Question

Найти y' и y''
y^2-x=cosy

Answer 1

2y•y’ - 1 = -sin y•y’
y’ = 1/(2y + sin y)

y” = -1/(2y + sin y)^2•(2y’ + cos y•y’) = -y’•(2 + cos y)/(2y + sin y)^2 = -1/(2y + sin y)•(2 + cos y)/(2y + sin y)^2 = -(2 + cos y)/(2y + sin y)^3