## Task Condition

Find the derivative.

$y=\arcsin \frac{1}{2 x+3}+2 \sqrt{x^{2}+3 x+2}, 2 x+3>0$
## Solution

$y^{\prime}=\left(\arcsin \frac{1}{2 x+3}+2 \sqrt{x^{2}+3 x+2}\right)^{\prime}=$

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

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

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

$$
\begin{aligned}
& =\frac{-1}{(2 x+3) \cdot \sqrt{x^{2}+x+2}}+\frac{4 x^{2}+12 x+9}{(2 x+3) \cdot \sqrt{x^{2}+3 x+2}}= \\
& =\frac{4 x^{2}+12 x+8}{(2 x+3) \cdot \sqrt{x^{2}+3 x+2}}=\frac{4\left(x^{2}+3 x+2\right)}{(2 x+3) \cdot \sqrt{x^{2}+3 x+2}}= \\
& =\frac{4 \sqrt{x^{2}+3 x+2}}{2 x+3}
\end{aligned}
$$

## Problem Kuznetsov Differentiation 14-29