2}}-4 \cdot\left(-x-x^{2}\right)}{\sqrt{-x-x^{2}} \cdot(2 x+1)^{2}}+ \\
& +\frac{-2 \cdot(2 x+1)^{2}-8 \cdot\left(-x-x^{2}\right)}{\sqrt{-x-x^{2}} \cdot(2 x+1)^{2}}= \\
& =\frac{2 x+1}{1+2 \sqrt{-x-x^{2}}} \cdot \frac{-4 x^{2}-4 x-1-2 \sqrt{-x-x^{2}}+4 x+4 x^{2}}{\sqrt{-x-x^{2}} \cdot(2 x+1)^{2}}+ \\
& +\frac{-8 x^{2}-8 x-2+8 x+8 x^{2}}{\sqrt{-x-x^{2}} \cdot(2 x+1)^{2}}= \\
& =\frac{2 x+1}{1+2 \sqrt{-x-x^{2}}} \cdot \frac{-\left(1+2 \sqrt{-x-x^{2}}\right)}{\sqrt{-x-x^{2}} \cdot(2 x+1)^{2}}+\frac{2}{\sqrt{-x-x^{2}} \cdot(2 x+1)^{2}}= \\
& =\frac{-2 x-1}{\sqrt{-x-x^{2}} \cdot(2 x+1)^{2}}-\frac{-2}{\sqrt{-x-x^{2}} \cdot(2 x+1)^{2}}= \\
& =-\frac{2 x+3}{\sqrt{-x-x^{2}} \cdot(2 x+1)^{2}}
\end{aligned}
$$

## Problem Kuznetsov Differentiation 13-10