Encontre as derivadas parciais partial zpartial x e partial zpartial y sabendo que z=f(u, v) com u=u(x, y) e v=v(x, y)
f(u, v)=uv^2, u(x, y)=sen(x)cos (y) e v(x, y)=3 x-2 y

Passo 1

O enunciado pede as derivadas parciais de
z = f(u,v)
com
f(u,v)=uv^2, u(x,y)=sin(x)cos(y), v(x,y)=3x-2y
O conteúdo principal é a regra da cadeia para funções de duas variáveis:
partial zpartial x = fu,partial upartial x + fv,partial vpartial x, partial zpartial y = fu,partial upartial y + fv,partial vpartial y

Passo 2

Calcular as derivadas parciais de f(u,v)=u,v^-2 em relação a u e a v:
fu = partialpartial ul(u,v^-2r) = v^-2, fv = partialpartial vl(u,v^-2r) = u,(-2),v^-3 = -2uv^3

Passo 3

Calcular partial u/partial x e partial v/partial x:
u(x,y)=sin(x)cos(y) ;Longrightarrow; partial upartial x = cos(x)cos(y),
v(x,y)=3x-2y ;Longrightarrow; partial vpartial x =3

Passo 4

Aplicar a regra da cadeia para partial z/partial x:
partial zpartial x = fu,partial upartial x + fv,partial vpartial x = 1v^2,cos(x)cos(y) + l(-2uv^3r),3
Substituindo u=dfracsin(x)cos(y) e v=3x-2y,
partial zpartial x = cos(x)cos(y),(3x-2y)^2 - 6,sin(x)cos(y),(3x-2y)^3

Passo 5

Calcular partial u/partial y e partial v/partial y:
u(x,y)=sin(x)cos(y) ;Longrightarrow; partial upartial y = sin(x),ddyl(sec(y)r) = sin(x),sin(y)cos^2(y) = sin(x),sin(y)cos^2(y),
v(x,y)=3x-2y ;Longrightarrow; partial vpartial y =-2

Passo 6

Aplicar a regra da cadeia para partial z/partial y:
partial zpartial y = fu,partial upartial y + fv,partial vpartial y = 1v^2,sin(x),sin(y)cos^2(y) + l(-2uv^3r)(-2)
Substituindo u e v
partial zpartial y = sin(x),sin(y)cos^2(y),(3x-2y)^2 + 4,sin(x)cos(y),(3x-2y)^3

Resposta

boxed partial zpartial x = cos(x)cos(y),(3x-2y)^2 - 6,sin(x)cos(y),(3x-2y)^3, partial zpartial y = sin(x),sin(y)cos^2(y),(3x-2y)^2 + 4,sin(x)cos(y),(3x-2y)^3

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