Use a regra da cadeia para encontrar partial zpartial s e partial zpartial t
z=x^2+y^2, x=t^2+s^2, y=3+s t

Passo 1

O enunciado define
z = x^2 + y^2, x = t^2 + s^2, y = 3 + s,t
e pede dfracpartial zpartial s e dfracpartial zpartial t.
O conteúdo aqui é a regra da cadeia para funções de várias variáveis:
partial zpartial s = partial zpartial x,partial xpartial s + partial zpartial y,partial ypartial s, partial zpartial t = partial zpartial x,partial xpartial t + partial zpartial y,partial ypartial t

Passo 2

Primeiro calculemos as derivadas parciais de z em relação a x e y:
partial zpartial x = partialpartial x(x^2+y^2) = 2x, partial zpartial y = partialpartial y(x^2+y^2) = 2y

Passo 3

Agora derivamos x e y em relação a s e aplicamos a cadeia para partial z/partial s:
partial xpartial s = partialpartial s(t^2+s^2) = 2s, partial ypartial s = partialpartial s(3+s t) = t
Logo
partial zpartial s = (2x),(2s) + (2y),(t) = 4s,x + 2t,y
Substituindo x=t^2+s^2 e y=3+st,
partial zpartial s = 4s,(t^2+s^2) + 2t,(3+st)

Passo 4

Derivamos x e y em relação a t e aplicamos para partial z/partial t:
partial xpartial t = partialpartial t(t^2+s^2) = 2t, partial ypartial t = partialpartial t(3+s t) = s
Logo
partial zpartial t = (2x),(2t) + (2y),(s) = 4t,x + 2s,y
Substituindo x=t^2+s^2 e y=3+st,
partial zpartial t = 4t,(t^2+s^2) + 2s,(3+st)

Resposta

partial zpartial s = 4s,(t^2+s^2) ;+; 2t,(3+st), partial zpartial t = 4t,(t^2+s^2) ;+; 2s,(3+st)

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