Question

Найти производную функции y=$frac{1}{sin3x}$,y=$frac{tgx-1}{tgx}$

Answer 1

Решение
1)  y = 1/sin3x
y` =(sin^(-1)3x)` = -(sin^(-2)3x) * cos3x * 3 = - 3cos3x / sin²3x
2)  y = (tgx - 1)/tgx = 1 - 1/tgx = 1 - tg^(-1)x
y` = 1/tg²xcos²x) =  = 1/sin²x