## Task Condition

Find the derivative.

$y=\frac{2 x-5}{4} \cdot \sqrt{5 x-4-x^{2}}+\frac{9}{4} \cdot \arcsin \sqrt{\frac{x-1}{3}}$
## Solution

$$
\begin{aligned}
& y^{\prime}=\left(\frac{2 x-5}{4} \cdot \sqrt{5 x-4-x^{2}}+\frac{9}{4} \cdot \arcsin \sqrt{\frac{x-1}{3}}\right)^{\prime}= \\
& =\frac{2}{4} \cdot \sqrt{5 x-4-x^{2}}+\frac{2 x-5}{4} \cdot \frac{1}{2 \sqrt{5 x-4-x^{2}}} \cdot(5-2 x)+\frac{9}{4} \cdot \frac{1}{\sqrt{1-\left(\sqrt{\frac{x-1}{3}}\right)^{2}}} \cdot \frac{1}{2 \sqrt{\frac{x-1}{3}}} \cdot \frac{1}{3}= \\
& =\frac{4\left(5 x-4-x^{2}\right)}{8 \sqrt{5 x-4-x^{2}}}-\frac{(5-2 x)^{2}}{8 \sqrt{5 x-4-x^{2}}}+\frac{9}{8 \sqrt{3-(x-1)}} \cdot \frac{1}{\sqrt{x-1}}= \\
& =\frac{20 x-16-4 x^{2}}{8 \sqrt{5 x-4-x^{2}}}-\frac{25-20 x+4 x^{2}}{8 \sqrt{5 x-4-x^{2}}}+\frac{9}{8 \sqrt{(4-x)(x-1)}}= \\
& =\frac{20 x-16-4 x^{2}-25+20 x-4 x^{2}}{8 \sqrt{5 x-4-x^{2}}}+\frac{9}{8 \sqrt{4 x-4-x^{2}+x}}= \\
& =\frac{40 x-41-8 x^{2}}{8 \sqrt{5 x-4-x^{2}}}+\frac{9}{8 \sqrt{4 x-4-x^{2}+x}}= \\
& =\frac{40 x-41-8 x^{2}}{8 \sqrt{5 x-4-x^{2}}}+\frac{9}{8 \sqrt{5 x-4-x^{2}}}= \\
& =\frac{40 x-32-8 x^{2}}{8 \sqrt{5 x-4-x^{2}}}=\frac{5 x-4-x^{2}}{\sqrt{5 x-4-x^{2}}}=\sqrt{5 x-4-x^{2}}
\end{aligned}
$$

Kuznetsov Differentiation Problem 10-20