Nos exercícios abaixo:
Calculo o divergente e o rotacional do campo \vec{F}(x, y, z)=\left(e^{x} y z, \cos (x y z), x y z\right).

Passo 1

O enunciado pede para calcular o divergente e o rotacional do campo \vec{F}(x,y,z) = \bigl(e^{x}yz,\;\cos(xyz),\;xyz\bigr).
Estamos em Cálculo Vetorial, usando derivadas parciais para obter
  1. o divergente \nabla\cdot\vec{F} e
  2. o rotacional \nabla\times\vec{F}.

Passo 2

Para o divergente usamos
\nabla\cdot\vec{F} = \frac{\partial}{\partial x}(F_{1}) + \frac{\partial}{\partial y}(F_{2}) + \frac{\partial}{\partial z}(F_{3}).
Aqui F_{1}=e^{x}yz, F_{2}=\cos(xyz) e F_{3}=xyz.
Calculando cada termo:
\frac{\partial}{\partial x}(e^{x}yz) = e^{x}yz,
\frac{\partial}{\partial y}\bigl(\cos(xyz)\bigr) = -\sin(xyz)\,\frac{\partial}{\partial y}(xyz) = -\sin(xyz)\,(xz),
\frac{\partial}{\partial z}(xyz) = xy.
Somando-os,
\nabla\cdot\vec{F} = e^{x}yz \;-\; xz\,\sin(xyz)\;+\;xy.

Passo 3

Para o rotacional usamos
\nabla\times\vec{F} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\\ \partial_x & \partial_y & \partial_z \\\ F_{1} & F_{2} & F_{3} \end{vmatrix},
ou seja, em componentes:
(\nabla\times\vec{F})_{1} = \frac{\partial F_{3}}{\partial y} - \frac{\partial F_{2}}{\partial z},
(\nabla\times\vec{F})_{2} = \frac{\partial F_{1}}{\partial z} - \frac{\partial F_{3}}{\partial x},
(\nabla\times\vec{F})_{3} = \frac{\partial F_{2}}{\partial x} - \frac{\partial F_{1}}{\partial y}.
Agora, cada derivada:
\frac{\partial F_{3}}{\partial y} = \frac{\partial (xyz)}{\partial y} = xz, \quad \frac{\partial F_{2}}{\partial z} = -\sin(xyz)\,\frac{\partial (xyz)}{\partial z} = -\sin(xyz)\,(xy),
\frac{\partial F_{1}}{\partial z} = \frac{\partial (e^{x}yz)}{\partial z} = e^{x}y, \quad \frac{\partial F_{3}}{\partial x} = \frac{\partial (xyz)}{\partial x} = yz,
\frac{\partial F_{2}}{\partial x} = -\sin(xyz)\,\frac{\partial (xyz)}{\partial x} = -\sin(xyz)\,(yz), \quad \frac{\partial F_{1}}{\partial y} = \frac{\partial (e^{x}yz)}{\partial y} = e^{x}z.
Substituindo:
(\nabla\times\vec{F})_{1} = xz \;-\;\bigl(-xy\,\sin(xyz)\bigr) = xz + xy\,\sin(xyz),
(\nabla\times\vec{F})_{2} = e^{x}y \;-\; yz,
(\nabla\times\vec{F})_{3} = -\,yz\,\sin(xyz)\;-\;e^{x}z.

Resposta

Divergente:
\nabla\cdot\vec{F} = e^{x}yz \;-\; xz\,\sin(xyz)\;+\;xy.
Rotacional:
\nabla\times\vec{F} = \bigl(xz + xy\,\sin(xyz),\;e^{x}y - yz,\;-\,yz\,\sin(xyz) - e^{x}z\bigr).

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