Problem condition

Find the derivative.
$y=6 \arcsin \frac{\sqrt{x}}{2}-\frac{6+x}{2} \cdot \sqrt{x(4-x)}$
## Solution

$$
\begin{aligned}
& y^{\prime}=\left(6 \arcsin \frac{\sqrt{x}}{2}-\frac{6+x}{2} \cdot \sqrt{x(4-x)}\right)^{\prime}= \\
& =6 \cdot \frac{1}{\sqrt{1-\left(\frac{\sqrt{x}}{2}\right)^{2}}} \cdot \frac{1}{4 \sqrt{x}}-\left(\frac{1}{2} \cdot \sqrt{x(4-x)}+\frac{6+x}{2} \cdot \frac{1}{2 \sqrt{x(4-x)}} \cdot(4-2 x)\right)= \\
& =6 \cdot \frac{2}{\sqrt{4-x}} \cdot \frac{1}{4 \sqrt{x}}-\left(\frac{1}{2} \cdot \frac{x(4-x)}{\sqrt{x(4-x)}}+\frac{6+x}{2} \cdot \frac{2-x}{\sqrt{x(4-x)}}\right)= \\
& =\frac{6}{2 \sqrt{x(4-x)}}-\frac{4 x-x^{2}}{2 \sqrt{x(4-x)}}-\frac{12-6 x+2 x-x^{2}}{2 \sqrt{x(4-x)}}= \\
& =\frac{6-4 x+x^{2}-12+6 x-2 x+x^{2}}{2 \sqrt{x(4-x)}}=\frac{2 x^{2}-6}{2 \sqrt{x(4-x)}}=\frac{x^{2}-3}{\sqrt{x(4-x)}}
\end{aligned}
$$

## Problem Kuznetsov Differentiation 10-16