Derivadas

por **manuela** » Qua Out 31, 2012 15:24

Seja F(u,v)= f(u+v, u-v) com f(2,0)= 1, $\frac{\partial f}{\partial x} (2,0)= -1, \frac{\partial f}{\partial y} (2,0)= 2, \frac{\partial ^2 f}{\partial x^2} (2,0)= 1, \frac{\partial ^2 f}{\partial y^2} (2,0)= 2, \frac{\partial ^2 f}{\partial x \partial y} (2,0)= \frac{\partial ^2 f}{\partial y \partial x} (2,0) = 3. Calcule \frac{\partial F}{\partial v} (1,1), \frac{\partial ^2 F}{\partial u \partial v} (1,1) e \frac{\partial ^2 F}{\partial v^2} (1,1).$

Não estou conseguindo resolver, alguém pode me ajudar?

por **young_jedi** » Qua Out 31, 2012 21:38

veja que

ou seja

e

dai tiramos as derivadas parciais

$\frac{\partial x}{\partial u}=1,\frac{\partial x}{\partial v}=1,\frac{\partial u}{\partial u}=1,\frac{\partial y}{\partial v}=-1$

portanto

$\frac{\partial F(1,1)}{\partial v}=\frac{\partial f}{\partial x}(2,0).\frac{\partial x}{\partial v}}(1,1)+\frac{\partial f}{\partial y}(2,0).\frac{\partial y}{\partial v}}(1,1)$

substituindo os valores

$\frac{\partial F(1,1)}{\partial v}=(-1).1+2.(-1)=-3$

para a segunda parte

$\frac{\partial^2 F(1,1)}{\partial v\partial u}=\left(\frac{\partial^2 f}{\partial x^2}(2,0).\frac{\partial x}{\partial u}}(1,1)+ \frac{\partial^2f}{\partial x.\partial y}(2,0).\frac{\partial y}{\partial u}(1,1)\right).\frac{\partial x}{\partial v}(1,1)+\frac{\partial f}{\partial x}(2,0).\frac{\partial^2x}{\partial v\partial u}(1,1)+ \\ \left(\frac{\partial^2 f}{\partial y \partial x}(2,0).\frac{\partial x}{\partial u}}(1,1)+ \frac{\partial^2f}{\partial y^2}(2,0).\frac{\partial y}{\partial u}(1,1)\right).\frac{\partial y}{\partial v}(1,1)+\frac{\partial f}{\partial y}(2,0).\frac{\partial^2y}{\partial v\partial u}(1,1)$

substituindo os valores voce encontra a resposta
e tente fazer o terceiro item

Derivadas

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