## Problem Statement

Find the derivative.

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

$$
\begin{aligned}
& y^{\prime}=\left(\frac{1}{2 a \sqrt{1+a^{2}}} \ln \frac{a+\sqrt{1+a^{2}} \operatorname{th} x}{a-\sqrt{1+a^{2}} \operatorname{th} x}\right)^{\prime}= \\
& =\frac{1}{2 a \sqrt{1+a^{2}}} \cdot \frac{a-\sqrt{1+a^{2}} \operatorname{th} x}{a+\sqrt{1+a^{2}} \operatorname{th} x} \cdot\left(\frac{a+\sqrt{1+a^{2}} \operatorname{th} x}{a-\sqrt{1+a^{2}} \operatorname{th} x}\right)^{\prime}= \\
& =\frac{1}{2 a \sqrt{1+a^{2}}} \cdot \frac{a-\sqrt{1+a^{2}} \operatorname{th} x}{a+\sqrt{1+a^{2}} \operatorname{th} x} \times \\
& \times \frac{\sqrt{1+a^{2}} \cdot \frac{1}{\mathrm{ch}^{2} x} \cdot\left(a-\sqrt{1+a^{2}} \operatorname{th} x\right)-\left(a+\sqrt{1+a^{2}} \operatorname{th} x\right) \cdot\left(-\sqrt{1+a^{2}} \cdot \frac{1}{\mathrm{ch}^{2} x}\right)}{\left(a-\sqrt{1+a^{2}} \operatorname{th} x\right)^{2}}= \\
& =\frac{1}{2 a \sqrt{1+a^{2}}} \cdot \frac{\sqrt{1+a^{2}} \cdot \frac{1}{\mathrm{ch}^{2} x} \cdot\left(a-\sqrt{1+a^{2}} \operatorname{th} x+a+\sqrt{1+a^{2}} \operatorname{th} x\right)}{\left(a+\sqrt{1+a^{2}} \operatorname{th} x\right)\left(a-\sqrt{1+a^{2}} \operatorname{th} x\right)}= \\
& =\frac{1}{2 a} \cdot \frac{\frac{1}{\operatorname{ch}^{2} x} \cdot 2 a}{a^{2}+\left(1+a^{2}\right) \cdot \operatorname{th}^{2} x}=\frac{1}{\operatorname{ch}^{2} x\left(a^{2}+\left(1+a^{2}\right) \cdot \operatorname{th}^{2} x\right)}= \\
& =\frac{1}{a^{2} \cdot \operatorname{ch}^{2} x+\left(1+a^{2}\right) \cdot \operatorname{sh}^{2} x}
\end{aligned}
$$

## Problem Kuznetsov Differentiation 11-7