Calcule as derivadas parciais de segunda ordem da função f.
f(x, y, z)=x e^{y}+y e^{z}+z e^{x};

Passo 1

O enunciado pede as derivadas parciais de segunda ordem de
f(x,y,z)=x e^{y} \;+\; y e^{z} \;+\; z e^{x}.
O conteúdo da matéria é cálculo de derivadas parciais iteradas. Vamos primeiro lembrar que, para obter uma segunda derivada mista, precisamos derivar duas vezes em variáveis possivelmente diferentes, sempre mantendo as outras constantes.

Passo 2

Calculamos as derivadas parciais de primeira ordem:
\frac{\partial f}{\partial x} = \underbrace{\frac{\partial}{\partial x}\bigl(xe^{y}\bigr)}_{e^{y}} + \underbrace{\frac{\partial}{\partial x}\bigl(ye^{z}\bigr)}_{0} + \underbrace{\frac{\partial}{\partial x}\bigl(ze^{x}\bigr)}_{z e^{x}} = e^{y} + z e^{x},
\frac{\partial f}{\partial y} = \underbrace{\frac{\partial}{\partial y}\bigl(xe^{y}\bigr)}_{x e^{y}} + \underbrace{\frac{\partial}{\partial y}\bigl(ye^{z}\bigr)}_{e^{z}} + \underbrace{\frac{\partial}{\partial y}\bigl(ze^{x}\bigr)}_{0} = x e^{y} + e^{z},
\frac{\partial f}{\partial z} = \underbrace{\frac{\partial}{\partial z}\bigl(xe^{y}\bigr)}_{0} + \underbrace{\frac{\partial}{\partial z}\bigl(ye^{z}\bigr)}_{y e^{z}} + \underbrace{\frac{\partial}{\partial z}\bigl(ze^{x}\bigr)}_{e^{x}} = y e^{z} + e^{x}.

Passo 3

Agora derivamos de novo em x para obter \frac{\partial^2 f}{\partial x^2}, em y e em z para as mistas envolvendo f_x:
\frac{\partial^2 f}{\partial x^2} = \frac{\partial}{\partial x}\bigl(e^{y} + z e^{x}\bigr) = \underbrace{\frac{\partial}{\partial x}e^{y}}_{0} + \underbrace{\frac{\partial}{\partial x}\bigl(z e^{x}\bigr)}_{z e^{x}} = z e^{x},
\frac{\partial^2 f}{\partial x \partial y} = \frac{\partial}{\partial y}\bigl(e^{y} + z e^{x}\bigr) = \underbrace{\frac{\partial}{\partial y}e^{y}}_{e^{y}} + \underbrace{\frac{\partial}{\partial y}\bigl(z e^{x}\bigr)}_{0} = e^{y},
\frac{\partial^2 f}{\partial x \partial z} = \frac{\partial}{\partial z}\bigl(e^{y} + z e^{x}\bigr) = \underbrace{\frac{\partial}{\partial z}e^{y}}_{0} + \underbrace{\frac{\partial}{\partial z}\bigl(z e^{x}\bigr)}_{e^{x}} = e^{x}.

Passo 4

Agora derivamos de novo em y para \frac{\partial^2 f}{\partial y^2} e em z para a mista f_{yz}:
\frac{\partial^2 f}{\partial y^2} = \frac{\partial}{\partial y}\bigl(x e^{y} + e^{z}\bigr) = \underbrace{\frac{\partial}{\partial y}\bigl(x e^{y}\bigr)}_{x e^{y}} + \underbrace{\frac{\partial}{\partial y}e^{z}}_{0} = x e^{y},
\frac{\partial^2 f}{\partial y \partial z} = \frac{\partial}{\partial z}\bigl(x e^{y} + e^{z}\bigr) = \underbrace{\frac{\partial}{\partial z}x e^{y}}_{0} + \underbrace{\frac{\partial}{\partial z}e^{z}}_{e^{z}} = e^{z}.

Passo 5

Por fim, derivamos em z para \frac{\partial^2 f}{\partial z^2}:
\frac{\partial^2 f}{\partial z^2} = \frac{\partial}{\partial z}\bigl(y e^{z} + e^{x}\bigr) = \underbrace{\frac{\partial}{\partial z}\bigl(y e^{z}\bigr)}_{y e^{z}} + \underbrace{\frac{\partial}{\partial z}e^{x}}_{0} = y e^{z}.

Resposta

\displaystyle f_{xx}=z e^{x},\quad f_{xy}=f_{yx}=e^{y},\quad f_{xz}=f_{zx}=e^{x},\quad f_{yy}=x e^{y},\quad f_{yz}=f_{zy}=e^{z},\quad f_{zz}=y e^{z}.

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