I have to solve the Question here, where I have to prove the relation
$$\nabla\cdot(f\textbf v) = f\nabla \cdot \textbf v+ (\nabla f)\cdot\textbf v$$
And I have the question:
Is there any difference between $\text{grad} f$ and $f \text{grad}$ ?
Thanks.
On
You just need to apply the product rule.
By definition we have:
$\nabla \cdot (f \bar{v}) = \frac{f\bar{v_x}}{\partial x} + \frac{f\bar{v_y}}{\partial y} + \frac{f\bar{v_z}}{\partial z} $
= $ \bar{v_x}\frac{\partial f}{\partial x} + \bar{v_y}\frac{\partial f}{\partial y} + \bar{v_z}\frac{\partial f}{\partial z} + f\frac{\partial v_x}{\partial x} + f\frac{\partial v_y}{\partial y} + f\frac{\partial v_z}{\partial z}$
= $(\nabla f) \cdot v + f\nabla \cdot v $
For your first question you ask Is there any difference between $\text{grad} f$ and $f \text{grad}$ ? The answer to that question is yes.
Let $f: \mathbb{R}^n \to \mathbb{R}$ be a function. So we have that
Where $f = f(\vec{x})$ for $\vec{x} = (x_1,x_2,\dots,x_n)$. So we have that $\text{grad}f$ is a vector in $\mathbb{R}^n$.
Now
Is an operator that acs in the vetors of the vector space $\mathbb{R}^n$. So they are quite different things, one is a vector in the space and the other is an operator that acs on the space.
I suppose that your question is, for a vector $\vec{v} \in \mathbb{R}^n$, you have to prove the equality
$$\nabla (f(\vec{x})\vec{v}) = f(\vec{x})\nabla\vec{v} + (\nabla f(\vec{x}))\vec{v}$$
Note that $\vec{v} \in \mathbb{R}^n$. Using Einstein summation notation
$$\nabla(f\vec{v}) = \partial_\color{red}{i}(fv^\color{red}{i}) = f\partial_\color{red}{i}v^\color{red}{i}+v^\color{red}{i}\partial_\color{red}{i}f = f(\nabla\vec{v})+v^\color{red}{i}(\nabla f)_\color{red}{i} = f(\nabla \vec{v})+\vec{v}\nabla f$$
Where $i = 1,2,3,\dots,n$ and $\partial _i := \partial/\partial x_i$ and we use that for two vectors $a,b \in\mathbb{R}^n$ we have that
$$\vec{a}\vec{b} = \sum_\color{red}{i}a_\color{red}{i}b_\color{red}{i} \stackrel{\text{Einstein}}{= }a_\color{red}{i}b^\color{red}{i}$$