## Problem Statement

Find the derivative.

$y=\frac{1}{2} \cdot \ln \frac{1+\sqrt{\tanh x}}{1-\sqrt{\tanh x}}-\arctan \sqrt{\tanh x}$
## Solution

$$
\begin{aligned}
& y^{\prime}=\left(\frac{1}{2} \cdot \ln \frac{1+\sqrt{\tanh x}}{1-\sqrt{\tanh x}}-\arctan \sqrt{\tanh x}\right)^{\prime}= \\
& =\frac{1}{2} \cdot \frac{1-\sqrt{\tanh x}}{1+\sqrt{\tanh x}} \cdot \frac{\frac{1}{2 \sqrt{\tanh x}} \cdot \frac{1}{\cosh^{2} x} \cdot(1-\sqrt{\tanh x})-(1+\sqrt{\tanh x}) \cdot\left(-\frac{1}{2 \sqrt{\tanh x}} \cdot \frac{1}{\cosh^{2} x}\right)}{(1-\sqrt{\tanh x})^{2}}- \\
& -\frac{1}{1+(\sqrt{\tanh x})^{2}} \cdot \frac{1}{2 \sqrt{\tanh x}} \cdot \frac{1}{\cosh^{2} x}= \\
& =\frac{\frac{1}{2 \sqrt{\tanh x}} \cdot \frac{1}{\cosh^{2} x} \cdot(1-\sqrt{\tanh x}+1+\sqrt{\tanh x})}{2(1+\sqrt{\tanh x}) \cdot(1-\sqrt{\tanh x})}-\frac{1}{1+\tanh x} \cdot \frac{1}{2 \sqrt{\tanh x}} \cdot \frac{1}{\cosh^{2} x}= \\
& =\frac{\frac{1}{\sqrt{\tanh x}} \cdot \frac{1}{\cosh^{2} x}}{2(1-\tanh x)}-\frac{1}{1+\tanh x} \cdot \frac{1}{2 \sqrt{\tanh x}} \cdot \frac{1}{\cosh^{2} x}= \\
& =\frac{1+\tanh x}{2 \sqrt{\tanh x} \cdot \cosh^{2} x \cdot(1-\tanh x) \cdot(1+\tanh x)}-\frac{1-\tanh x}{2 \sqrt{\tanh x} \cdot \cosh^{2} x \cdot(1-\tanh x)(1+\tanh x)}= \\
& =\frac{1+\tanh x-1+\tanh x}{2 \sqrt{\tanh x} \cdot \cosh^{2} x \cdot\left(1-\tanh^{2} x\right)}=\frac{2 \tanh x}{2 \sqrt{\tanh x} \cdot\left(\cosh^{2} x-\sinh^{2} x\right)}= \\
& =\frac{\tanh x}{\sqrt{\tanh x} \cdot 1}=\sqrt{\tanh x}
\end{aligned}
$$

## Problem Kuznetsov Differentiation 11-3