Диференциране

[Гост]

Може ли помощ с тази задача?

[Добромир Глухаров]

$f(x,y,z)=\varphi(xy,\frac{y}{z})=\varphi(u,v)$

$\frac{\partial f}{\partial x}=\frac{\partial \varphi}{\partial u}\cdot\frac{\partial u}{\partial x}+\frac{\partial \varphi}{\partial v}\cdot\frac{\partial v}{\partial x}=\frac{\partial \varphi}{\partial u}\cdot(xy)'_x+\frac{\partial \varphi}{\partial v}\cdot(\frac{y}{z})'_x=y\frac{\partial \varphi}{\partial u}+0$

$\frac{\partial f}{\partial y}=\frac{\partial \varphi}{\partial u}\cdot\frac{\partial u}{\partial y}+\frac{\partial \varphi}{\partial v}\cdot\frac{\partial v}{\partial y}=\frac{\partial \varphi}{\partial u}\cdot(xy)'_y+\frac{\partial \varphi}{\partial v}\cdot(\frac{y}{z})'_y=x\frac{\partial \varphi}{\partial u}+\frac{1}{z}\frac{\partial \varphi}{\partial v}$

$\frac{\partial f}{\partial z}=\frac{\partial \varphi}{\partial u}\cdot\frac{\partial u}{\partial z}+\frac{\partial \varphi}{\partial v}\cdot\frac{\partial v}{\partial z}=\frac{\partial \varphi}{\partial u}\cdot(xy)'_z+\frac{\partial \varphi}{\partial v}\cdot(\frac{y}{z})'_z=0-\frac{y}{z^2}\frac{\partial \varphi}{\partial v}$

$-x\frac{\partial f}{\partial x}+y\frac{\partial f}{\partial y}+z\frac{\partial f}{\partial z}=-xy\frac{\partial \varphi}{\partial u}+xy\frac{\partial \varphi}{\partial u}+\frac{y}{z}\frac{\partial \varphi}{\partial v}-\frac{zy}{z^2}\frac{\partial \varphi}{\partial v}=0$