Use a regra da cadeia nos exercícios abaixo:
Se w=u v+v^{2} ; u=x \sin y, v=y \sin x determine \frac{\partial w}{\partial x} \mathrm{e} \frac{\partial w}{\partial y}

Passo 1

Beleza, o enunciado diz que
w = u\,v + v^{2},\quad u = x \sin y,\quad v = y \sin x
e pede \frac{\partial w}{\partial x} e \frac{\partial w}{\partial y}. O conteúdo é regra da cadeia para funções de duas variáveis.

Passo 2

Para achar \frac{\partial w}{\partial x} usamos a regra da cadeia:
\frac{\partial w}{\partial x} = \frac{\partial w}{\partial u}\,\frac{\partial u}{\partial x} + \frac{\partial w}{\partial v}\,\frac{\partial v}{\partial x}.

Passo 3

Como w = u\,v + v^{2}, temos
\frac{\partial w}{\partial u} = v, \quad \frac{\partial w}{\partial v} = u + 2v.
Também
u = x\sin y \quad\Longrightarrow\quad \frac{\partial u}{\partial x} = \sin y,
v = y\sin x \quad\Longrightarrow\quad \frac{\partial v}{\partial x} = y\cos x.

Passo 4

Substituindo tudo em \frac{\partial w}{\partial x},
\frac{\partial w}{\partial x} = v\,\sin y + (u+2v)\,y\cos x.
E voltando a x,y:
\frac{\partial w}{\partial x} = y\sin x\;\sin y + \bigl(x\sin y + 2y\sin x\bigr)\,y\cos x.

Passo 5

Para \frac{\partial w}{\partial y} aplicamos a mesma ideia:
\frac{\partial w}{\partial y} = \frac{\partial w}{\partial u}\,\frac{\partial u}{\partial y} + \frac{\partial w}{\partial v}\,\frac{\partial v}{\partial y}.

Passo 6

Já temos \frac{\partial w}{\partial u}=v e \frac{\partial w}{\partial v}=u+2v. Como
u = x\sin y \quad\Longrightarrow\quad \frac{\partial u}{\partial y} = x\cos y,
v = y\sin x \quad\Longrightarrow\quad \frac{\partial v}{\partial y} = \sin x,
substituindo dá
\frac{\partial w}{\partial y} = v\,x\cos y + (u+2v)\,\sin x
e, em x,y:
\frac{\partial w}{\partial y} = y\sin x\;x\cos y + \bigl(x\sin y + 2y\sin x\bigr)\,\sin x.

Resposta

\displaystyle \frac{\partial w}{\partial x} = y\sin x\,\sin y + (x\sin y + 2y\sin x)\,y\cos x, \displaystyle \frac{\partial w}{\partial y} = y\sin x\,x\cos y + (x\sin y + 2y\sin x)\,\sin x.

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