Encontre a derivada parcial indicada usando a regra da cadeia:
u=sen(x y) ; x=2 z e^t ; y=t^2 e^-z ; partial upartial t ; partial upartial z

Passo 1

O enunciado define
u=sin(xy),
com as substituições
x=2z e^t e y=t^2e^-z,
e pede as derivadas parciais partial upartial t e partial upartial z.
O conteúdo aqui é a regra da cadeia em funções de várias variáveis:
partial upartial t = partial upartial x,partial xpartial t + partial upartial y,partial ypartial t,
partial upartial z = partial upartial x,partial xpartial z + partial upartial y,partial ypartial z

Passo 2

Primeiro achamos
partial upartial x = cos(xy),partial (xy)partial x = cos(xy),y,
partial upartial y = cos(xy),partial (xy)partial y = cos(xy),x

Passo 3

Agora as derivadas de x e y em relação a t:
partial xpartial t = partial (2z e^t)partial t = 2z e^t,
partial ypartial t = partial (t^2e^-z)partial t = 2t,e^-z
Substituindo na regra da cadeia:
partial upartial t = cos(xy)l(ycdot 2z e^t ;+; xcdot 2t e^-zr)

Passo 4

Agora as derivadas de x e y em relação a z:
partial xpartial z = partial (2z e^t)partial z = 2e^t,
partial ypartial z = partial (t^2e^-z)partial z = -,t^2e^-z
Aplicando a regra da cadeia:
partial upartial z = cos(xy)l(ycdot 2e^t ;+; xcdot(-t^2e^-z)r)

Resposta

partial upartial t = cosl(2z t^2e^,t-zr),l(2z t^2e^,t-z+4z t,e^,t-zr),
partial upartial z = cosl(2z t^2e^,t-zr),l(2t^2e^,t-z-2z t^2e^,t-zr)

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