## Task Condition

Find the derivative.

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

$y^{\prime}=\left(\ln \sqrt[4]{\frac{1+2 x}{1-2 x}}\right)^{\prime}=\sqrt[4]{\frac{1-2 x}{1+2 x}} \cdot\left(\sqrt[4]{\frac{1+2 x}{1-2 x}}\right)^{\prime}=$

$$
\begin{aligned}
& =\sqrt[4]{\frac{1-2 x}{1+2 x}} \cdot \frac{1}{4} \cdot \sqrt[4]{\left(\frac{1-2 x}{1+2 x}\right)^{3}} \cdot\left(\frac{1+2 x}{1-2 x}\right)^{\prime}= \\
& =\frac{1}{4} \cdot \frac{1-2 x}{1+2 x} \cdot \frac{2 \cdot(1-2 x)-(1+2 x) \cdot(-2)}{(1-2 x)^{2}}= \\
& =\frac{1}{4(1+2 x)} \cdot \frac{2-4 x+2+4 x}{1-2 x}=\frac{1}{1+2 x} \cdot \frac{1}{1-2 x}=\frac{1}{1-4 x^{2}}
\end{aligned}
$$

## Problem Kuznetsov Differentiation 8-11