Usando a regra da cadeia para z=f(x, y) e w=f(x, y, z), calcule:
partial zpartial t, partial zpartial s e partial wpartial t, partial wpartial s e partial wpartial r
w=x y+y z+z x, x=t r, y=s t, z=t s

Passo 1

O enunciado define
x = t,r, y = s,t, z = t,s
e
w = x,y + y,z + z,x
Queremos as derivadas parciais
partial zpartial t, partial zpartial s e
partial wpartial t, partial wpartial s, partial wpartial r.

Passo 2

Como z aparece explicitamente em termos de t e s pela fórmula z(t,s)=t,s, basta derivar:
partial zpartial t = partial (t,s)partial t = s, partial zpartial s = partial (t,s)partial s = t

Passo 3

Para w(x,y,z)=x,y + y,z + z,x calculamos primeiro
partial wpartial x = y + z, partial wpartial y = x + z, partial wpartial z = x + y

Passo 4

Usando a regra da cadeia para partial wpartial t:
partial wpartial t = partial wpartial x,partial xpartial t + partial wpartial y,partial ypartial t + partial wpartial z,partial zpartial t
Como partial xpartial t=r, partial ypartial t=s e partial zpartial t=s, obtemos
partial wpartial t = (y+z),r ;+;(x+z),s;+;(x+y),s = r,(y+z) + s,(2x + y + z)

Passo 5

Para partial wpartial s temos partial xpartial s=0, partial ypartial s=t, partial zpartial s=t. Então
partial wpartial s = 0cdot(y+z) + (x+z),t + (x+y),t = t,(2x + y + z)

Passo 6

Para partial wpartial r x depende de r, com partial xpartial r=t e os outros nulos. Logo
partial wpartial r = (y+z),t

Resposta

partial zpartial t = s, partial zpartial s = t,
partial wpartial t = r,(y+z) ;+; s,(2x+y+z), partial wpartial s = t,(2x+y+z), partial wpartial r = t,(y+z)
Em que x=t,r, y=s,t e z=t,s.

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