Задача 72555 найти дефференциал первого и второго...
Условие
Решение
Дифференциал 1 порядка.
[m]\frac{dz}{dx} = -\frac{1}{\sin^2(x^3y)} \cdot 3x^2y = -\frac{3x^2y}{\sin^2(x^3y)}[/m]
[m]\frac{dz}{dy} = -\frac{1}{\sin^2(x^3y)} \cdot x^3 = -\frac{x^3}{\sin^2(x^3y)}[/m]
Полный дифференциал 1 порядка:
[m]dz = -\frac{3x^2y}{\sin^2(x^3y)}dx -\frac{x^3}{\sin^2(x^3y)}dy[/m]
Дифференциал 2 порядка:
[m]\frac{d^2z}{dx^2} = -\frac{6xy \sin^2(x^3y) - 3x^2y \cdot 2\sin(x^3y) \cos(x^3y) \cdot 3x^2y}{\sin^4(x^3y)} = -\frac{6xy \sin(x^3y) - 18x^4y^2 \cdot \cos(x^3y)}{\sin^3(x^3y)} [/m]
[m]\frac{d^2z}{dxdy} = -\frac{3x^2 \sin^2(x^3y) - 3x^2y \cdot 2\sin(x^3y) \cos(x^3y) \cdot x^3}{\sin^4(x^3y)} = -\frac{3x^2 \sin(x^3y) - 6x^5y \cdot \cos(x^3y)}{\sin^3(x^3y)} [/m]
[m]\frac{d^2z}{dy^2} = -\frac{0-x^3 \cdot 2\sin(x^3y) \cos(x^3y) \cdot x^3}{\sin^4(x^3y)} = \frac{2x^6 \cos(x^3y)}{\sin^3(x^3y)}[/m]
Полный дифференциал 2 порядка.
[m]d^2z = \frac{d^2z}{dx^2} dx^2 + 2\frac{d^2z}{dxdy} dxdy + \frac{d^2z}{dy^2} dy^2[/m]
[m]d^2z = -\frac{6xy \sin(x^3y) - 18x^4y^2 \cdot \cos(x^3y)}{\sin^3(x^3y)}dx^2
- \frac{6x^2 \sin(x^3y) - 12x^5y \cdot \cos(x^3y)}{\sin^3(x^3y)}dxdy + \frac{2x^6 \cos(x^3y)}{\sin^3(x^3y)}dy^2[/m]