Utilize a Regra da Cadeia para determinar as derivadas parciais indicadas.
6) t=z cdot sec (x cdot y), x=u cdot v, y=v cdot w, z=w cdot u ; partial tpartial u, partial tpartial v, partial tpartial w.

Passo 1

O enunciado define
t = z ,sec(xy), x = u,v, y = v,w, z = w,u
e pede as derivadas parciais partial tpartial u, partial tpartial v e partial tpartial w.
O conteúdo é aplicação da Regra da Cadeia em funções compostas de várias variáveis.

Passo 2

Primeiro, vamos achar as derivadas parciais de t em relação a x,y,z.
Como sec(u)'!=!sec(u)tan(u), temos
partial tpartial z = sec(xy),
partial tpartial x = z ,ddxl[sec(xy)r] = z,sec(xy)tan(xy),ddx(xy) = z,y,sec(xy)tan(xy),
partial tpartial y = z,sec(xy)tan(xy),ddy(xy) = z,x,sec(xy)tan(xy)

Passo 3

Agora, as derivadas de x,y,z em relação a u,v,w:
partial xpartial u=v, partial xpartial v=u, partial xpartial w=0,
partial ypartial u=0, partial ypartial v=w, partial ypartial w=v,
partial zpartial u=w, partial zpartial v=0, partial zpartial w=u

Passo 4

Aplicando a Regra da Cadeia:
partial tpartial u = partial tpartial x,partial xpartial u + partial tpartial y,partial ypartial u + partial tpartial z,partial zpartial u,
partial tpartial v = partial tpartial x,partial xpartial v + partial tpartial y,partial ypartial v + partial tpartial z,partial zpartial v,
partial tpartial w = partial tpartial x,partial xpartial w + partial tpartial y,partial ypartial w + partial tpartial z,partial zpartial w
Substituindo cada termo:
partial tpartial u = l(z,y,sec(xy)tan(xy)r),v + l(z,x,sec(xy)tan(xy)r),0 + l(sec(xy)r),w,
partial tpartial v = l(z,y,sec(xy)tan(xy)r),u + l(z,x,sec(xy)tan(xy)r),w + l(sec(xy)r),0,
partial tpartial w = l(z,y,sec(xy)tan(xy)r),0 + l(z,x,sec(xy)tan(xy)r),v + l(sec(xy)r),u

Resposta

partial tpartial u = z,y,v,sec(xy)tan(xy) + w,sec(xy),
partial tpartial v = z,y,u,sec(xy)tan(xy) + z,x,w,sec(xy)tan(xy),
partial tpartial w = z,x,v,sec(xy)tan(xy) + u,sec(xy)

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