Exercício 6. Utilize a regra da cadeia para determinar as derivadas parciais indicadas.
T=v2 u+v, u=p q raizr, v=p raizq r. Encontre
partial Tpartial p, partial Tpartial q, partial Tpartial r

Passo 1

O enunciado dá
T = v2u + v, u = p,qraizr, v = praizq,r
e pede partial Tpartial p, partial Tpartial q, partial Tpartial r.
Vamos usar a regra da cadeia na forma
partial Tpartial x = partial Tpartial u,partial upartial x + partial Tpartial v,partial vpartial x,
para x=p,q,r.

Passo 2

Calculamos primeiro as derivadas parciais de u e v:
partial upartial p = qraizr, partial upartial q = praizr, partial upartial r = p,q2raizr,
partial vpartial p = raizq,r, partial vpartial q = p,r2raizq,r = p,raizr2raizq, partial vpartial r = p,q2raizq,r = p,raizq2raizr

Passo 3

Agora as derivadas de T em relação a u e v:
partial Tpartial u = 0cdot(2u+v) - vcdot2(2u+v)^2 = -2v(2u+v)^2, partial Tpartial v = (2u+v)cdot1 - vcdot1(2u+v)^2 = 2u(2u+v)^2

Passo 4

Aplicando a cadeia para x=p:
partial Tpartial p = -2v(2u+v)^2,partial upartial p + 2u(2u+v)^2,partial vpartial p
Substituindo partial upartial p=qraiz r e partial vpartial p=raizq,r
e lembrando que u=pqraiz r, v=praizq,r, obtém-se zero após simplificar, logo
partial Tpartial p = 0

Passo 5

Para x=q:
partial Tpartial q = -2v(2u+v)^2,praizr + 2u(2u+v)^2,praizr2raizq = -2v,praiz r + u,praiz r/raiz q(2u+v)^2
Fazendo u=pqraiz r, v=praizq,r e simplificando,
partial Tpartial q = -1raizq,l(2raizq+1r)^2

Passo 6

Para x=r:
partial Tpartial r = -2v(2u+v)^2,p,q2raiz r + 2u(2u+v)^2,praizq2raiz r
Substituindo e simplificando, nota-se que os termos se cancelam, resultando em
partial Tpartial r = 0

Resposta

partial Tpartial p=0, partial Tpartial q=-1raizq,l(2raizq+1r)^2, partial Tpartial r=0.

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