Question

y=ln sin^2(x-1) найти пооизводную сложной функции

Answer 1

Ответ:

$y = ln( \sin {}^{2} (x - 1) )$

$y' = \frac{1}{ \sin {}^{2} (x - 1) } \times ( \sin {}^{2} (x - 1)) ' \times (\sin(x - 1)) ) '= \\ = \frac{1}{ \sin {}^{2} (x - 1) } \times 2 \sin(x - 1) \cos(x - 1) = \\ = \frac{2 \cos(x - 1) }{ \sin(x - 1) } = 2ctg(x - 1)$