Question

Помогите решить:
y": y=cos(4x^2-1)

Answer 1

y' = -sin(4x^2 -1) * (8x) = - 8x*sin(4x^2 -1)
y'' = (-8x*sin(4x^2 -1))' = -8*sin(4x^2 -1) - 8x*8x*cos(4x^2 -1)