Функции многих переменных Задача №2

Решите задачу:

$z=cos(y^2x^3-9) \\\\ \frac{dz}{dx} =-sin(y^2x^3-9)*(y^2x^3-9)'_x=-sin(y^2x^3-9)*(3x^2y^2)= \\ =-3x^2y^2sin(y^2x^3-9) \\\\ \frac{d^2z}{dxdy} =(-3x^2y^2sin(y^2x^3-9))'_y=-3x^2(y^2sin(y^2x^3-9))'_y= \\ =-3x^2(2ysin(y^2x^3-9)+y^2cos(y^2x^3-9)*2yx^3)= \\ =-3x^2(2ysin(y^2x^3-9)+2x^3y^3cos(y^2x^3-9))= \\ =\boxed{-6x^2y(sin(y^2x^3-9)+x^3y^2cos(y^2x^3-9))}$
$\frac{d^2z}{dxdy} (1;3)=-6*1^2*3(sin(3^2*1^3-9)+1^3*3^2cos(3^2*1^3-9))= \\ =-6*3(sin(3^2-9)+3^2cos(3^2-9))= \\ =-18(sin(9-9)+9cos(9-9))=-18(sin0+9cos0)= \\ -18(0+9)=-162$