Nos seguintes exercícios, calcule D_{11} f(x, y), D_{22} f(x, y) e prove que D_{12} f(x, y) e D_{21} f(x, y) são iguais.
f(x, y)=e^{-x / y}+\ln \frac{x}{y}

Passo 1

O enunciado pede para a função
f(x,y)=e^{-x/y}+\ln\frac{x}{y}
calcular as segundas derivadas parciais \,D_{11}f(x,y), D_{22}f(x,y) e provar que as derivadas mistas D_{12}f(x,y) e D_{21}f(x,y) coincidem. O tema aqui é derivação parcial de funções de duas variáveis, usando a regra da cadeia e a linearidade da derivada.

Passo 2

Calcular a derivada parcial de primeira ordem em x:
\frac{\partial f}{\partial x} =\frac{\partial}{\partial x}\Bigl(e^{-x/y}\Bigr) +\frac{\partial}{\partial x}\Bigl(\ln\tfrac{x}{y}\Bigr).
Pela regra da cadeia,
\frac{\partial}{\partial x}\Bigl(e^{-x/y}\Bigr) =e^{-x/y}\cdot\frac{\partial}{\partial x}\Bigl(-\tfrac{x}{y}\Bigr) =e^{-x/y}\cdot\Bigl(-\tfrac1y\Bigr) =-\frac{1}{y}\,e^{-x/y},
e
\frac{\partial}{\partial x}\Bigl(\ln\tfrac{x}{y}\Bigr) =\frac{1}{x}.
Logo,
D_{1}f(x,y) =-\frac{1}{y}\,e^{-x/y}+\frac{1}{x}.

Passo 3

Agora derivamos \frac{\partial f}{\partial x} em x para obter D_{11}f:
D_{11}f(x,y) =\frac{\partial}{\partial x}\Bigl(-\tfrac{1}{y}e^{-x/y}+\tfrac1x\Bigr).
Temos
\frac{\partial}{\partial x}\Bigl(-\tfrac{1}{y}e^{-x/y}\Bigr) =-\frac{1}{y}\,e^{-x/y}\cdot\Bigl(-\tfrac1y\Bigr) =\frac{1}{y^2}e^{-x/y},
\frac{\partial}{\partial x}\Bigl(\tfrac1x\Bigr) =-\frac{1}{x^2}.
Portanto
D_{11}f(x,y) =\frac{1}{y^2}e^{-x/y}-\frac{1}{x^2}.

Passo 4

Calcular a derivada parcial de primeira ordem em y:
\frac{\partial f}{\partial y} =\frac{\partial}{\partial y}\Bigl(e^{-x/y}\Bigr) +\frac{\partial}{\partial y}\Bigl(\ln\tfrac{x}{y}\Bigr).
Pela cadeia,
\frac{\partial}{\partial y}\Bigl(e^{-x/y}\Bigr) =e^{-x/y}\cdot\frac{\partial}{\partial y}\Bigl(-\tfrac{x}{y}\Bigr) =e^{-x/y}\cdot\Bigl(x\,\tfrac{1}{y^2}\Bigr) =\frac{x}{y^2}e^{-x/y},
e
\frac{\partial}{\partial y}\Bigl(\ln\tfrac{x}{y}\Bigr) =\frac{\partial}{\partial y}(\ln x-\ln y) =-\frac1y.
Logo
D_{2}f(x,y) =\frac{x}{y^2}e^{-x/y}-\frac{1}{y}.

Passo 5

Derivar \frac{\partial f}{\partial y} em y para obter D_{22}f:
D_{22}f(x,y) =\frac{\partial}{\partial y}\Bigl(\frac{x}{y^2}e^{-x/y}-\frac1y\Bigr).
Para o termo exponencial use produto u=\tfrac{x}{y^2}, v=e^{-x/y}:
u'=-2\,\frac{x}{y^3},\quad v'=e^{-x/y}\cdot\frac{x}{y^2}.
Então
\frac{\partial}{\partial y}\Bigl(\tfrac{x}{y^2}e^{-x/y}\Bigr) =u'v+u\,v' =\Bigl(-\frac{2x}{y^3}+\frac{x^2}{y^4}\Bigr)e^{-x/y}.
E
\frac{\partial}{\partial y}\Bigl(-\tfrac1y\Bigr) =\frac{1}{y^2}.
Assim
D_{22}f(x,y) =\Bigl(\frac{x^2}{y^4}-\frac{2x}{y^3}\Bigr)e^{-x/y} +\frac{1}{y^2}.

Passo 6

Calcular a mista D_{12}f=\frac{\partial}{\partial y}(D_{1}f):
D_{12}f(x,y) =\frac{\partial}{\partial y}\Bigl(-\tfrac{1}{y}e^{-x/y}+\tfrac1x\Bigr).
O termo \tfrac1x some e, para -\tfrac1y\,e^{-x/y},
\frac{\partial}{\partial y} \Bigl(-\tfrac1y\,e^{-x/y}\Bigr) =e^{-x/y}\Bigl(\tfrac{1}{y^2}-\tfrac{x}{y^3}\Bigr).
Logo
D_{12}f(x,y) =e^{-x/y}\Bigl(\frac{1}{y^2}-\frac{x}{y^3}\Bigr).

Passo 7

Calcular a mista D_{21}f=\frac{\partial}{\partial x}(D_{2}f):
D_{21}f(x,y) =\frac{\partial}{\partial x}\Bigl(\frac{x}{y^2}e^{-x/y}-\frac1y\Bigr).
O termo -\tfrac1y some e, aplicando produto em \tfrac{x}{y^2}e^{-x/y}:
\frac{\partial}{\partial x} \Bigl(\tfrac{x}{y^2}e^{-x/y}\Bigr) =\frac{1}{y^2}e^{-x/y} +\frac{x}{y^2}\Bigl(-\frac1y\Bigr)e^{-x/y} =e^{-x/y}\Bigl(\frac{1}{y^2}-\frac{x}{y^3}\Bigr).
Portanto
D_{21}f(x,y) =e^{-x/y}\Bigl(\frac{1}{y^2}-\frac{x}{y^3}\Bigr).

Passo 8

Comparando
D_{12}f(x,y) =e^{-x/y}\Bigl(\frac{1}{y^2}-\frac{x}{y^3}\Bigr) \quad\text{e}\quad D_{21}f(x,y) =e^{-x/y}\Bigl(\frac{1}{y^2}-\frac{x}{y^3}\Bigr),
vemos que D_{12}f(x,y)=D_{21}f(x,y), como exige o Teorema de Clairaut sempre que f é suficientemente suave.

Resposta

D_{11}f(x,y)=\displaystyle \frac{1}{y^2}e^{-x/y}-\frac{1}{x^2}. D_{22}f(x,y)=\displaystyle \Bigl(\frac{x^2}{y^4}-\frac{2x}{y^3}\Bigr)e^{-x/y}+\frac{1}{y^2}.
E
D_{12}f(x,y)=D_{21}f(x,y) =e^{-x/y}\Bigl(\frac{1}{y^2}-\frac{x}{y^3}\Bigr).

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