Question

Найдите значение производной функции в указанной точке.

y = arccos³sinx, x0=π/6

Answer 1

Ответ:

$y' = 3 {arccos}^{2} ( \sin(x)) \times ( - \frac{1}{ \sqrt{1 - \sin^{2} (x) } } ) \times \cos(x) = \\ = - 3 {arccos}^{2} ( \sin(x)) \times \frac{1}{ \cos(x) } \times \cos(x) = \\ = - 3 {arccos}^{2} ( \sin(x)) \\ \\ y'( \frac{\pi}{6} ) = - 3 {arccos}^{2} ( \sin( \frac{\pi}{6} ) ) = - 3 {arccos}^{2} ( \frac{1}{2} ) = \\ = - 3 {( \frac{\pi}{3} )}^{2} = - 3 \times \frac{\pi}{9} = - \frac{\pi}{3}$