6) $$y = \ln \sqrt[5]{\frac{1-x^2}{1+x^2}}$$;

6) $$y = \ln \sqrt[5]{\frac{1-x^2}{1+x^2}}$$;

Преобразуем функцию, используя свойства логарифмов:

$$y = \ln \left(\frac{1-x^2}{1+x^2}\right)^{\frac{1}{5}} = \frac{1}{5} \ln \left(\frac{1-x^2}{1+x^2}\right) = \frac{1}{5} \left[\ln(1-x^2) - \ln(1+x^2)\right]$$

Теперь найдем производную:

$$y' = \frac{1}{5} \left[\frac{(1-x^2)'}{1-x^2} - \frac{(1+x^2)'}{1+x^2}\right] = \frac{1}{5} \left[\frac{-2x}{1-x^2} - \frac{2x}{1+x^2}\right]$$

$$y' = \frac{1}{5} \cdot (-2x) \left[\frac{1}{1-x^2} + \frac{1}{1+x^2}\right] = -\frac{2x}{5} \left[\frac{1+x^2 + 1-x^2}{(1-x^2)(1+x^2)}\right] = -\frac{2x}{5} \cdot \frac{2}{1-x^4}$$

$$y' = -\frac{4x}{5(1-x^4)}$$

Ответ: $$y' = -\frac{4x}{5(1-x^4)}$$