Dado z=\operatorname{sen}\left(3 x^{2}+y^{2}\right), x=u^{2}-v^{2}, y=2 u v. Encontre \frac{\partial z}{\partial u} e \frac{\partial z}{\partial v}.
Dado w=x^{2}+y^{2}+z^{2}, x=2 u-3 v, y=u+v, z=u-v. Encontre \frac{\partial w}{\partial u} e \frac{\partial w}{\partial v}.
Dado z=e^{\tan (3 x-6 y)}, x=t^{2}-s, y=s^{2}-t. Encontre \frac{\partial z}{\partial t} e \frac{\partial z}{\partial s}.
Dado w=e^{x^{2}+y^{2}+z^{2}}, x=p^{2}-q t, y=t q, z=p^{2}+t. Encontre \frac{\partial w}{\partial p} e \frac{\partial w}{\partial t}.
Dado w=y^{2}+4 x^{2}-z, x=u^{2}-v^{2}, y=3 u v, z=2 u-v. Encontre \frac{\partial w}{\partial u} \mathrm{e} \frac{\partial w}{\partial v}.

Passo 1

Vou listar claramente cada subitem do enunciado e as informações fornecidas (sem resolver ainda):
  1. Dados:
  • z=\sin\!\bigl(3x^{2}+y^{2}\bigr),
  • x=u^{2}-v^{2},
  • y=2uv.
    Pediram \dfrac{\partial z}{\partial u} e \dfrac{\partial z}{\partial v}.
  1. Dados:
  • w=x^{2}+y^{2}+z^{2},
  • x=2u-3v,
  • y=u+v,
  • z=u-v.
    Pediram \dfrac{\partial w}{\partial u} e \dfrac{\partial w}{\partial v}.
  1. Dados:
  • z=e^{\tan(3x-6y)},
  • x=t^{2}-s,
  • y=s^{2}-t.
    Pediram \dfrac{\partial z}{\partial t} e \dfrac{\partial z}{\partial s}.
  1. Dados:
  • w=e^{x^{2}+y^{2}+z^{2}},
  • x=p^{2}-qt,
  • y=tq,
  • z=p^{2}+t.
    Pediram \dfrac{\partial w}{\partial p} e \dfrac{\partial w}{\partial t}.
  1. Dados:
  • w=y^{2}+4x^{2}-z,
  • x=u^{2}-v^{2},
  • y=3uv,
  • z=2u-v.
    Pediram \dfrac{\partial w}{\partial u} e \dfrac{\partial w}{\partial v}.

Passo 2

Vou resolver o item 1 passo a passo. Regra utilizada: regra da cadeia para derivadas parciais. Defino \varphi=3x^{2}+y^{2} e z=\sin\varphi. A regra da cadeia diz:
\frac{\partial z}{\partial u}=\frac{\partial z}{\partial x}\frac{\partial x}{\partial u}+\frac{\partial z}{\partial y}\frac{\partial y}{\partial u}, \qquad \frac{\partial z}{\partial v}=\frac{\partial z}{\partial x}\frac{\partial x}{\partial v}+\frac{\partial z}{\partial y}\frac{\partial y}{\partial v}.
Agora calculo cada derivada parcial necessária.

Passo 3

Calculo as derivadas parciais internas para o item 1:
  • Como z=\sin\varphi, \dfrac{\partial z}{\partial \varphi}=\cos\varphi. Aplicando \varphi=3x^{2}+y^{2}:
\frac{\partial z}{\partial x}=\frac{\partial z}{\partial \varphi}\frac{\partial \varphi}{\partial x} =\cos(3x^{2}+y^{2})\cdot(6x)=6x\cos(3x^{2}+y^{2}).
\frac{\partial z}{\partial y}=\frac{\partial z}{\partial \varphi}\frac{\partial \varphi}{\partial y} =\cos(3x^{2}+y^{2})\cdot(2y)=2y\cos(3x^{2}+y^{2}).
Derivadas de x e y em relação a u,v:
\frac{\partial x}{\partial u}=2u,\quad \frac{\partial x}{\partial v}=-2v,\quad \frac{\partial y}{\partial u}=2v,\quad \frac{\partial y}{\partial v}=2u.

Passo 4

Agora monto as derivadas pedidas (item 1) usando as fórmulas da cadeia:
\frac{\partial z}{\partial u} =6x\cos(3x^{2}+y^{2})\cdot 2u \;+\; 2y\cos(3x^{2}+y^{2})\cdot 2v.
Fatorando \cos(3x^{2}+y^{2}):
\frac{\partial z}{\partial u} =\cos(3x^{2}+y^{2})\bigl(12xu+4yv\bigr).
Substituo x=u^{2}-v^{2} e y=2uv e simplifico:
12xu+4yv=12u(u^{2}-v^{2})+4(2uv)v=12u^{3}-12uv^{2}+8uv^{2}=12u^{3}-4uv^{2}=4u(3u^{2}-v^{2}).
Portanto
\frac{\partial z}{\partial u}=4u(3u^{2}-v^{2})\cos\!\bigl(3(u^{2}-v^{2})^{2}+(2uv)^{2}\bigr).

Passo 5

Faço o mesmo para \dfrac{\partial z}{\partial v} (item 1):
\frac{\partial z}{\partial v} =6x\cos(3x^{2}+y^{2})\cdot(-2v) \;+\; 2y\cos(3x^{2}+y^{2})\cdot 2u =\cos(3x^{2}+y^{2})\bigl(-12xv+4yu\bigr).
Substituindo x=u^{2}-v^{2},\,y=2uv:
-12xv+4yu=-12v(u^{2}-v^{2})+4(2uv)u=-12u^{2}v+12v^{3}+8u^{2}v= -4u^{2}v+12v^{3}=4v(3v^{2}-u^{2}).
Portanto
\frac{\partial z}{\partial v}=4v(3v^{2}-u^{2})\cos\!\bigl(3(u^{2}-v^{2})^{2}+(2uv)^{2}\bigr).

Passo 6

Vou resolver o item 2. Função w=x^{2}+y^{2}+z^{2}. Regra da cadeia (três variáveis internas):
\frac{\partial w}{\partial u}=w_{x}x_{u}+w_{y}y_{u}+w_{z}z_{u},\qquad \frac{\partial w}{\partial v}=w_{x}x_{v}+w_{y}y_{v}+w_{z}z_{v}.
Calculo w_x,w_y,w_z e as derivadas de x,y,z.

Passo 7

Derivadas necessárias (item 2):
w_x=2x,\quad w_y=2y,\quad w_z=2z.
Derivadas internas:
x_u=2,\; x_v=-3,\quad y_u=1,\; y_v=1,\quad z_u=1,\; z_v=-1.
Agora calculo \dfrac{\partial w}{\partial u}:
\frac{\partial w}{\partial u}=2x\cdot 2 + 2y\cdot 1 + 2z\cdot 1 =4x+2y+2z.
Substituindo x=2u-3v,\;y=u+v,\;z=u-v:
4(2u-3v)+2(u+v)+2(u-v)=8u-12v+2u+2v+2u-2v=12u-12v=12(u-v).

Passo 8

Calculo \dfrac{\partial w}{\partial v} (item 2):
\frac{\partial w}{\partial v}=2x\cdot(-3)+2y\cdot 1 +2z\cdot(-1)= -6x+2y-2z.
Substituindo x,y,z:
-6(2u-3v)+2(u+v)-2(u-v)=-12u+18v+2u+2v-2u+2v=-12u+22v.
Portanto \dfrac{\partial w}{\partial v}=-12u+22v.

Passo 9

Vou resolver o item 3. Função z=e^{\tan(3x-6y)}. Novamente regra da cadeia:
Defino \psi=\tan(3x-6y) e z=e^{\psi}. Então
z_x = e^{\psi}\psi_x,\quad z_y=e^{\psi}\psi_y,
e \psi_x=\operatorname{sec}^{2}(3x-6y)\cdot 3,\\ \psi_y=\operatorname{sec}^{2}(3x-6y)\cdot(-6).
As derivadas de x,y em função de t,s: x_t=2t,\\ x_s=-1,\\ y_t=-1,\\ y_s=2s.

Passo 10

Calculo \dfrac{\partial z}{\partial t} (item 3):
\frac{\partial z}{\partial t}=z_x x_t + z_y y_t = e^{\psi}\operatorname{sec}^{2}(3x-6y)\bigl(3\cdot 2t + (-6)\cdot(-1)\bigr).
Simplifico o parêntese: 3\cdot 2t + (-6)(-1)=6t+6=6(t+1). Logo
\frac{\partial z}{\partial t}=6(t+1)\,e^{\tan(3x-6y)}\operatorname{sec}^{2}(3x-6y).
Substituo x=t^{2}-s,\\ y=s^{2}-t se quiser a expressão totalmente em t,s. O argumento é
3x-6y=3(t^{2}-s)-6(s^{2}-t).

Passo 11

Calculo \dfrac{\partial z}{\partial s} (item 3):
\frac{\partial z}{\partial s}=z_x x_s + z_y y_s = e^{\psi}\operatorname{sec}^{2}(3x-6y)\bigl(3\cdot(-1) + (-6)\cdot(2s)\bigr).
Simplificando: 3(-1)+(-6)(2s)=-3-12s=-3(1+4s). Assim
\frac{\partial z}{\partial s}=-3(1+4s)\,e^{\tan(3x-6y)}\operatorname{sec}^{2}(3x-6y),
com x,y substituídos conforme dado se necessário.

Passo 12

Vou resolver o item 4. Função w=e^{x^{2}+y^{2}+z^{2}}. Defino \eta=x^{2}+y^{2}+z^{2}, então w=e^{\eta} e
w_x=2x e^{\eta},\quad w_y=2y e^{\eta},\quad w_z=2z e^{\eta}.
Cálculo as derivadas parciais de x,y,z em relação a p,t:
x_p=2p,\\ x_t=-q;\quad y_p=0,\\ y_t=q;\quad z_p=2p,\\ z_t=1.

Passo 13

Calculo \dfrac{\partial w}{\partial p} (item 4):
\frac{\partial w}{\partial p}=w_x x_p + w_y y_p + w_z z_p = e^{\eta}\bigl(2x\cdot 2p + 2y\cdot 0 + 2z\cdot 2p\bigr) =4p\,e^{\eta}(x+z).
Substituindo x=p^{2}-qt,\\ z=p^{2}+t:
x+z=2p^{2}+t(1-q),
portanto
\frac{\partial w}{\partial p}=4p\bigl(2p^{2}+t(1-q)\bigr)\,e^{x^{2}+y^{2}+z^{2}}.

Passo 14

Calculo \dfrac{\partial w}{\partial t} (item 4):
\frac{\partial w}{\partial t}=w_x x_t + w_y y_t + w_z z_t = e^{\eta}\bigl(2x(-q)+2y(q)+2z(1)\bigr) =2e^{\eta}\bigl(-qx+qy+z\bigr).
Substituindo x=p^{2}-qt,\\ y=tq,\\ z=p^{2}+t:
-qx+qy+z=-q(p^{2}-qt)+q(tq)+(p^{2}+t)= -q p^{2}+q^{2}t + q^{2}t + p^{2}+t.
Agrupando:
= p^{2}(1-q) + t(1+2q^{2}).
Logo
\frac{\partial w}{\partial t}=2\bigl(p^{2}(1-q)+t(1+2q^{2})\bigr)\,e^{x^{2}+y^{2}+z^{2}}.

Passo 15

Vou resolver o item 5. Função w=y^{2}+4x^{2}-z. Regra da cadeia:
\frac{\partial w}{\partial u}=w_x x_u + w_y y_u + w_z z_u, \qquad \frac{\partial w}{\partial v}=w_x x_v + w_y y_v + w_z z_v.
Calculo w_x,w_y,w_z e as derivadas internas.

Passo 16

Derivadas necessárias (item 5):
w_x=8x,\quad w_y=2y,\quad w_z=-1.
Derivadas de x,y,z:
x_u=2u,\;x_v=-2v,\quad y_u=3v,\;y_v=3u,\quad z_u=2,\;z_v=-1.

Passo 17

Calculo \dfrac{\partial w}{\partial u} (item 5):
\frac{\partial w}{\partial u}=8x\cdot 2u + 2y\cdot 3v + (-1)\cdot 2 =16xu + 6yv -2.
Substituindo x=u^{2}-v^{2},\\ y=3uv:
16u(u^{2}-v^{2}) + 6(3uv)v -2 =16u^{3}-16uv^{2}+18uv^{2}-2=16u^{3}+2uv^{2}-2.
Pode também escrever como 2(8u^{3}+uv^{2}-1).

Passo 18

Calculo \dfrac{\partial w}{\partial v} (item 5):
\frac{\partial w}{\partial v}=8x\cdot(-2v) + 2y\cdot 3u + (-1)\cdot(-1) =-16xv +6yu +1.
Substituindo x=u^{2}-v^{2},\\ y=3uv:
-16v(u^{2}-v^{2}) + 6(3uv)u +1 = -16u^{2}v +16v^{3}+18u^{2}v +1 =2u^{2}v+16v^{3}+1.

Resposta

  1. \displaystyle \frac{\partial z}{\partial u}=4u(3u^{2}-v^{2})\cos\!\bigl(3(u^{2}-v^{2})^{2}+(2uv)^{2}\bigr)
\displaystyle \frac{\partial z}{\partial v}=4v(3v^{2}-u^{2})\cos\!\bigl(3(u^{2}-v^{2})^{2}+(2uv)^{2}\bigr)
  1. \displaystyle \frac{\partial w}{\partial u}=12(u-v)
\displaystyle \frac{\partial w}{\partial v}=-12u+22v
  1. \displaystyle \frac{\partial z}{\partial t}=6(t+1)\,e^{\tan(3x-6y)}\operatorname{sec}^{2}(3x-6y) com x=t^{2}-s,\\ y=s^{2}-t
\displaystyle \frac{\partial z}{\partial s}=-3(1+4s)\,e^{\tan(3x-6y)}\operatorname{sec}^{2}(3x-6y) com x=t^{2}-s,\\ y=s^{2}-t
  1. \displaystyle \frac{\partial w}{\partial p}=4p\bigl(2p^{2}+t(1-q)\bigr)\,e^{x^{2}+y^{2}+z^{2}} com x=p^{2}-qt,\\ y=tq,\\ z=p^{2}+t
\displaystyle \frac{\partial w}{\partial t}=2\bigl(p^{2}(1-q)+t(1+2q^{2})\bigr)\,e^{x^{2}+y^{2}+z^{2}} com x=p^{2}-qt,\\ y=tq,\\ z=p^{2}+t
  1. \displaystyle \frac{\partial w}{\partial u}=16u^{3}+2uv^{2}-2
\displaystyle \frac{\partial w}{\partial v}=2u^{2}v+16v^{3}+1

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