## Task Condition

Find the derivative.

$$
y=\frac{1}{4 \sqrt{5}} \ln \frac{2+\sqrt{5} \tanh x}{2-\sqrt{5} \cdot \tanh x}
$$
## Solution

$$
\begin{aligned}
& y^{\prime}=\left(\frac{1}{4 \sqrt{5}} \ln \frac{2+\sqrt{5} \tanh x}{2-\sqrt{5} \cdot \tanh x}\right)^{\prime}=\frac{1}{4 \sqrt{5}} \cdot \frac{2-\sqrt{5} \tanh x}{2+\sqrt{5} \cdot \tanh x} \cdot\left(\frac{2+\sqrt{5} \tanh x}{2-\sqrt{5} \cdot \tanh x}\right)^{\prime}= \\
& =\frac{1}{4 \sqrt{5}} \cdot \frac{2-\sqrt{5} \tanh x}{2+\sqrt{5} \cdot \tanh x} \cdot \frac{\frac{\sqrt{5}}{\cosh^{2} x} \cdot(2-\sqrt{5} \cdot \tanh x)-(2+\sqrt{5} \tanh x) \cdot\left(-\frac{\sqrt{5}}{\cosh^{2} x}\right)}{(2-\sqrt{5} \cdot \tanh x)^{2}}= \\
& =\frac{1}{4 \sqrt{5}} \cdot \frac{1}{2+\sqrt{5} \cdot \tanh x} \cdot \frac{\frac{4 \sqrt{5}}{\cosh^{2} x}}{2-\sqrt{5} \cdot \tanh x}=\frac{1}{\left(4-5 \tanh^{2} x\right) \cosh^{2} x}= \\
& =\frac{1}{4 \cosh^{2} x-5 \sinh^{2} x}=\frac{1}{4-\sinh^{2} x}
\end{aligned}
$$

Find the derivative.

$y=(\arctan x)^{\frac{1}{2} \cdot \ln (\arctan x)}$

## Solution

$y=(\arctan x)^{\frac{1}{2} \cdot \ln (\arctan x)}$

$\ln y=\ln (\arctan x)^{\frac{1}{2} \cdot \ln (\arctan x)}=\frac{1}{2} \cdot \ln (\arctan x) \cdot \ln (\arctan x)=\frac{1}{2} \cdot \ln ^{2}(\arctan x)$

$\frac{y^{\prime}}{y}=\left(\frac{1}{2} \cdot \ln ^{2}(\arctan x)\right)^{\prime}=\frac{1}{2} \cdot 2 \ln (\arctan x) \cdot \frac{1}{\arctan x} \cdot \frac{1}{1+x^{2}}=$

$=\frac{\ln (\arctan x)}{\arctan x \cdot\left(1+x^{2}\right)}$

$y^{\prime}=y \cdot \frac{\ln (\arctan x)}{\arctan x \cdot\left(1+x^{2}\right)}=(\arctan x)^{\frac{1}{2} \cdot \ln (\arctan x)} \cdot \frac{\ln (\arctan x)}{\arctan x \cdot\left(1+x^{2}\right)}$

## Problem Kuznetsov Differentiation 12-1