Use a Regra da Cadeia para achar partial zpartial s e partial zpartial t.
z=sen(theta) cos (phi), theta=s t^2, phi=s^2 t

Passo 1

O enunciado define
z = sin(theta)cos(phi),
com theta = s,t^2 e phi = s^2t.
Queremos encontrar partial zpartial s e partial zpartial t usando a regra da cadeia para funções de várias variáveis:
partial zpartial s = partial zpartial theta,partial thetapartial s + partial zpartial phi,partial phipartial s, partial zpartial t = partial zpartial theta,partial thetapartial t + partial zpartial phi,partial phipartial t

Passo 2

Calculamos as derivadas parciais de z em relação a theta e phi:
partial zpartial theta = partialpartial thetal[sin(theta)cos(phi)r] = cos(theta),cos(phi),
partial zpartial phi = partialpartial phil[sin(theta)cos(phi)r] = -,sin(theta),sin(phi)

Passo 3

Calculamos as derivadas de theta(s,t) e phi(s,t):
partial thetapartial s = partialpartial s(s,t^2) = t^2, partial thetapartial t = partialpartial t(s,t^2) = 2,s,t,
partial phipartial s = partialpartial s(s^2t) = 2,s,t, partial phipartial t = partialpartial t(s^2t) = s^2

Passo 4

Agora aplicamos a regra da cadeia para partial zpartial s:
partial zpartial s = cos(theta)cos(phi)cdot t^2 ;+;l(-sin(theta)sin(phi)r)cdot(2,s,t)
e substituímos theta = s,t^2, phi = s^2t:
partial zpartial s = t^2,cosl(s,t^2r),cosl(s^2tr) ;-;2,s,t,sinl(s,t^2r),sinl(s^2tr)

Passo 5

De modo análogo, para partial zpartial t temos:
partial zpartial t = cos(theta)cos(phi)cdot(2,s,t) ;+;l(-sin(theta)sin(phi)r)cdot s^2
e, trocando theta e phi:
partial zpartial t = 2,s,t,cosl(s,t^2r),cosl(s^2tr) ;-;s^2,sinl(s,t^2r),sinl(s^2tr)

Resposta

As derivadas parciais são:
partial zpartial s = t^2,cosl(s,t^2r),cosl(s^2tr) - 2,s,t,sinl(s,t^2r),sinl(s^2tr),
partial zpartial t = 2,s,t,cosl(s,t^2r),cosl(s^2tr) - s^2,sinl(s,t^2r),sinl(s^2tr)

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