Question

ПОМОГИТЕ ПОЖАЛУЙСТА!!! Нужно найти производную:
$y = ln\sqrt[3]{ \sin2x }$

Answer 1

$y = ln\sqrt[3]{ \sin2x} \\ y' = \frac{1}{\sqrt[3]{ \sin2x}} \times (\sqrt[3]{ \sin2x})' = \\ = \frac{1}{\sqrt[3]{ \sin2x}} \times {(sin2x)}^{ \frac{1}{3} } )' = \\ = \frac{1}{\sqrt[3]{ \sin2x}} \times \frac{1}{3} {(sin2x)}^{ \frac{1}{3} - 1} = \\ = \frac{1}{\sqrt[3]{ \sin2x}} \times \frac{1}{3} {(sin2x)}^{ - \frac{2}{3}} \times (sin2x)' = \\ = \frac{1}{\sqrt[3]{ \sin2x}} \times \frac{1}{3} {(sin2x)}^{ - \frac{2}{3}} \times 2cos2x = \\ = \frac{2}{3} cos2x \times\frac{1}{\sqrt[3]{ \sin2x}} \times \frac{1}{\sqrt[3]{ {(sin2x)}^{2} }} = \\ = \frac{2cos2x}{3sin2x} = \frac{2}{3} ctg2x$