How to show that $\nabla \cdot (\nabla f) = \nabla ^{2} f$ for this function? I've been given a homework question that asks me to show that $\nabla \cdot (\nabla f) = \nabla ^{2} f$, where $f(x,y,z)=xyz-z^{2}sin(y)$. However, $\nabla \cdot (\nabla f)$ is a scalar quantity, whilst $\nabla ^{2} f$ is a vector quantity, so they cannot be equal.
What am I missing here?
Thanks.
 A: The Laplacian operator, $\nabla ^2$ can be defined to operate on both scalar fields and vector fields.  If $f$ is a scalar field, then 
$$\bbox[5px,border:2px solid #C0A000]{\nabla ^2 f(\vec r)\equiv \nabla \cdot \nabla f(\vec r)}$$
which in Cartesian coordinates can be written
$$\begin{align}
\nabla \cdot \nabla f(\vec r)&=\sum_{i=1}^{3}\sum_{j=1}^{3}\left(\hat x_i\frac{\partial}{\partial x_i}\right)\cdot \hat x_j\frac{\partial f(\vec r)}{\partial x_j}\\\\
&=\sum_{i=1}^{3}\frac{\partial^2f(\vec r)}{\partial x_i^2}
\end{align}$$
If $\vec f(\vec r)$ is a vector field, then the 
$$\bbox[5px,border:2px solid #C0A000]{\nabla^2 \vec f(\vec r)\equiv \nabla \left(\nabla \cdot \vec f(\vec r)\right)-\nabla \times \nabla \times \vec f(\vec r)}$$
which in Cartesian coordinates can be written
$$\begin{align}
\nabla (\nabla \cdot \vec f(\vec r)-\nabla \times \nabla \times \vec f(\vec r)
&=\sum_{i=1}^{3}\sum_{j=1}^{3}\hat x_i\frac{\partial}{\partial x_i}\frac{\partial f_j(\vec r)}{\partial x_j}-\sum_{i=1}^{3}\sum_{j=1}^{3}\sum_{k=1}^{3}\left(\hat x_i\times(\hat x_j\times \hat x_k)\frac{\partial^2 f_k(\vec r)}{\partial x_i \partial x_j}\right) \\\\
&=\sum_{i=1}^{3}\frac{\partial^2 \vec f(\vec r)}{\partial x_i^2}
\end{align}$$
We note that while in Cartesian coordinates, the forms for the Laplacian on a scalar and a vector are identical, this is not the case in other curvilinear coordinates.
A: The notation $\nabla^2 f$ stands for the Laplacian of $f$.  It has the value(in three dimensions):
$$\nabla^2f = \frac{\partial^2f}{\partial x^2} + \frac{\partial^2f}{\partial y^2} + \frac{\partial^2f}{\partial z^2}$$
It is also sometime written as $\Delta f$.
A: You have to understand the difference between the gradient $\nabla f$of a function f which is the vector $(\partial_xf,\partial_yf,\partial_zf)$, the divergence of a vector X=(U,V,W) which is $div(X)=\partial_xU+\partial_yV+\partial_zW$ which is a scalar
$div(\nabla f)=\partial_x(\partial_xf)+\partial_y(\partial_yf)+\partial_z(\partial_zf)=\partial^2_xf+\partial^2_yf+\partial^2_zf=\nabla^2f$
I believe the confusion come from the fact that some authors write $div(X)=\nabla.X$
