Найти производные dz/du, dz/dv функции z=x^2*y^2 где x=u+v , y=u/v я так понял по формуле: как это расписать?

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Ответ:

Dogi2019

02.10.2020 17:05

$\dfrac{\partial z}{\partial u} =2 \cdot x \cdot y^{2} + \dfrac{ 2 \cdot x^{2} \cdot y}{v}$

$\dfrac{\partial z}{\partial v} =2 \cdot x \cdot y^{2} - \dfrac{ 2 \cdot x^{2} \cdot y \cdot u}{v^{2}}$

Пошаговое объяснение:

$\dfrac{\partial z}{\partial u} = \dfrac{\partial z}{\partial x} \cdot \dfrac{\partial x}{\partial u} + \dfrac{\partial z}{\partial y} \cdot \dfrac{\partial y}{\partial u} = \dfrac{\partial (x^{2} \cdot y^{2} )}{\partial x} \cdot \dfrac{\partial (u+v)}{\partial u} + \dfrac{\partial (x^{2} \cdot y^{2} )}{\partial y} \cdot \dfrac{\partial (\dfrac{u}{v} )}{\partial u} =$

$= \dfrac{\partial (x^{2} )}{\partial x} \cdot y^{2}\cdot \dfrac{\partial (u+v)}{\partial u} + x^{2} \cdot\dfrac{\partial (y^{2} )}{\partial y} \cdot \dfrac{1}{v} \cdot \dfrac{\partial u}{\partial u} =$

$= (2 \cdot x) \cdot y^{2} \cdot (1+0)+ x^{2} \cdot (2 \cdot y) \cdot \dfrac{1}{v} \cdot 1 =2 \cdot x \cdot y^{2} + \dfrac{ 2 \cdot x^{2} \cdot y}{v}$

$\dfrac{\partial z}{\partial v} = \dfrac{\partial z}{\partial x} \cdot \dfrac{\partial x}{\partial v} + \dfrac{\partial z}{\partial y} \cdot \dfrac{\partial y}{\partial v} = \dfrac{\partial (x^{2} \cdot y^{2} )}{\partial x} \cdot \dfrac{\partial (u+v)}{\partial v} + \dfrac{\partial (x^{2} \cdot y^{2} )}{\partial y} \cdot \dfrac{\partial (\dfrac{u}{v} )}{\partial v} =$

$= \dfrac{\partial (x^{2} )}{\partial x} \cdot y^{2}\cdot \dfrac{\partial (u+v)}{\partial v} + x^{2} \cdot\dfrac{\partial (y^{2} )}{\partial y} \cdot u \cdot \dfrac{\partial (v^{-1})}{\partial v} =$

$= (2 \cdot x) \cdot y^{2} \cdot (0+1)+ x^{2} \cdot (2 \cdot y) \cdot u \cdot (-1) \cdot v^{-2} =2 \cdot x \cdot y^{2} - \dfrac{ 2 \cdot x^{2} \cdot y \cdot u}{v^{2}}$