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Part of problem 4 on page 65.
Let $w$ be a scalar field and let $\vec{\varphi}$ be a vector field. Show that
$$\nabla \cdot (w\vec{\varphi}) = \vec{\varphi}\cdot \nabla w + w \nabla \cdot \vec{\varphi}.$$
$\textbf{Theorem}$. Suppose $\omega(x,y,z)$ is a scalar field and $\vec{\varphi}(x,y,z)$ is a vector field. Then,
$$\nabla\cdot\left(\omega\vec{\varphi}\right)=\vec{\varphi}\cdot\nabla\omega+\omega\nabla\cdot\vec{\varphi}.$$
$\textit{Proof.}$
$$\nabla\cdot\left(\omega\vec{\varphi}\right)=\left\langle \frac{\partial}{\partial x},\frac{\partial}{\partial y}, \frac{\partial}{\partial z}\right\rangle\cdot\omega\left\langle\varphi_x,\varphi_y,\varphi_z\right\rangle$$
$$=\left\langle \frac{\partial}{\partial x},\frac{\partial}{\partial y}, \frac{\partial}{\partial z}\right\rangle\cdot\left\langle\omega\varphi_x,\omega\varphi_y,\omega\varphi_z\right\rangle$$
$$=\frac{\partial(\omega\varphi_x)}{\partial x}+\frac{\partial(\omega\varphi_y)}{\partial y}+\frac{\partial(\omega\varphi_z)}{\partial z}$$
$$=\left(\frac{\partial\omega}{\partial x}\varphi_x+\frac{\partial\varphi_x}{\partial x}\omega\right)+\left(\frac{\partial\omega}{\partial y}\varphi_y+\frac{\partial\varphi_y}{\partial y}\omega\right)+\left(\frac{\partial\omega}{\partial z}\varphi_z+\frac{\partial\varphi_z}{\partial z}\omega\right)$$
$$=\left(\varphi_x\frac{\partial\omega}{\partial x}+\varphi_y\frac{\partial\omega}{\partial y}+\varphi_z\frac{\partial\omega}{\partial z}\right)+\omega\left(\frac{\partial\varphi_x}{\partial x}+\frac{\partial\varphi_y}{\partial y}+\frac{\partial\varphi_z}{\partial z}\right)$$
$$=\vec{\varphi}\cdot\nabla\omega+\omega\nabla\cdot\vec{\varphi}. $$
$$ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \square$$
@dualrey This looks great! For those reading along - this is what we call a *componentwise proof*.
Let $w$ be a scalar field and let $\vec{\varphi}$ be a vector field. Then
$$\nabla\cdot\left(w\vec{\varphi}\right) = \left\langle \frac{\partial}{\partial x},\frac{\partial}{\partial y}, \frac{\partial}{\partial z}\right\rangle\cdot w\left\langle\varphi_x,\varphi_y,\varphi_z\right\rangle.$$
Since $w$ is a scalar, we multiply all the vector components of $\varphi$ by $w$ to get
$$\left\langle \frac{\partial}{\partial x},\frac{\partial}{\partial y}, \frac{\partial}{\partial z}\right\rangle\cdot\left\langle w\varphi_x,w\varphi_y,w\varphi_z \right\rangle,$$
which we can rewrite as
$$\left(\hat{\imath}\frac{\partial}{\partial x} + \hat{\jmath}\frac{\partial}{\partial y} + \hat{k}\frac{\partial}{\partial z}\right)\cdot\left(w\varphi_x\hat{\imath} + w\varphi_y\hat{\jmath} + w\varphi_z\hat{k}\right).$$
Recall that for dot product multiplication,
$$\hat{\imath}\cdot \hat{\imath} = \hat{\jmath}\cdot\hat{\jmath} = \hat{k}\cdot\hat{k} = 1$$
and
$$\hat{\imath}\cdot\hat{\jmath} = \hat{\imath}\cdot\hat{k} = \hat{\jmath}\cdot\hat{\imath} = \hat{\jmath}\cdot\hat{k} = \hat{k}\cdot\hat{\imath} = \hat{k}\cdot\hat{\jmath} = 0.$$
Hence, our original product is equal to
$$\frac{\partial(w\varphi_x)}{\partial x}+\frac{\partial(w\varphi_y)}{\partial y}+\frac{\partial(w\varphi_z)}{\partial z}$$
Applying the product rule, we obtain
$$\left(\frac{\partial w}{\partial x}\varphi_x+\frac{\partial\varphi_x}{\partial x}w\right)+\left(\frac{\partial w}{\partial y}\varphi_y+\frac{\partial\varphi_y}{\partial y}w\right)+\left(\frac{\partial w}{\partial z}\varphi_z+\frac{\partial\varphi_z}{\partial z}w\right),$$
which we rewrite by factoring out $w$:
$$\left(\varphi_x\frac{\partial w}{\partial x}+\varphi_y\frac{\partial w}{\partial y}+\varphi_z\frac{\partial w}{\partial z}\right)+w\left(\frac{\partial\varphi_x}{\partial x}+\frac{\partial\varphi_y}{\partial y}+\frac{\partial\varphi_z}{\partial z}\right).$$
Finally, we apply the definitions of the dot product and del operator in reverse to obtain our final form of the expression,
$$\vec{\varphi}\cdot\nabla w + w\nabla\cdot\vec{\varphi}.$$
Thus,
Comments
$\textbf{Theorem}$. Suppose $\omega(x,y,z)$ is a scalar field and $\vec{\varphi}(x,y,z)$ is a vector field. Then,
$$\nabla\cdot\left(\omega\vec{\varphi}\right)=\vec{\varphi}\cdot\nabla\omega+\omega\nabla\cdot\vec{\varphi}.$$
$\textit{Proof.}$
$$\nabla\cdot\left(\omega\vec{\varphi}\right)=\left\langle \frac{\partial}{\partial x},\frac{\partial}{\partial y}, \frac{\partial}{\partial z}\right\rangle\cdot\omega\left\langle\varphi_x,\varphi_y,\varphi_z\right\rangle$$
$$=\left\langle \frac{\partial}{\partial x},\frac{\partial}{\partial y}, \frac{\partial}{\partial z}\right\rangle\cdot\left\langle\omega\varphi_x,\omega\varphi_y,\omega\varphi_z\right\rangle$$
$$=\frac{\partial(\omega\varphi_x)}{\partial x}+\frac{\partial(\omega\varphi_y)}{\partial y}+\frac{\partial(\omega\varphi_z)}{\partial z}$$
$$=\left(\frac{\partial\omega}{\partial x}\varphi_x+\frac{\partial\varphi_x}{\partial x}\omega\right)+\left(\frac{\partial\omega}{\partial y}\varphi_y+\frac{\partial\varphi_y}{\partial y}\omega\right)+\left(\frac{\partial\omega}{\partial z}\varphi_z+\frac{\partial\varphi_z}{\partial z}\omega\right)$$
$$=\left(\varphi_x\frac{\partial\omega}{\partial x}+\varphi_y\frac{\partial\omega}{\partial y}+\varphi_z\frac{\partial\omega}{\partial z}\right)+\omega\left(\frac{\partial\varphi_x}{\partial x}+\frac{\partial\varphi_y}{\partial y}+\frac{\partial\varphi_z}{\partial z}\right)$$
$$=\vec{\varphi}\cdot\nabla\omega+\omega\nabla\cdot\vec{\varphi}. $$
$$ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \square$$
@dualrey This looks great! For those reading along - this is what we call a *componentwise proof*.
Let $w$ be a scalar field and let $\vec{\varphi}$ be a vector field. Then
$$\nabla\cdot\left(w\vec{\varphi}\right) = \left\langle \frac{\partial}{\partial x},\frac{\partial}{\partial y}, \frac{\partial}{\partial z}\right\rangle\cdot w\left\langle\varphi_x,\varphi_y,\varphi_z\right\rangle.$$
Since $w$ is a scalar, we multiply all the vector components of $\varphi$ by $w$ to get
$$\left\langle \frac{\partial}{\partial x},\frac{\partial}{\partial y}, \frac{\partial}{\partial z}\right\rangle\cdot\left\langle w\varphi_x,w\varphi_y,w\varphi_z \right\rangle,$$
which we can rewrite as
$$\left(\hat{\imath}\frac{\partial}{\partial x} + \hat{\jmath}\frac{\partial}{\partial y} + \hat{k}\frac{\partial}{\partial z}\right)\cdot\left(w\varphi_x\hat{\imath} + w\varphi_y\hat{\jmath} + w\varphi_z\hat{k}\right).$$
Recall that for dot product multiplication,
$$\hat{\imath}\cdot \hat{\imath} = \hat{\jmath}\cdot\hat{\jmath} = \hat{k}\cdot\hat{k} = 1$$
and
$$\hat{\imath}\cdot\hat{\jmath} = \hat{\imath}\cdot\hat{k} = \hat{\jmath}\cdot\hat{\imath} = \hat{\jmath}\cdot\hat{k} = \hat{k}\cdot\hat{\imath} = \hat{k}\cdot\hat{\jmath} = 0.$$
Hence, our original product is equal to
$$\frac{\partial(w\varphi_x)}{\partial x}+\frac{\partial(w\varphi_y)}{\partial y}+\frac{\partial(w\varphi_z)}{\partial z}$$
Applying the product rule, we obtain
$$\left(\frac{\partial w}{\partial x}\varphi_x+\frac{\partial\varphi_x}{\partial x}w\right)+\left(\frac{\partial w}{\partial y}\varphi_y+\frac{\partial\varphi_y}{\partial y}w\right)+\left(\frac{\partial w}{\partial z}\varphi_z+\frac{\partial\varphi_z}{\partial z}w\right),$$
which we rewrite by factoring out $w$:
$$\left(\varphi_x\frac{\partial w}{\partial x}+\varphi_y\frac{\partial w}{\partial y}+\varphi_z\frac{\partial w}{\partial z}\right)+w\left(\frac{\partial\varphi_x}{\partial x}+\frac{\partial\varphi_y}{\partial y}+\frac{\partial\varphi_z}{\partial z}\right).$$
Finally, we apply the definitions of the dot product and del operator in reverse to obtain our final form of the expression,
$$\vec{\varphi}\cdot\nabla w + w\nabla\cdot\vec{\varphi}.$$
Thus,
$$\nabla \cdot (w\vec{\varphi}) = \vec{\varphi}\cdot \nabla w + w \nabla \cdot \vec{\varphi}.$$