Question

Дифференциальное исчисление функции нескольких переменных. Задание на скриншотах.

Answer 1

Ответ:

1) $z = ctg(y - x)$ :

а) $\frac{dz}{dx} = \frac{1}{sin^2(y- x) } , \frac{dz}{dy} = -\frac{1}{sin^2(y- x) }$

б) $\frac{d^2z}{dx^2} = \frac{2cos(y-x)}{sin^3(y-x)} , \frac{d^2z}{dy^2} = \frac{2cos(y-x)}{sin^3(y-x)}, \frac{d^2z}{dxdy} = \frac{d^2z}{dydx} = -\frac{2cos(y-x)}{sin^3(y-x)}$

2) $z = x^2 + y^2 + xy -2x - y + 2$ :

а) $\frac{dz}{dx} = 2x + y - 2, \frac{dz}{dy} = 2y + x -1$

б) $\frac{d^2z}{dx^2} = 2, \frac{d^2z}{dy^2} = 2, \frac{d^2z}{dxdy} = \frac{d^2z}{dydx} = 1,$

Пошаговое объяснение:

   Частная производная — это предел отношения приращения функции по выбранной переменной к приращению этой переменной, при стремлении этого приращения к нулю.

1) $z = ctg(y - x)$ :

а) $\frac{dz}{dx} = \frac{d(ctg(y - x))}{dx} = \frac{1}{sin^2(y- x) } , \frac{dz}{dx} = \frac{d(ctg(y - x))}{dy} = - \frac{1}{sin^2(y - x) }$

б)

$\frac{d^2z}{dx^2} = \frac{d}{dx}(\frac{dz}{dx}) = \frac{d}{dx}(\frac{1}{sin^2(y-x)}) =-\frac{\frac{d}{dx}(sin^2(y-x))}{(sin^2(y-x))^2} =-\frac{2sin(y-x)\frac{d}{dx}(sin(y-x))}{sin^4(y-x)} =-\frac{2cos(y-x)\frac{d}{dx}((y-x)}{sin^3(y-x)} =-\frac{2cos(y-x)(-1)}{sin^3(y-x)} = \frac{2cos(y-x)}{sin^3(y-x)}$

$\frac{d^2z}{dy^2} = \frac{d}{dy}(\frac{dz}{dy}) = \frac{d}{dy}(-\frac{1}{sin^2(y-x)}) =\frac{\frac{d}{dy}(sin^2(y-x))}{(sin^2(y-x))^2} =\frac{2sin(y-x)\frac{d}{dy}(sin(y-x))}{sin^4(y-x)} =\frac{2cos(y-x)\frac{d}{dy}((y-x)}{sin^3(y-x)} =\frac{2cos(y-x)}{sin^3(y-x)}$

$\frac{d^2z}{dxdy} = \frac{d}{dx}(\frac{dz}{dy}) = \frac{d}{dx}(-\frac{1}{sin^2(y-x)}) =\frac{\frac{d}{dx}(sin^2(y-x))}{(sin^2(y-x))^2} =\frac{2sin(y-x)\frac{d}{dx}(sin(y-x))}{sin^4(y-x)} =\frac{2cos(y-x)\frac{d}{dx}((y-x)}{sin^3(y-x)} =\frac{2cos(y-x)(-1)}{sin^3(y-x)} = -\frac{2cos(y-x)}{sin^3(y-x)}$

$\frac{d^2z}{dydx} = \frac{d}{dy}(\frac{dz}{dx}) = \frac{d}{dy}(\frac{1}{sin^2(y-x)}) =-\frac{\frac{d}{dy}(sin^2(y-x))}{(sin^2(y-x))^2} =-\frac{2sin(y-x)\frac{d}{dy}(sin(y-x))}{sin^4(y-x)} =-\frac{2cos(y-x)\frac{d}{dy}((y-x)}{sin^3(y-x)} =-\frac{2cos(y-x)}{sin^3(y-x)}$

2) $z = x^2 + y^2 + xy -2x - y + 2$ :

а)

$\frac{dz}{dx} =\frac{d(2x^2 + 2y^2 +xy -2x -y +2)}{dx}= \frac{d(2x^2)}{dx}+\frac{d(2y^2)}{dx}+\frac{d(xy)}{dx}+\frac{d(-2x))}{dx}+\frac{d(-y)}{dx}+\frac{d(2)}{dx} =\\ \\= 2x + 0 + y - 2 + 0 + 0 = 2x + y - 2$

$\frac{dz}{dy} =\frac{d(2x^2 + 2y^2 +xy -2x -y +2)}{dy}= \frac{d(2x^2)}{dy}+\frac{d(2y^2)}{dy}+\frac{d(xy)}{dy}+\frac{d(-2x))}{dy}+\frac{d(-y)}{dy}+\frac{d(2)}{dy} =\\ \\= 0 + 2y + x +0 - 1 + 0 = 2y + x - 1$

б) $\frac{d^2z}{dx^2} = \frac{d}{dx} (\frac{dz}{dx} ) = \frac{d}{dx}(2x + y - 2) =\frac{d(2x)}{dx} +\frac{d(y)}{dx} +\frac{d(-2)}{dx} = 2 + 0 + 0 = 2$

$\frac{d^2z}{dy^2} = \frac{d}{dy} (\frac{dz}{dy} ) = \frac{d}{dy}(2y + x - 1) =\frac{d(2y)}{dy} +\frac{d(x)}{dy} +\frac{d(-1)}{dy} = 2 + 0 + 0 = 2$

$\frac{d^2z}{dxdy} = \frac{d}{dx} (\frac{dz}{dy} ) = \frac{d}{dx}(2y + x - 1) =\frac{d(2y)}{dx} +\frac{d(x)}{dx} +\frac{d(-1)}{dx} = 0 + 1 + 0 = 1$

$\frac{d^2z}{dydx} = \frac{d}{dy} (\frac{dz}{dx} ) = \frac{d}{dy}(2x + y - 2) =\frac{d(2x)}{dy} +\frac{d(y)}{dy} +\frac{d(-2)}{dy} = 0 + 1 + 0 = 1$