## Problem Statement

Find the derivative.

$$
y=\ln \sqrt[3]{\frac{x-1}{x+1}}-\frac{1}{2}\left(\frac{1}{2}+\frac{1}{x^{2}-1}\right) \operatorname{arctan} x
$$
## Solution

$$
\begin{aligned}
& y^{\prime}=\left(\ln \sqrt[3]{\frac{x-1}{x+1}}-\frac{1}{2}\left(\frac{1}{2}+\frac{1}{x^{2}-1}\right) \operatorname{arctg} x\right)^{\prime}= \\
& =\sqrt[3]{\frac{x+1}{x-1}} \cdot \frac{1}{3} \cdot\left(\frac{x-1}{x+1}\right)^{-\frac{2}{3}} \cdot \frac{1 \cdot(x+1)-(x-1) \cdot 1}{(x+1)^{2}}- \\
& -\frac{1}{2} \cdot\left(-\frac{1}{\left(x^{2}-1\right)^{2}} \cdot 2 x \cdot \operatorname{arctg} x+\left(\frac{1}{2}+\frac{1}{x^{2}-1}\right) \cdot \frac{1}{1+x^{2}}\right)= \\
& =\sqrt[3]{\frac{x+1}{x-1}} \cdot \frac{1}{3} \cdot\left(\frac{x+1}{x-1}\right)^{\frac{2}{3}} \cdot \frac{2}{(x+1)^{2}}+\frac{x \cdot \operatorname{arctg} x}{\left(x^{2}-1\right)^{2}}-\frac{1}{4\left(1+x^{2}\right)}-\frac{1}{2\left(x^{4}-1\right)}= \\
& =\frac{1}{3} \cdot \frac{x+1}{x-1} \cdot \frac{2}{(x+1)^{2}}+\frac{x \cdot \operatorname{arctg} x}{\left(x^{2}-1\right)^{2}}-\frac{x^{2}-1}{4\left(x^{4}-1\right)}-\frac{2}{4\left(x^{4}-1\right)}= \\
& =\frac{1}{3} \cdot \frac{2}{x^{2}-1}+\frac{x \cdot \operatorname{arctg} x}{\left(x^{2}-1\right)^{2}}-\frac{x^{2}-3}{4\left(x^{4}-1\right)}= \\
& =\frac{8\left(x^{2}+1\right)}{12\left(x^{4}-1\right)}+\frac{x \cdot \operatorname{arctg} x}{\left(x^{2}-1\right)^{2}}-\frac{3 x^{2}-9}{12\left(x^{4}-1\right)}= \\
& =\frac{8 x^{2}+8-3 x^{2}+9}{12\left(x^{4}-1\right)}+\frac{x \cdot \operatorname{arctg} x}{\left(x^{2}-1\right)^{2}}=\frac{5 x^{2}+17}{12\left(x^{4}-1\right)}+\frac{x \cdot \operatorname{arctg} x}{\left(x^{2}-1\right)^{2}}
\end{aligned}
$$

## Problem Kuznetsov Differentiation 14-19