## Problem Statement

Find the derivative.

$y=\frac{1}{2} \operatorname{th} x+\frac{1}{4 \sqrt{2}} \ln \frac{1+\sqrt{2} \operatorname{th} x}{1-\sqrt{2} \operatorname{th} x}$
## Solution

$$
\begin{aligned}
& y^{\prime}=\left(\frac{1}{2} \tanh x+\frac{1}{4 \sqrt{2}} \ln \frac{1+\sqrt{2} \tanh x}{1-\sqrt{2} \tanh x}\right)^{\prime}= \\
& =\frac{1}{2} \cdot \frac{1}{\cosh^{2} x}+\frac{1}{4 \sqrt{2}} \cdot \frac{1-\sqrt{2} \tanh x}{1+\sqrt{2} \tanh x} \cdot\left(\frac{1+\sqrt{2} \tanh x}{1-\sqrt{2} \tanh x}\right)^{\prime}= \\
& =\frac{1}{2 \cosh^{2} x}+\frac{1}{4 \sqrt{2}} \cdot \frac{1-\sqrt{2} \tanh x}{1+\sqrt{2} \tanh x} \cdot \frac{\sqrt{2} \cdot \frac{1}{\cosh^{2} x} \cdot(1-\sqrt{2} \tanh x)-(1+\sqrt{2} \tanh x) \cdot\left(-\sqrt{2} \cdot \frac{1}{\cosh^{2} x}\right)}{(1-\sqrt{2} \tanh x)^{2}}= \\
& =\frac{1}{2 \cosh^{2} x}+\frac{1}{4} \cdot \frac{1}{1+\sqrt{2} \tanh x} \cdot \frac{(1-\sqrt{2} \tanh x)+(1+\sqrt{2} \tanh x)}{(1-\sqrt{2} \tanh x) \cdot \cosh^{2} x}= \\
& =\frac{1}{2 \cosh^{2} x}+\frac{1}{2\left(1-2 \tanh^{2} x\right) \cdot \cosh^{2} x}=\frac{1}{2 \cosh^{2} x}+\frac{1}{2\left(\cosh^{2} x-2 \sinh^{2} x\right)}= \\
& =\frac{1}{2 \cosh^{2} x}+\frac{1}{2\left(1-\sinh^{2} x\right)}=\frac{1-\sinh^{2} x}{2 \cosh^{2} x\left(1-\sinh^{2} x\right)}+\frac{\cosh^{2} x}{2 \cosh^{2} x\left(1-\sinh^{2} x\right)}= \\
& =\frac{1-\sinh^{2} x+\cosh^{2} x}{2 \cosh^{2} x\left(1-\sinh^{2} x\right)}=\frac{2}{2 \cosh^{2} x\left(1-\sinh^{2} x\right)}=\frac{1}{\cosh^{2} x\left(1-\sinh^{2} x\right)}
\end{aligned}
$$

## Problem Kuznetsov Differentiation 11-5