Considere as funções
z=e^x cdot sen(y), x=r^2+s^2 e y=2 r s
(a) Calcule partial zpartial s(r, s).
(b) Calcule partial^2 zpartial r partial s(r, s)=partialpartial r(partial zpartial s)(r, s).

Passo 1

Beleza, o exercício é sobre derivadas parciais e regra da cadeia. A gente tem
z = e^xsin(y), x = r^2+s^2, y=2rs
e precisa calcular:
(a) partial zpartial s(r,s)
(b) partial^2 zpartial rpartial s(r,s)
O que vamos usar:
  • Derivada de e^x é e^x.
  • Derivada de sin(y) é cos(y),y'.
  • Regra da cadeia para funções compostas:
    partial zpartial s = zx,xs + zy,ys.
  • Derivada parcial trata as outras variáveis como constantes.

Passo 2

Queremos partial zpartial s. Pela regra da cadeia:
partial zpartial s = partial zpartial x,partial xpartial s + partial zpartial y,partial ypartial s
  1. zx = partialpartial x(e^xsin y)=e^xsin y
  2. zy = partialpartial y(e^xsin y)=e^xcos y
  3. xs = partialpartial s(r^2+s^2)=2s
  4. ys = partialpartial s(2rs)=2r
Substituindo:
partial zpartial s = e^xsin ycdot2s ;+; e^xcos ycdot2r =2e^xl(ssin y + rcos yr)

Passo 3

Agora partial^2 zpartial rpartial s =partialpartial rl(2e^x,(ssin y + rcos y)r)
Chama u=2e^x e v=ssin y+rcos y. Então
partialpartial r(uv)=u',v+u,v'
  • u'=2e^xcdot xr=2e^xcdot2r=4r,e^x
  • Para v'= partialpartial r(ssin y) + partialpartial r(rcos y):
    • partialpartial r(ssin y) = scos ycdot yr = scos ycdot2s =2s^2cos y
    • partialpartial r(rcos y) = 1cdotcos y + r(-sin y),yr = cos y -2rssin y
      Logo v'=2s^2cos y +cos y -2rssin y=(2s^2+1)cos y-2rssin y
Montando tudo:
partial^2 zpartial rpartial s =4r,e^x,(ssin y+rcos y) +2e^xl((2s^2+1)cos y-2rssin yr)
Fatorando 2e^x e simplificando os termos de seno:
partial^2 zpartial rpartial s =2e^xl(2r(ssin y+rcos y)+(2s^2+1)cos y-2rssin yr) =2e^x(2x+1)cos y,
pois 2r(ssin y)-2rssin y=0 e x=r^2+s^2. Substituindo x e y:
partial^2 zpartial rpartial s =2e^r^2+s^2l(2(r^2+s^2)+1r)cos(2rs)

Resposta

(a) partial zpartial s(r,s) =2e^r^2+s^2l(ssin(2rs)+rcos(2rs)r)
(b) partial^2 zpartial rpartial s(r,s) =2e^r^2+s^2l(2(r^2+s^2)+1r)cos(2rs)

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