What is the first derivative of a plane equation using point-normal form? Given a plane defined in its point-normal form: $$f(\mathbf r)=(\mathbf r-\mathbf p)\cdot \mathbf n=0$$


*

*with $\mathbf r=$ any point in $\mathbb{R^3}$

*$\mathbf p=$ one point of the plane 

*$\mathbf n=$ normal of the plane.


What is its derivative regarding $ \mathbf x$:
$$f'(\mathbf x)={\delta f \over \delta \mathbf x} = \:?$$

This is what I found so far. Writing the dot-product in its component form produces: 
$$f(\mathbf x)=\sum_{i}^{} (x_i- p_i) \cdot n_i$$
Now I can see, that the scalar $n_i$ can be multiplied into the parentheses to get:
$$f(\mathbf x)=\sum_{i}^{} ( n_ix_i-n_ip_i )$$
Deriving the scalar equation in the sum on its own:
$$g(x)=n_ix_i-n_ip_i \Rightarrow g'(x)={\delta g \over \delta x}=n_i $$
Inserting that into our sum results in:
$$f'(\mathbf x)=\sum_{i}^{} (n_i) = \vec{1} \cdot \mathbf n = n_x + n_y + n_z$$ 
Is this correct? If not where is my error?
 A: Yes, you are correct. Only 'but' is that the vector should be on the $n_i$'s, not on the $1$ (and you don't really need the $1$ there), so the last equation should read:
$$f'(x)=\sum_{i}^{} (n_i) = \vec{n}.$$
Think that the function $f$ has planes parallel to this one as its level sets. This means that $f$ doesn't change on the planes, and so the direction in which is grows the most is the one perpendicular to the planes. I hope you find this intuition helpful in the future.
Addendum: remaking calculations with more rigorous notation.
We have the plane defined as the zero set of a function $f:\mathbb R^3\to \mathbb R$. This is, $$f(x,y,z) := (\vec x - \vec p)\cdot\vec n = [(x,y,z) - (p_x,p_y,p_z)]\cdot (n_x,n_y,n_z).$$
The gradient of $f$ is the vector whose $n$-th coordinate is the derivative of $f$ in the direction of that coordinate. It is a vector field $\nabla f:\mathbb R^3\to \mathbb R^3$ associating to each point in the space a vector that is tangent to it. In this case, where
$$f(x,y,z) = (x-p_x)n_x + (y-p_y)n_y + (z-p_z)n_z,
$$
the gradient shall be
$$\nabla f (x,y,z) = \left(\frac{\partial f}{\partial x},\frac{\partial f}{\partial y},\frac{\partial f}{\partial z}\right) = (n_x,n_y,n_z) = \vec n.
$$
