## Task Condition

Find the derivative.

$$
y=\ln \frac{1+\sqrt{-3-4 x-x^{2}}}{-x-2}-\frac{2}{x+2} \sqrt{-3-4 x-x^{2}}
$$
## Solution

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

$-2 \cdot \frac{\frac{1}{2 \sqrt{-3-4 x-x^{2}}} \cdot(-4-2 x) \cdot(x+2)-\sqrt{-3-4 x-x^{2}} \cdot 1}{(x+2)^{2}}=$

$=\frac{x+2}{1+\sqrt{-3-4 x-x^{2}}} \cdot \frac{(x+2)^{2}-\sqrt{-3-4 x-x^{2}}-\left(-3-4 x-x^{2}\right)}{(x+2)^{2} \cdot \sqrt{-3-4 x-x^{2}}}+$

$+2 \cdot \frac{(x+2)^{2}+\left(-3-4 x-x^{2}\right)}{(x+2)^{2} \cdot \sqrt{-3-4 x-x^{2}}}=$

$=\frac{1}{1+\sqrt{-3-4 x-x^{2}}} \cdot \frac{2 x^{2}+8 x+7-\sqrt{-3-4 x-x^{2}}}{(x+2) \cdot \sqrt{-3-4 x-x^{2}}}+$

$+\frac{2}{(x+2)^{2} \cdot \sqrt{-3-4 x-x^{2}}}=$

$=\frac{2 x^{2}+8 x+7-\sqrt{-3-4 x-x^{2}}+2+2 \sqrt{-3-4 x-x^{2}}}{(x+2) \cdot \sqrt{-3-4 x-x^{2}} \cdot\left(1+\sqrt{-3-4 x-x^{2}}\right)}=$

$=\frac{2 x^{2}+8 x+9+\sqrt{-3-4 x-x^{2}}}{(x+2) \cdot \sqrt{-3-4 x-x^{2}} \cdot\left(1+\sqrt{-3-4 x-x^{2}}\right)}
$$

## Problem Kuznetsov Differentiation 13-20