Proof that $\nabla(ab) = a\nabla b + b\nabla a$ I am trying to prove the following are equivalent:
$a(x,y,z)$ and $b(x,y,z)$
$$\nabla(ab) = a\nabla b + b\nabla a$$
So looking at the left side: $\nabla(ab)= \cfrac{\partial ab}{\partial x} + \cfrac{\partial ab}{\partial y}+\cfrac{\partial ab}{\partial z} \dots (1)$
And the right: $a\nabla b + b \nabla a=a\left(\cfrac{\partial b}{\partial x} + \cfrac{\partial b}{\partial y}+\cfrac{\partial b}{\partial z}\right)+b\left(\cfrac{\partial a}{\partial x} + \cfrac{\partial a}{\partial y}+\cfrac{\partial a}{\partial z}\right) 
\dots (2)$
Now I am not sure where to go from here. Do I apply a product rule to (1), I don't know if this applies to partial derivatives. Do I use the chain rule on (1)? What can I even do with (2)?
Thanks for any tips. 
 A: $$\nabla(ab) = \left(
  \begin{array}{c}
     \frac{\partial(ab)}{\partial x}\\
     \frac{\partial(ab)}{\partial y}\\
     \frac{\partial(ab)}{\partial z}\\
  \end{array}
\right) = \left(
  \begin{array}{c}
     a\frac{\partial b}{\partial x}+b\frac{\partial a}{\partial x}\\
     a\frac{\partial b}{\partial y}+b\frac{\partial a}{\partial y}\\
     a\frac{\partial b}{\partial z}+b\frac{\partial a}{\partial z}\\
  \end{array}
\right) = a\left(
  \begin{array}{c}
     \frac{\partial b}{\partial x}\\
     \frac{\partial b}{\partial y}\\
     \frac{\partial b}{\partial z}\\
  \end{array}
\right) + b\left(
  \begin{array}{c}
     \frac{\partial a}{\partial x}\\
     \frac{\partial a}{\partial y}\\
     \frac{\partial a}{\partial z}\\
  \end{array}
\right) = a\nabla b + b\nabla a.$$
A: $$\vec \nabla (ab)=\left(\frac{\partial (ab)}{\partial x}, \frac{\partial (ab)}{\partial y}, \frac{\partial (ab)}{\partial z}\right) \tag{by definition}$$
$$=\left( a \frac{\partial b}{\partial x}+b\frac{\partial a}{\partial x}, \ \  a\frac{\partial b}{\partial y}+b\frac{\partial a}{\partial y}, \ \ a \frac{\partial b}{\partial z}+b\frac{\partial a}{\partial z} \right) \tag{product rule}$$
$$=a\left(\frac{\partial b}{\partial x}, \frac{\partial  b}{\partial y}, \frac{\partial b}{\partial z} \right)+b\left(\frac{\partial a}{\partial x}, \frac{\partial  a}{\partial y}, \frac{\partial a}{\partial z} \right) \tag{factorise}$$
$$=: a \vec \nabla b + b \vec \nabla a \quad \square$$
A: First note that, using Cartesian coordinates,
$$
\begin{eqnarray}
\nabla &=& \hat{\bf x} \frac{\partial}{\partial x} + \hat{\bf y} \frac{\partial}{\partial y} + \hat{\bf z} \frac{\partial}{\partial z}
 \\
&=& \hat{\bf e}_i \partial_i
\end{eqnarray}
$$
where $\hat{\bf e}_i$ are the unit vectors, $\partial_i = \partial/\partial x_i$, and a sum over $i$ is implied (Einstein summation convention). Then
$$
\begin{eqnarray}
\nabla\left(ab\right) &=& \hat{\bf e}_i \partial_i \left(ab\right) \\
&=& a \ \hat{\bf e}_i \partial_i b + b \ \hat{\bf e}_i \partial_i a \\
&=& a \nabla b + b \nabla a
\end{eqnarray}
$$
