Es sei F(x,y)=2x+yy−2x,x=2u−3v,y=u+2v{\displaystyle F(x,y)={\frac {2x+y}{y-2x}},\;x=2u-3v,\;y=u+2v}. Man berechne ∂F∂u{\displaystyle {\frac {\partial F}{\partial u}}} und ∂F∂v{\displaystyle {\frac {\partial F}{\partial v}}} für u=2,v=1{\displaystyle u=2,v=1}.
einsetzen von x=2⋅u−3⋅v{\displaystyle x=2\cdot u-3\cdot v} und y=u+2⋅v{\displaystyle y=u+2\cdot v} in F(x){\displaystyle F(x)}:
F(x)=2⋅(2u−3v)+(u+2v)(u+2v)−2⋅(2u−3v)=4u−6v+u+2vu+2v−4u+6v=5u−4v8v−3u{\displaystyle F(x)={\frac {2\cdot (2u-3v)+(u+2v)}{(u+2v)-2\cdot (2u-3v)}}={\frac {4u-6v+u+2v}{u+2v-4u+6v}}={\frac {5u-4v}{8v-3u}}}
Bilden der partiellen Ableitungen (Quotientenregel):
∂F∂u=5⋅(8v−3u)−(−3)⋅(5u−4v)(8v−3u)2=−15u+40v−(−15u+12v)(8v−3u)2=28v(8v−3u)2{\displaystyle {\frac {\partial F}{\partial u}}={\frac {5\cdot (8v-3u)-(-3)\cdot (5u-4v)}{(8v-3u)^{2}}}={\frac {-15u+40v-(-15u+12v)}{(8v-3u)^{2}}}={\frac {28v}{(8v-3u)^{2}}}}
∂F∂v=(−4)⋅(8v−3u)−8⋅(5u−4v)(8v−3u)2=−32v+12u−(40u−32v)(8v−3u)2=−28u(8v−3u)2{\displaystyle {\frac {\partial F}{\partial v}}={\frac {(-4)\cdot (8v-3u)-8\cdot (5u-4v)}{(8v-3u)^{2}}}={\frac {-32v+12u-(40u-32v)}{(8v-3u)^{2}}}={\frac {-28u}{(8v-3u)^{2}}}}
einsetzen von u=2{\displaystyle u=2} und v=1{\displaystyle v=1}:
∂F∂u=28⋅1(8⋅1−3⋅2)2=284=7{\displaystyle {\frac {\partial F}{\partial u}}={\frac {28\cdot 1}{(8\cdot 1-3\cdot 2)^{2}}}={\frac {28}{4}}=7}
∂F∂v=−28⋅24=−14{\displaystyle {\frac {\partial F}{\partial v}}={\frac {-28\cdot 2}{4}}=-14}
SS07 Beispiel 91
und 
in 
:
und 
:
