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)=cos (x)sen(y).

Passo 1

O enunciado pede para calcular as derivadas parciais de z = f(u,v) onde
f(u,v) = uv^2, u(x,y) = sin(x)cos(y), v(x,y) = cos(x)sin(y)
O conteúdo é aplicação da regra da cadeia para funções de duas variáveis.

Passo 2

Pela regra da cadeia, temos
partial zpartial x = partial fpartial u,partial upartial x + partial fpartial v,partial vpartial x,
partial zpartial y = partial fpartial u,partial upartial y + partial fpartial v,partial vpartial y

Passo 3

Calculamos as derivadas parciais intermediárias:
partial fpartial u = 1v^2, partial fpartial v = -2,uv^3,
partial upartial x = partialpartial xl(sin(x)sec(y)r) = cos(x)sec(y), partial upartial y = sin(x),partialpartial yl(sec(y)r) = sin(x)sec(y)tan(y),
partial vpartial x = partialpartial xl(cos(x)csc(y)r) = -sin(x)csc(y), partial vpartial y = cos(x),partialpartial yl(csc(y)r) = -cos(x)csc(y)cot(y)

Passo 4

Substituindo em partial z/partial x:
partial zpartial x = 1v^2l(cos(x)sec(y)r) + l(-2,uv^3r)l(-sin(x)csc(y)r)
Escrevendo sec(y)=1/cos(y) e csc(y)=1/sin(y),
partial zpartial x = cos(x)v^2cos(y) + 2,u,sin(x)v^3sin(y)
Como u=sin(x)/cos(y) e v=cos(x)/sin(y), obtemos
partial zpartial x = sin^2(y)cos(x)cos(y) + 2,sin^2(x)sin^2(y)cos^3(x)cos(y)

Passo 5

Substituindo em partial z/partial y:
partial zpartial y = 1v^2l(sin(x)sec(y)tan(y)r) + l(-2,uv^3r)l(-cos(x)csc(y)cot(y)r)
Usando tan(y)=sin(y)/cos(y) e cot(y)=cos(y)/sin(y),
partial zpartial y = sin(x)sin^3(y)cos^2(x)cos^2(y) + 2,sin(x)sin(y)cos^2(x)

Resposta

partial zpartial x = sin^2(y)cos(x),cos(y) + 2,sin^2(x),sin^2(y)cos^3(x),cos(y),
partial zpartial y = sin(x),sin^3(y)cos^2(x),cos^2(y) + 2,sin(x),sin(y)cos^2(x)

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