Questão 4. (6pts) Usando a Regra da Cadeia para funções de várias variáveis, ache \frac{\partial f}{\partial u} e \frac{\partial f}{\partial v} como função de u e v da função
f(x, y)=x^{2} \operatorname{sen}(x y), \quad x=v^{3}+u^{2} \operatorname{sen}(v), \quad y=u^{3}+\ln \left(u^{2} v^{3}\right) .

Passo 1

O enunciado dá as funções e pede as derivadas parciais de f em relação a u e a v, escritas como funções de u e v. As informações fornecidas são:
  • f(x,y)=x^{2}\,\sin(xy).
  • x(v,u)=v^{3}+u^{2}\sin(v).
  • y(u,v)=u^{3}+\ln\!\bigl(u^{2}v^{3}\bigr).
O que é pedido:
  • Achar \dfrac{\partial f}{\partial u} expressa somente em termos de u e v.
  • Achar \dfrac{\partial f}{\partial v} expressa somente em termos de u e v.

Passo 2

Vamos usar a Regra da Cadeia para funções de várias variáveis. Se f=f(x,y) com x=x(u,v) e y=y(u,v), então
\frac{\partial f}{\partial u}=\frac{\partial f}{\partial x}\,\frac{\partial x}{\partial u}+\frac{\partial f}{\partial y}\,\frac{\partial y}{\partial u},
\frac{\partial f}{\partial v}=\frac{\partial f}{\partial x}\,\frac{\partial x}{\partial v}+\frac{\partial f}{\partial y}\,\frac{\partial y}{\partial v}.
Explicando: \dfrac{\partial f}{\partial x} é a derivada parcial de f tratanto y como constante; \dfrac{\partial x}{\partial u} é a derivada parcial de x em relação a u, etc. A fórmula soma as contribuições via as variações de x e de y.

Passo 3

Calcular \dfrac{\partial f}{\partial x} e \dfrac{\partial f}{\partial y} para f(x,y)=x^{2}\sin(xy).
Primeiro \dfrac{\partial f}{\partial x}: aqui f é produto de x^{2} por \sin(xy), então uso a regra do produto e, dentro de \sin(xy), a regra da cadeia.
\frac{\partial f}{\partial x} = \frac{\partial}{\partial x}\bigl(x^{2}\bigr)\cdot\sin(xy) + x^{2}\cdot\frac{\partial}{\partial x}\bigl(\sin(xy)\bigr).
\frac{\partial}{\partial x}\bigl(x^{2}\bigr)=2x.
\frac{\partial}{\partial x}\bigl(\sin(xy)\bigr)=\cos(xy)\cdot\frac{\partial}{\partial x}(xy)=\cos(xy)\cdot y.
Portanto
\frac{\partial f}{\partial x}=2x\sin(xy)+x^{2}y\cos(xy).
Agora \dfrac{\partial f}{\partial y}: trate x como constante e use a regra da cadeia sobre \sin(xy).
\frac{\partial f}{\partial y}=x^{2}\cdot\frac{d}{dy}\bigl(\sin(xy)\bigr)=x^{2}\cos(xy)\cdot\frac{\partial}{\partial y}(xy)=x^{2}\cos(xy)\cdot x.
Logo
\frac{\partial f}{\partial y}=x^{3}\cos(xy).

Passo 4

Calcular as derivadas parciais de x e y em relação a u e v.
Para x(v,u)=v^{3}+u^{2}\sin(v):
\frac{\partial x}{\partial u}=\frac{\partial}{\partial u}\bigl(v^{3}\bigr)+\frac{\partial}{\partial u}\bigl(u^{2}\sin(v)\bigr)=0+2u\sin(v).
\frac{\partial x}{\partial v}=\frac{\partial}{\partial v}\bigl(v^{3}\bigr)+\frac{\partial}{\partial v}\bigl(u^{2}\sin(v)\bigr)=3v^{2}+u^{2}\cos(v).
Para y(u,v)=u^{3}+\ln\!\bigl(u^{2}v^{3}\bigr) use a regra da cadeia para o logaritmo. Uma forma útil é decompor \ln(u^{2}v^{3})=\ln(u^{2})+\ln(v^{3})=2\ln|u|+3\ln|v| (assumindo domínio onde os log são definidos), mas podemos fazer direto pela cadeia:
\frac{\partial y}{\partial u}=\frac{\partial}{\partial u}(u^{3})+\frac{\partial}{\partial u}\bigl(\ln(u^{2}v^{3})\bigr)=3u^{2}+\frac{1}{u^{2}v^{3}}\cdot\frac{\partial}{\partial u}\bigl(u^{2}v^{3}\bigr).
\frac{\partial}{\partial u}\bigl(u^{2}v^{3}\bigr)=2u\,v^{3},
logo
\frac{\partial y}{\partial u}=3u^{2}+\frac{2u\,v^{3}}{u^{2}v^{3}}=3u^{2}+\frac{2}{u}.
Agora \dfrac{\partial y}{\partial v}:
\frac{\partial y}{\partial v}=\frac{\partial}{\partial v}(u^{3})+\frac{\partial}{\partial v}\bigl(\ln(u^{2}v^{3})\bigr)=0+\frac{1}{u^{2}v^{3}}\cdot\frac{\partial}{\partial v}\bigl(u^{2}v^{3}\bigr).
\frac{\partial}{\partial v}\bigl(u^{2}v^{3}\bigr)=u^{2}\cdot 3v^{2},
logo
\frac{\partial y}{\partial v}=\frac{3u^{2}v^{2}}{u^{2}v^{3}}=\frac{3}{v}.
Resumindo as derivadas parciais de x e y:
\frac{\partial x}{\partial u}=2u\sin(v),\qquad \frac{\partial x}{\partial v}=3v^{2}+u^{2}\cos(v),
\frac{\partial y}{\partial u}=3u^{2}+\frac{2}{u},\qquad \frac{\partial y}{\partial v}=\frac{3}{v}.

Passo 5

Agora aplicamos a Regra da Cadeia para \dfrac{\partial f}{\partial u}:
\frac{\partial f}{\partial u}=\frac{\partial f}{\partial x}\,\frac{\partial x}{\partial u}+\frac{\partial f}{\partial y}\,\frac{\partial y}{\partial u}.
Substituindo \dfrac{\partial f}{\partial x}, \dfrac{\partial f}{\partial y}, \dfrac{\partial x}{\partial u} e \dfrac{\partial y}{\partial u},
\frac{\partial f}{\partial u}=\bigl(2x\sin(xy)+x^{2}y\cos(xy)\bigr)\cdot\bigl(2u\sin(v)\bigr)+\bigl(x^{3}\cos(xy)\bigr)\cdot\bigl(3u^{2}+\tfrac{2}{u}\bigr).

Passo 6

Finalmente, substituir x e y pelas expressões em u e v para obter \dfrac{\partial f}{\partial u} somente em termos de u e v. Lembrando que
x=v^{3}+u^{2}\sin(v),\qquad y=u^{3}+\ln\!\bigl(u^{2}v^{3}\bigr).
Portanto
\frac{\partial f}{\partial u}= \Bigl(2\bigl(v^{3}+u^{2}\sin(v)\bigr)\sin\bigl((v^{3}+u^{2}\sin(v))\,(u^{3}+\ln(u^{2}v^{3}))\bigr)
\qquad+\bigl(v^{3}+u^{2}\sin(v)\bigr)^{2}\bigl(u^{3}+\ln(u^{2}v^{3})\bigr)\cos\bigl((v^{3}+u^{2}\sin(v))\,(u^{3}+\ln(u^{2}v^{3}))\bigr)\Bigr)\cdot(2u\sin(v))
\qquad+\bigl(v^{3}+u^{2}\sin(v)\bigr)^{3}\cos\bigl((v^{3}+u^{2}\sin(v))\,(u^{3}+\ln(u^{2}v^{3}))\bigr)\cdot\Bigl(3u^{2}+\frac{2}{u}\Bigr).
Essa é a expressão de \dfrac{\partial f}{\partial u} como função de u e v.

Passo 7

Agora fazemos \dfrac{\partial f}{\partial v} usando a mesma ideia:
\frac{\partial f}{\partial v}=\frac{\partial f}{\partial x}\,\frac{\partial x}{\partial v}+\frac{\partial f}{\partial y}\,\frac{\partial y}{\partial v}.
Substituindo as derivadas já calculadas:
\frac{\partial f}{\partial v}=\bigl(2x\sin(xy)+x^{2}y\cos(xy)\bigr)\cdot\bigl(3v^{2}+u^{2}\cos(v)\bigr)+\bigl(x^{3}\cos(xy)\bigr)\cdot\frac{3}{v}.

Passo 8

Substituir x=v^{3}+u^{2}\sin(v) e y=u^{3}+\ln(u^{2}v^{3}) em \dfrac{\partial f}{\partial v} para obter a forma final em u e v:
\frac{\partial f}{\partial v}= \Bigl(2\bigl(v^{3}+u^{2}\sin(v)\bigr)\sin\bigl((v^{3}+u^{2}\sin(v))\,(u^{3}+\ln(u^{2}v^{3}))\bigr)
\qquad+\bigl(v^{3}+u^{2}\sin(v)\bigr)^{2}\bigl(u^{3}+\ln(u^{2}v^{3})\bigr)\cos\bigl((v^{3}+u^{2}\sin(v))\,(u^{3}+\ln(u^{2}v^{3}))\bigr)\Bigr)
\qquad\cdot\bigl(3v^{2}+u^{2}\cos(v)\bigr) +\bigl(v^{3}+u^{2}\sin(v)\bigr)^{3}\cos\bigl((v^{3}+u^{2}\sin(v))\,(u^{3}+\ln(u^{2}v^{3}))\bigr)\cdot\frac{3}{v}.

Resposta

\frac{\partial f}{\partial u}= \Bigl(2\bigl(v^{3}+u^{2}\sin(v)\bigr)\sin\bigl((v^{3}+u^{2}\sin(v))\,(u^{3}+\ln(u^{2}v^{3}))\bigr)
\qquad+\bigl(v^{3}+u^{2}\sin(v)\bigr)^{2}\bigl(u^{3}+\ln(u^{2}v^{3})\bigr)\cos\bigl((v^{3}+u^{2}\sin(v))\,(u^{3}+\ln(u^{2}v^{3}))\bigr)\Bigr)\cdot(2u\sin(v))
\qquad+\bigl(v^{3}+u^{2}\sin(v)\bigr)^{3}\cos\bigl((v^{3}+u^{2}\sin(v))\,(u^{3}+\ln(u^{2}v^{3}))\bigr)\cdot\Bigl(3u^{2}+\frac{2}{u}\Bigr),
\frac{\partial f}{\partial v}= \Bigl(2\bigl(v^{3}+u^{2}\sin(v)\bigr)\sin\bigl((v^{3}+u^{2}\sin(v))\,(u^{3}+\ln(u^{2}v^{3}))\bigr)
\qquad+\bigl(v^{3}+u^{2}\sin(v)\bigr)^{2}\bigl(u^{3}+\ln(u^{2}v^{3})\bigr)\cos\bigl((v^{3}+u^{2}\sin(v))\,(u^{3}+\ln(u^{2}v^{3}))\bigr)\Bigr)
\qquad\cdot\bigl(3v^{2}+u^{2}\cos(v)\bigr) +\bigl(v^{3}+u^{2}\sin(v)\bigr)^{3}\cos\bigl((v^{3}+u^{2}\sin(v))\,(u^{3}+\ln(u^{2}v^{3}))\bigr)\cdot\frac{3}{v}.

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