Question

Как вычислить производную функции в квадрате?

Answer 1

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