Calcule partial zpartial u e partial zpartial v :
z=sen theta cos phi, theta=u t^2, phi=v^2 t;

Passo 1

O enunciado dá a função composta
z = sentheta ,cosphi,
com
theta = u,t^2, phi = v^2,t
O conteúdo aqui é a regra da cadeia para derivadas parciais. Vamos calcular partial zpartial u e partial zpartial v tratando t como constante.

Passo 2

Para partial zpartial u usamos a regra da cadeia:
partial zpartial u = partial zpartial theta,partial thetapartial u + partial zpartial phi,partial phipartial u
Calculamos cada termo:
partial zpartial theta = partialpartial thetal(sentheta cosphir) = costheta;cosphi,
partial zpartial phi = partialpartial phil(sentheta cosphir) = sentheta;(-sinphi) = -sentheta,sinphi,
partial thetapartial u = partialpartial u(u,t^2) = t^2, partial phipartial u = 0 (phi não depende de u) ã
Substituindo:
partial zpartial u = l(costhetacosphir),t^2 + l(-senthetasinphir),0 = t^2,costheta,cosphi
Voltando às variáveis originais:
partial zpartial u = t^2,cosl(u,t^2r),cosl(v^2tr)

Passo 3

Para partial zpartial v aplicamos de novo a regra da cadeia:
partial zpartial v = partial zpartial theta,partial thetapartial v + partial zpartial phi,partial phipartial v
Já sabemos partial zpartial theta=costhetacosphi e partial zpartial phi=-senthetasinphi. Como theta não depende de v,
partial thetapartial v=0, partial phipartial v = partialpartial v(v^2t) = 2v,t
Logo
partial zpartial v = costhetacosphicdot 0 + l(-sentheta,sinphir),2v,t = -2v,t,sentheta,sinphi
Substituindo theta=u,t^2 e phi=v^2t:
partial zpartial v = -2,v,t;senl(u,t^2r);sinl(v^2tr)

Resposta

partial zpartial u = t^2,cosl(u,t^2r),cosl(v^2tr), partial zpartial v = -2,v,t;senl(u,t^2r);sinl(v^2tr)

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