## Task Condition

Find the derivative.

$$
y=\frac{1}{2} \cdot(x-4) \sqrt{8 x-x^{2}-7}-9 \arccos \sqrt{\frac{x-1}{6}}
$$
## Solution

$$
\begin{aligned}
& y^{\prime}=\left(\frac{1}{2} \cdot(x-4) \sqrt{8 x-x^{2}-7}-9 \arccos \sqrt{\frac{x-1}{6}}\right)^{\prime}= \\
& =\frac{1}{2} \cdot\left(\sqrt{8 x-x^{2}-7}+(x-4) \cdot \frac{1}{2 \sqrt{8 x-x^{2}-7}} \cdot(8-2 x)\right)-9 \cdot \frac{-1}{\sqrt{1-\frac{x-1}{6}}} \cdot \frac{1}{2 \sqrt{\frac{x-1}{6}}} \cdot \frac{1}{6}= \\
& =\frac{1}{2} \cdot\left(\sqrt{8 x-x^{2}-7}-\frac{(x-4)^{2}}{\sqrt{8 x-x^{2}-7}}\right)+\frac{9}{2 \sqrt{(7-x)(x-1)}}= \\
& =\frac{1}{2} \cdot \sqrt{8 x-x^{2}-7}-\frac{(x-4)^{2}}{2 \sqrt{8 x-x^{2}-7}}+\frac{9}{2 \sqrt{8 x-x^{2}-7}}= \\
& =\frac{1}{2} \cdot \sqrt{8 x-x^{2}-7}+\frac{-x^{2}+8 x-16+9}{2 \sqrt{8 x-x^{2}-7}}=
\end{aligned}
$$

$=\frac{1}{2} \cdot \sqrt{8 x-x^{2}-7}+\frac{\sqrt{8 x-x^{2}-7}}{2}=\sqrt{8 x-x^{2}-7}$

## Problem Kuznetsov Differentiation 10-8