## Task Condition

Find the derivative.

$$
y=\frac{1}{18 \sqrt{2}} \ln \frac{1+\sqrt{2} \operatorname{coth} x}{1-\sqrt{2} \operatorname{coth} x}
$$
## Solution

$y^{\prime}=\left(\frac{1}{18 \sqrt{2}} \ln \frac{1+\sqrt{2} \operatorname{cth} x}{1-\sqrt{2} \operatorname{cth} x}\right)^{\prime}=\frac{1}{18 \sqrt{2}} \cdot \frac{1-\sqrt{2} \operatorname{cth} x}{1+\sqrt{2} \operatorname{cth} x} \cdot\left(\frac{1+\sqrt{2} \operatorname{cth} x}{1-\sqrt{2} \operatorname{cth} x}\right)^{\prime}=$

$=\frac{1}{18 \sqrt{2}} \cdot \frac{1-\sqrt{2} \operatorname{cth} x}{1+\sqrt{2} \operatorname{cth} x} \cdot \frac{\sqrt{2} \cdot \frac{-1}{\operatorname{sh}^{2} x} \cdot(1-\sqrt{2} \operatorname{cth} x)-(1+\sqrt{2} \operatorname{cth} x) \cdot\left(-\sqrt{2} \cdot \frac{-1}{\operatorname{sh}^{2} x}\right)}{(1-\sqrt{2} \operatorname{cth} x)^{2}}=$

$=\frac{1}{18 \sqrt{2}} \cdot \frac{\frac{-2 \sqrt{2}}{\operatorname{sh}^{2} x}}{(1+\sqrt{2} \operatorname{cth} x) \cdot(1-\sqrt{2} \operatorname{cth} x)}=-\frac{1}{9} \cdot \frac{1}{\operatorname{sh}^{2} x \cdot\left(1-2 \cdot \operatorname{cth}^{2} x\right)}=$

$=-\frac{1}{9} \cdot \frac{1}{\operatorname{sh}^{2} x-2 \cdot \operatorname{ch}^{2} x}=\frac{1}{9 \cdot\left(1+\operatorname{ch}^{2} x\right)}$

## Problem Kuznetsov Differentiation 11-8