Zu zeigen:$$w={ \left( \frac { x-y+z }{ x+y-z } \right) }^{ h }\Rightarrow x\frac { \partial { w } }{ \partial x } +y\frac { \partial w }{ \partial y } +z\frac { \partial w }{ \partial z } =0$$ Setze:$$w={ u }^{ h }\quad mit\quad u=\frac { x-y+z }{ x+y-z }$$Es ist$$\frac { \partial { u } }{ \partial x } =\frac { \partial { \frac { (x-y+z) }{ (x+y-z) } } }{ \partial x } =\frac { (x+y-z)-(x-y+z) }{ (x+y-z)^{ 2 } } =\frac { 2y-2z }{ (x+y-z)^{ 2 } }$$$$\frac { \partial { u } }{ \partial y } =\frac { \partial { \frac { (x-y+z) }{ (x+y-z) } } }{ \partial y } =\frac { -(x+y-z)-(x-y+z) }{ (x+y-z)^{ 2 } } =\frac { -2x }{ (x+y-z)^{ 2 } }$$$$\frac { \partial { u } }{ \partial z } =\frac { \partial { \frac { (x-y+z) }{ (x+y-z) } } }{ \partial z } =\frac { (x+y-z)+(x-y+z) }{ (x+y-z)^{ 2 } } =\frac { 2x }{ (x+y-z)^{ 2 } }$$Setze zur Übersichtlichkeit:$$v=x+y-z$$Dann ist nach Kettenregel (innere Ableitung * äußere Ableitung):$$\frac { \partial w }{ \partial x } =\frac { \partial { u }^{ h } }{ \partial x } =\frac { \partial { u } }{ \partial x } h{ u }^{ h-1 }=\frac { 2y-2z }{ v^{ 2 } } h{ u }^{ h-1 }$$$$\frac { \partial w }{ \partial y } =\frac { \partial { u }^{ h } }{ \partial y } =\frac { \partial { u } }{ \partial y } h{ u }^{ h-1 }=\frac { -2x }{ v^{ 2 } } h{ u }^{ h-1 }$$$$\frac { \partial w }{ \partial z } =\frac { \partial { u }^{ h } }{ \partial z } =\frac { \partial { u } }{ \partial z } h{ u }^{ h-1 }=\frac { 2x }{ v^{ 2 } } h{ u }^{ h-1 }$$und somit:$$x\frac { \partial w }{ \partial x } +y\frac { \partial w }{ \partial y } +z\frac { \partial w }{ \partial z } =\left( x\frac { 2y-2z }{ v^{ 2 } } +y\frac { -2x }{ v^{ 2 } } +z\frac { 2x }{ v^{ 2 } } \right) h{ u }^{ h-1 }$$$$=(2xy-2xz-2xy+2xz)\frac { h{ u }^{ h-1 } }{ v^{ 2 } }$$$$=0\frac { h{ u }^{ h-1 } }{ v^{ 2 } }$$$$=0$$
