## Task Condition

Find the derivative.
$y=2 \arcsin \frac{2}{3 x+1}+\sqrt{9 x^{2}+6 x-3}, 3 x+1>0$
## Solution

$$
\begin{aligned}
& y^{\prime}=\left(2 \arcsin \frac{2}{3 x+1}+\sqrt{9 x^{2}+6 x-3}\right)^{\prime}= \\
& =2 \cdot \frac{1}{\sqrt{1-\left(\frac{2}{3 x+1}\right)^{2}}} \cdot\left(-\frac{2}{(3 x+1)^{2}} \cdot 3\right)+\frac{1}{2 \sqrt{9 x^{2}+6 x-3}} \cdot(18 x+6)= \\
& =-\frac{2(3 x+1)}{\sqrt{(3 x+1)^{2}-2^{2}}} \cdot \frac{6}{(3 x+1)^{2}}+\frac{9 x+3}{\sqrt{9 x^{2}+6 x-3}}= \\
& =-\frac{12}{(3 x+1) \sqrt{9 x^{2}+6 x-3}}+\frac{3(3 x+1)}{\sqrt{9 x^{2}+6 x-3}}= \\
& =\frac{-12+3(3 x+1)^{2}}{(3 x+1) \sqrt{9 x^{2}+6 x-3}}=\frac{-12+27 x^{2}+18 x+3}{(3 x+1) \sqrt{9 x^{2}+6 x-3}}= \\
& =\frac{27 x^{2}+18 x-9}{(3 x+1) \sqrt{9 x^{2}+6 x-3}}=\frac{3\left(9 x^{2}+6 x-3\right)}{(3 x+1) \sqrt{9 x^{2}+6 x-3}}= \\
& =\frac{3 \sqrt{9 x^{2}+6 x-3}}{3 x+1}
\end{aligned}
$$

Problem Kuznetsov Differentiation 14-15