## Task Condition

Find the derivative.

$$
y=4 \arcsin \frac{4}{2 x+3}+\sqrt{4 x^{2}+12 x-7}, 2 x+3>0
$$
## Solution

$$
\begin{aligned}
& y^{\prime}=\left(4 \arcsin \frac{4}{2 x+3}+\sqrt{4 x^{2}+12 x-7}\right)^{\prime}= \\
& =4 \cdot \frac{1}{\sqrt{1-\left(\frac{4}{2 x+3}\right)^{2}}} \cdot\left(-\frac{4}{(2 x+3)^{2}} \cdot 2\right)+\frac{1}{2 \sqrt{4 x^{2}+12 x-7}} \cdot(8 x+12)= \\
& =-\frac{4(2 x+3)}{\sqrt{(2 x+3)^{2}-4^{2}}} \cdot \frac{8}{(2 x+3)^{2}}+\frac{2(2 x+3)}{\sqrt{4 x^{2}+12 x-7}}= \\
& =-\frac{32}{(2 x+3) \sqrt{4 x^{2}+12 x-7}}+\frac{2(2 x+3)}{\sqrt{4 x^{2}+12 x-7}}= \\
& =\frac{-32+2(2 x+3)^{2}}{(2 x+3) \sqrt{4 x^{2}+12 x-7}}=\frac{-32+8 x^{2}+24 x+18}{(2 x+3) \sqrt{4 x^{2}+12 x-7}}= \\
& =\frac{8 x^{2}+24 x-14}{(2 x+3) \sqrt{4 x^{2}+12 x-7}}=\frac{2\left(4 x^{2}+12 x-7\right)}{(2 x+3) \sqrt{4 x^{2}+12 x-7}}=
\end{aligned}
$$

$$
=\frac{2 \sqrt{4 x^{2}+12 x-7}}{2 x+3}
$$

## Problem Kuznetsov Differentiation 14-14