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=log 5(x y+y z+z x), x=t^2 r, y=s t^2, z=t^2 s

Passo 1

O enunciado define
w = log5l(xy + yz + zxr), x = t^2r, y = s,t^2, z = t^2s
e pede as derivadas parciais
partial zpartial t, partial zpartial s e
partial wpartial t, partial wpartial s, partial wpartial r.
O conteúdo principal é a regra da cadeia para funções compostas de várias variáveis.

Passo 2

Como z = t^2s, basta derivar em relação a t e a s:
partial zpartial t = partialpartial tl(t^2sr) = 2t,s, partial zpartial s = partialpartial sl(t^2sr) = t^2

Passo 3

Derivamos x,y,z em relação a t,s,r:
partial xpartial t = 2t,r, partial xpartial s = 0, partial xpartial r = t^2,
partial ypartial t = 2t,s, partial ypartial s = t^2, partial ypartial r = 0,
partial zpartial t = 2t,s, partial zpartial s = t^2, partial zpartial r = 0

Passo 4

Derivamos w=log5(u) com u=xy+yz+zx. Usando
ddulog5u = 1uln 5
e partial u/partial x = y+z (simetricamente para y,z), obtemos
partial wpartial x = y+z(xy+yz+zx),ln 5, partial wpartial y = x+z(xy+yz+zx),ln 5, partial wpartial z = x+y(xy+yz+zx),ln 5

Passo 5

Aplicando a regra da cadeia
partial wpartial t = partial wpartial xpartial xpartial t + partial wpartial ypartial ypartial t + partial wpartial zpartial zpartial t,
e análogas para s e r, temos:
partial wpartial t = y+z(xy+yz+zx)ln 5,(2tr) + x+z(xy+yz+zx)ln 5,(2ts) + x+y(xy+yz+zx)ln 5,(2ts),
partial wpartial s = y+z(xy+yz+zx)ln 5,0 + x+z(xy+yz+zx)ln 5,(t^2) + x+y(xy+yz+zx)ln 5,(t^2),
partial wpartial r = y+z(xy+yz+zx)ln 5,(t^2) + x+z(xy+yz+zx)ln 5,0 + x+y(xy+yz+zx)ln 5,0

Resposta

partial zpartial t = 2t,s, partial zpartial s = t^2,
partial wpartial t = 2tr,(y+z) + 2ts,(x+z) + 2ts,(x+y)(xy+yz+zx)ln 5,
partial wpartial s = t^2,(x+z + x+y)(xy+yz+zx)ln 5, partial wpartial r = t^2,(y+z)(xy+yz+zx)ln 5

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