## Condition of the problem

Find the differential $d y$

$$
y=x \sqrt{4-x^{2}}+a \cdot \arcsin \frac{x}{2}
$$
## Solution

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