Question

Вычислите частные производные первого и второго порядков:

для функции

z=e^xy(x^2+y^2)

найти dz/dx; dz/dy

Answer 1

Ответ:

$z = {e}^{xy} ( {x}^{2} + {y}^{2} )$

$\frac{dz}{dx} = {e}^{xy} \times y( {x}^{2} + {y}^{2} ) + 2x {e}^{xy} = \\ = {e}^{xy} ( {x}^{2} y + {y}^{3} + 2x)$

$\frac{dz}{dy} = {e}^{xy} \times x( {x}^{2} + {y}^{2} ) + 2y {e}^{xy} = \\ = {e}^{xy} ( {x}^{3} + x {y}^{2} + 2y)$

$\frac{ {d}^{2} z}{ {dx}^{2} } = {e}^{xy} \times y( {x}^{2} y + {y}^{3} + 2x) + (2xy + 2) {e}^{xy} = \\ = {e}^{xy} ( {x}^{2} {y}^{2} + {y}^{4} + 2xy + 2xy + 2) = \\ = {e}^{xy} ( {y}^{x} + {x}^{2} {y}^{2} + 4 xy + 2)$

$\frac{ {d}^{2} z}{ {dy}^{2} } = {e}^{xy} \times x( {x}^{3} + x {y}^{2} + 2y) + {e}^{xy} (2xy + 2) = \\ = {e}^{xy} ( {x}^{4} + {x}^{2} {y}^{2} + 2xy + 2xy + 2) = \\ = {e}^{xy} ( {x}^{4} + {x}^{2} {y}^{2} + 4xy + 2)$

$\frac{ {d}^{2} z}{dxdy} = \frac{ {d}^{2}z }{dydx} = {e}^{xy} \times x( {x}^{2} y + {y}^{3} + 2x) + {e}^{xy} ( {x}^{2} + 3 {y}^{2} ) = \\ = {e}^{xy} ( {x}^{3} y + {y}^{3} x + 2 {x}^{2} + {x}^{2} + 3 {y}^{2} ) = \\ = {e}^{xy} ( {y}^{3} x + {x}^{3} y + 3 {x}^{2} + 3 {y}^{2} )$