## Problem Statement

Find the derivative.

$$
y=\frac{1}{2} \cdot \sqrt{\frac{1}{x^{2}}-1}-\frac{\arccos x}{2 x^{2}}
$$
## Solution

$y^{\prime}=\left(\frac{1}{2} \cdot \sqrt{\frac{1}{x^{2}}-1}-\frac{\arccos x}{2 x^{2}}\right)^{\prime}=\left(\frac{1}{2} \cdot \frac{\sqrt{x^{2}-1}}{x}-\frac{\arccos x}{2 x^{2}}\right)^{\prime}=$

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

Problem Kuznetsov Differentiation 10-15