Use a regra da cadeia para encontrar partial zpartial s e partial zpartial t
z=sen alpha cos (alpha+beta), alpha=2 s+3 t, beta=raizt s,

Passo 1

O enunciado define
z = sin(alpha)cos(alpha+beta), alpha = 2s + 3t, beta = raizts,
e pede partial z/partial s e partial z/partial t.
O conteúdo é a regra da cadeia para derivadas parciais: primeiro encontramos partial z/partial alpha e partial z/partial beta, depois partial alpha/partial s, partial beta/partial s, partial alpha/partial t, partial beta/partial t e combinamos.

Passo 2

Derivamos z(alpha,beta)=sin(alpha)cos(alpha+beta) em relação a alpha e beta usando produto e cadeia:
partial zpartial alpha = cos(alpha)cos(alpha+beta) -sin(alpha)sin(alpha+beta),
partial zpartial beta = sin(alpha),ddbetal[cos(alpha+beta)r] = -sin(alpha)sin(alpha+beta)

Passo 3

Calculamos as derivadas de alpha e beta em relação a s:
partial alphapartial s=2, partial betapartial s =ddsl[(ts)^1/2r] =tfrac12(t,s)^-1/2,t =t2raizts

Passo 4

Aplicamos a regra da cadeia para partial z/partial s:
partial zpartial s = partial zpartial alpha,partial alphapartial s + partial zpartial beta,partial betapartial s
Substituindo,
partial zpartial s =2l[cos(alpha)cos(alpha+beta)-sin(alpha)sin(alpha+beta)r] -t2raizts;sin(alpha)sin(alpha+beta)

Passo 5

Agora as derivadas de alpha e beta em relação a t:
partial alphapartial t=3, partial betapartial t =12(t,s)^-1/2,s =s2raizts

Passo 6

Aplicamos a regra da cadeia para partial z/partial t:
partial zpartial t = partial zpartial alpha,partial alphapartial t + partial zpartial beta,partial betapartial t,
ou seja,
partial zpartial t =3l[cos(alpha)cos(alpha+beta)-sin(alpha)sin(alpha+beta)r] -s2raizts;sin(alpha)sin(alpha+beta)

Resposta

partial zpartial s =2l[cos(alpha)cos(alpha+beta)-sin(alpha)sin(alpha+beta)r] -t2raizts;sin(alpha)sin(alpha+beta),
partial zpartial t =3l[cos(alpha)cos(alpha+beta)-sin(alpha)sin(alpha+beta)r] -s2raizts;sin(alpha)sin(alpha+beta)

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