derivative of a gradient just a very simple question: 

I have a continuous and differentiable function $\phi$ defined on $R^3$ with its gradient $\nabla\phi$.

Could you please tell me if it makes sense to take a derivative of this gradient with respect to $\phi$? 

So:

$\frac{\partial}{\partial{\phi}} \nabla \phi = ?$

Would it make sense to say $\frac{\partial}{\partial{\phi}} \nabla \phi = \frac{\partial}{\partial{\phi}}(\phi_{x}, \phi_{y},\phi_{z}) = (\phi_{xx}, \phi_{yy},\phi_{zz}) $?

Thanks for your time,
Matt
Well from my point of view, you are thinking something but your notation of expression is describing other thing !!!! Like, what does it mean by $\nabla \phi $ ? We know in multi-variable calculus we can not express the derivative of a multi-variable function with just one scaler valued function.we need the help of vector valued function or expression .and here the $\nabla \phi$ comes into action.so derivative of $\phi$ is not a single expression (scalar valued function). So, by the derivative of $\phi$ you are just meaning $\nabla \phi $. 

Now coming to your question. here, what does it mean by $\partial \phi $ ?
It is not a appropriate notation here,because here you wanting to say how $\nabla \phi$ changes as I change $\phi$ a little bit.Now,the problem is,this $\phi$ is not a single variable rather a scalar valued function.So you have to work on variable after variable of $\phi$ and component after component of $\nabla \phi$ (As,now it is a vector valued function) you can't express the whole expression as
$$\color{red}{\frac{\partial}{\partial{\phi}} \nabla \phi = \frac{\partial}{\partial{\phi}}(\phi_{x}, \phi_{y},\phi_{z}) = (\phi_{xx}, \phi_{yy},\phi_{zz})}$$ 
rather.if  $$\nabla \phi=
    \begin{bmatrix}
    \phi_x \\
    \phi_y \\
    \phi_z \\
    \end{bmatrix}
$$
then you should compute $\frac{\partial \phi}{\partial{x}}$ and $\frac{\partial}{\partial{x}}\nabla \phi$ individually.
And finally after dividing them then we may  say something like that you want. But that would yields,
$$\frac{\partial}{\partial{\phi}} \nabla \phi=\frac{1}{\phi_x} \begin{bmatrix}
    \phi_{xx} \\
    \phi_{yx} \\
    \phi_{zx} \\
    \end{bmatrix}\text{(differentiating partially with respect to $x$)}$$

But it is not a completed answer though.you have to follow the procedure in respect of y and z also,to get a compact answer.It will be somehow a rigorous computation.
Well from my point of view, you are thinking something but your notation of expression is describing other thing !!!! Like, what does it mean by $\nabla \phi $ ? We know in multi-variable calculus we can not express the derivative of a multi-variable function with just one scaler valued function.we need the help of vector valued function or expression .and here the $\nabla \phi$ comes into action.so derivative of $\phi$ is not a single expression (scalar valued function). So, by the derivative of $\phi$ you are just meaning $\nabla \phi $. 

Now coming to your question. here, what does it mean by $\partial \phi $ ?
It is not a appropriate notation here,because here you wanting to say how $\nabla \phi$ changes as I change $\phi$ a little bit.Now,the problem is,this $\phi$ is not a single variable rather a scalar valued function.So you have to work on variable after variable of $\phi$ and component after component of $\nabla \phi$ (As,now it is a vector valued function) you can't express the whole expression as
$$\color{red}{\frac{\partial}{\partial{\phi}} \nabla \phi = \frac{\partial}{\partial{\phi}}(\phi_{x}, \phi_{y},\phi_{z}) = (\phi_{xx}, \phi_{yy},\phi_{zz})}$$ 
rather.if  $$\nabla \phi=
    \begin{bmatrix}
    \phi_x \\
    \phi_y \\
    \phi_z \\
    \end{bmatrix}
$$
then you should compute $\frac{\partial \phi}{\partial{x}}$ and $\frac{\partial}{\partial{x}}\nabla \phi$ individually.
And finally after dividing them then we may  say something like that you want. But that would yields,
$$\frac{\partial}{\partial{\phi}} \nabla \phi=\frac{1}{\phi_x} \begin{bmatrix}
    \phi_{xx} \\
    \phi_{yx} \\
    \phi_{zx} \\
    \end{bmatrix}\text{(differentiating partially with respect to $x$)}$$

But it is not a completed answer though.you have to follow the procedure in respect of y and z also,to get a compact answer.It will be somehow a rigorous computation.