Encontrar
partial^2 fpartial x^2, partial^2 fpartial y^2, partial^2 fpartial x partial y, partial^3 fpartial x^2 partial y e partial^3 fpartial z^3
das seguintes funções:
f(x, y, z)=(e^y sen x, e^x sen y, z)

Passo 1

O enunciado pede para calcular derivadas parciais de ordem superior da função vetorial
vec f(x,y,z) = l(e^ysin x,;e^xsin y,;zr).
Lembrando que, para uma função vetorial, as derivadas parciais são feitas componente a componente.

Passo 2

Cálculo de partial^2vec fpartial x^2.
Compondo cada segundo derivada em x:
  1. Para f1(x,y,z)=e^ysin x:
partial f1partial x = e^ycos x, partial^2 f1partial x^2 = -e^ysin x
  1. Para f2(x,y,z)=e^xsin y:
partial f2partial x = e^xsin y, partial^2 f2partial x^2 = e^xsin y
  1. Para f3(x,y,z)=z:
partial f3partial x=0, partial^2 f3partial x^2=0
Logo,
partial^2vec fpartial x^2 =l(-e^ysin x,;e^xsin y,;0r)

Passo 3

Cálculo de partial^2vec fpartial y^2.
  1. Para f1=e^ysin x:
partial f1partial y = e^ysin x, partial^2 f1partial y^2 = e^ysin x
  1. Para f2=e^xsin y:
partial f2partial y = e^xcos y, partial^2 f2partial y^2 = -,e^xsin y
  1. Para f3=z:
partial f3partial y=0, partial^2 f3partial y^2=0
Assim,
partial^2vec fpartial y^2 =l(e^ysin x,;-e^xsin y,;0r)

Passo 4

Cálculo de partial^2vec fpartial xpartial y.
  1. Para f1=e^ysin x:
partialpartial x!l(partial f1partial yr) = partialpartial xl(e^ysin xr) = e^ycos x
  1. Para f2=e^xsin y:
partialpartial x!l(partial f2partial yr) = partialpartial xl(e^xcos yr) = e^xcos y
  1. Para f3=z:
partialpartial x!l(partial f3partial yr) = partialpartial x(0)=0
Portanto,
partial^2vec fpartial xpartial y =l(e^ycos x,;e^xcos y,;0r)

Passo 5

Cálculo de partial^3vec fpartial x^2partial y.
Basta derivar em y o resultado de partial^2vec f/partial x^2:
  1. A partir de -e^ysin x:
partialpartial yl(-e^ysin xr) = -,e^ysin x
  1. A partir de e^xsin y:
partialpartial yl(e^xsin yr) = e^xcos y
  1. A partir de 0:
partialpartial y(0)=0
Logo,
partial^3vec fpartial x^2partial y =l(-e^ysin x,;e^xcos y,;0r)

Passo 6

Cálculo de partial^3vec fpartial z^3.
Como só a terceira componente f3=z depende de z, e
ddz(z)=1, d^2dz^2(z)=0, d^3dz^3(z)=0,
as três derivadas em z deixam
partial^3vec fpartial z^3 =(0,0,0)

Resposta

partial^2vec fpartial x^2 =l(-e^ysin x,;e^xsin y,;0r)
partial^2vec fpartial y^2 =l(e^ysin x,;-e^xsin y,;0r)
partial^2vec fpartial xpartial y =l(e^ycos x,;e^xcos y,;0r)
partial^3vec fpartial x^2partial y =l(-e^ysin x,;e^xcos y,;0r)
partial^3vec fpartial z^3 =(0,0,0)

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