## Task Condition

Find the derivative.

$y=\sqrt{(3-x)(2+x)}+5 \arcsin \sqrt{\frac{x+2}{5}}$
## Solution

$$
\begin{aligned}
& y^{\prime}=\left(\sqrt{(3-x)(2+x)}+5 \arcsin \sqrt{\frac{x+2}{5}}\right)^{\prime}= \\
& =\frac{1}{2 \sqrt{(3-x)(2+x)}} \cdot(-1 \cdot(2+x)+(3-x) \cdot 1)+5 \cdot \frac{1}{\sqrt{1-\left(\sqrt{\frac{x+2}{5}}\right)^{2}}} \cdot \frac{1}{2 \sqrt{\frac{x+2}{5}}} \cdot \frac{1}{5}=
\end{aligned}
$$

$$
\begin{aligned}
& =\frac{1-2 x}{2 \sqrt{(3-x)(2+x)}}+\frac{\sqrt{5}}{\sqrt{5-(x+2)}} \cdot \frac{\sqrt{5}}{2 \sqrt{x+2}}= \\
& =\frac{1-2 x}{2 \sqrt{(3-x)(2+x)}}+\frac{5}{\sqrt{3-x} \cdot 2 \sqrt{x+2}}= \\
& =\frac{6-2 x}{2 \sqrt{(3-x)(2+x)}}=\frac{3-x}{\sqrt{(3-x)(2+x)}}=\sqrt{\frac{3-x}{2+x}}
\end{aligned}
$$

## Problem Kuznetsov Differentiation 14-23