Help with derivation steps (vector-calculus) Can someone explain the steps listed below? (taken from a textbook)
For context, $ \rho = const. ; \phi = \phi (x,y,z) $
$$\rho \nabla\phi \space \frac \partial {\partial{x}} (\nabla \phi) = \frac \rho 2 \space \frac \partial {\partial{x}} (\nabla \phi \space \cdot \space \nabla \phi )   $$
Thank you!
 A: You should look at those operations as they are taken point-wise.
Here's what's done on the left side of the equation:

*

*They firstly compute the gradient, which is a vector of partial derivatives.

*Then they take a partial derivative of the gradient w.r.t. x. They do it point-wise, so they take every single element of the gradient and compute the partial derivative of it. So what they get after those 2 steps is $\Big( \frac{\partial^2 \phi}{\partial x^2}, \frac{\partial^2 \phi}{\partial x \partial y}, \frac{\partial^2 \phi}{\partial x \partial z} \Big)$.

*After that they mutliply the result by the gradient of the $\phi$ also point-wise and get $\Big( \frac{\partial \phi}{\partial x} \cdot \frac{\partial^2 \phi}{\partial x^2}, \frac{\partial \phi}{\partial y} \cdot \frac{\partial^2 \phi}{\partial x \partial y}, \frac{\partial \phi}{\partial z} \cdot \frac{\partial^2 \phi}{\partial x \partial z} \Big)$

*They multiply everything by $\rho$
On the right part they are doing something similar. $\frac{1}{2}$ there is just to cancel out the the 2 multiplier, which is going to come out according to the rule $(f(x)^2)' = 2 f(x) f'(x)$
A: Use product rule for dot products.
$$
\frac{\partial }{\partial x} \left( \nabla \phi \cdot \nabla \phi\right) = \nabla \phi \cdot \frac{\partial }{\partial x} \nabla \phi + \frac{\partial }{\partial x} \left(\nabla \phi \right) \cdot \nabla \phi 
= 2 \nabla \phi \cdot \frac{\partial }{\partial x} \left(\nabla \phi \right) 
$$
