# Derivative with respect to vector of a function depending on vectors

I've been trying to understand this concept for hours without any success. I found similar questions on this forum (Derivative with respect to a vector is a gradient?) but I still don't understand.

Let's consider a multiple particle system with $$n$$ particles. I'm running multiple times into the notation $$\vec{F_k}=- \frac{\partial}{\partial \vec{x_k}} V(\vec{x_1},\vec{x_2},..., \vec{x_n})$$ where $$F_k$$ denotes the resulting force acting on the particle $$k$$.

I assume the right hand side is an $$\mathbb{R}^3$$ vector because the left hand side is.

I came up with one example: Let $$f:\mathbb{R}^n \rightarrow \mathbb{R}, f(x_1, ... ,x_n)=3x_1+{x_2}^2x_4-5x_3+...+x_n$$ Then (because the variables are real) I know: $$\frac{\partial}{\partial x_1}f(x_1,...x_n)=3 \quad \frac{\partial}{\partial x_2}f(x_1,...x_n)=2x_2x_4 \quad \frac{\partial}{\partial x_3}f(x_1,...x_n)=-5$$ Now I guess one could define $$\vec{a}=(x_1,x_2,x_3)^T$$, then $$\frac{\partial}{\partial \vec{a}}f(x_1,...x_n)=(3,2x_2x_4,-5)^T$$

Even if this is correct, I'm struggling to apply this idea to the potential $$V$$ above because the variables there aren't scalars. What if for example $$V(\vec{x_1},\vec{x_2},..., \vec{x_n})=\vec{x_1}^3$$ ? Raising a vector to the 3rd power doesn't make any sense to me. Should I think of V as a function $$V(x_{11},x_{12},x_{13},...,x_{n1},x_{n2},x_{n3}): \mathbb{R}^{3n}\rightarrow \mathbb{R}$$?

I'm sorry about the long question but I hope to have at least clarified my problems and possibly helped you to help me.

I think the notation $$\frac{\partial}{\partial{\vec{x_k}}}$$ stand for the gradient $$\vec{\nabla}$$. The gradient operator act on scalar funtion like $$V$$ and gives a vector : $$\vec{\nabla} V= \begin{pmatrix} \frac{\partial}{\partial_{x_1}}V\\ \dots \\ \frac{\partial}{\partial_{x_n}}V \end{pmatrix}$$ Here you describe $$V$$ as a function of multiple vectors but i think is like you use one vector that contains all the $$\vec{x_i}$$ coefficients $$V(x_{11},x_{12},...,x_{n3})$$. You give the example of $$V=\vec{x_1}^3$$ in this case $$V$$ depend only on $$x_1$$ and either $$V=x_1^2 \vec{x_1}$$ and $$V$$ is a vector or $$V=|\vec{x_1}|^3$$ and this is a scalar.

I hope this helps you.

If you're talking about newtonian mechanics, then $$\mathbf{F} = - \nabla V$$. I'm not sure why you're indexing $$\mathbf{F}$$, either it's a vector $$\mathbf{F}$$ or you're referring to the components $$F_k$$. The components are then $$F_k = \left[-\nabla V \right]_k = -\frac{\partial V}{\partial x_k}$$. In general if you have a scalar function $$f: \mathbb{R}^n \to \mathbb{R}$$ then you can define the directional derivative just like a normal derivative as \begin{align*} \partial_{\mathbf{v}} f\big|_{\mathbf{x_0}}& = \lim_{h \to 0} \frac{f(\mathbf{x}_0 + h \mathbf{v}) - f(\mathbf{x}_0)}{h} \\ &= \mathbf{v} \cdot \nabla f \end{align*} When you have multiple variables then the slope of the function can obviously be different if you're "walking" in different directions.

The notation $$\mathbf{x}^3$$ is usually meant as repeated scalar products, therefore $$\mathbf{x}^2 = ||\mathbf{x}||^2 = x^2 + y^2 + z^2$$ is actually just a scalar but $$\mathbf{x}^3 = \mathbf{x}^2 \mathbf{x} = (x^2 + y^2 + z^2) \begin{pmatrix} x \\ y \\ z \end{pmatrix}$$ is then a vector. Taking the derivative along $$\mathbf{v} = (1, 0, 1)$$ would then be \begin{align*} \partial_{\mathbf{v}} \mathbf{x}^3 &= \mathbf{v} \cdot \nabla \mathbf{x}^3 \\ &= \mathbf{v} \cdot \begin{pmatrix} 3x^2 + y^2 + z^2 \\ x^2 + 3y^2 + z^2 \\ x^2 + y^2 + 3z^2 \end{pmatrix} \\ &= 1(3x^2 + y^2 + z^2) + 0 (x^2 + 3y^2 + z^2) + 1 (x^2 + y^2 + 3z^2) \end{align*}

To illustrate it with multiple particles, let's look at the Kepler problem and ignore the constants for a second, then $$V(\mathbf{r}_1, \mathbf{r}_2) = \frac{1}{|\mathbf{r}_1 - \mathbf{r}_2|} = \left((x_1 - x_2)^2 + (y_1 - y_2)^2 + (z_1 + z_2)^2 \right)^{-1/2} = V(x_1, y_1, z_1, x_2, y_2, z_2)$$. The force on the first particle is then \begin{align*} \mathbf{F}_1(\mathbf{r}_1, \mathbf{r}_2) &= - \frac{\partial V}{\partial \mathbf{r}_1} = -\nabla_{\mathbf{r}_1} V = (1, 1, 1, 0, 0, 0) \nabla V \\ &= - \begin{pmatrix} \partial_{x_1} V \\ \partial_{y_1} V \\ \partial_{z_1} V \end{pmatrix} \\ &= -\frac{1}{2}\frac{1}{((x_1 - x_2)^2 + (y_1 - y_2)^2 + (z_1 - z_2)^2 )^{3/2}} \begin{pmatrix} 2(x_1 - x_2) \\ 2(y_1 - y_2) \\ 2(z_1 - z_2) \end{pmatrix} \\ &= - \frac{\mathbf{r}_1 - \mathbf{r}_2}{|\mathbf{r}_1 - \mathbf{r}_2|^3} \end{align*} Note that $$(1, 1, 1, 0, 0, 0)$$ points along $$\mathbf{r}_1$$. The last line is just Newton's law of gravity $$\mathbf{F}_G = \frac{1}{r^2}\mathbf{e}_r$$. So your comment is correct.

• I think I'm getting there. I consider a multiple particle system in newtonian mechanics, $F_k$ denotes the resulting force on particle $k$. As 2piOmega suggested, is it correct to think of $V=V(x_{11},x_{12},x_{13},...,x_{n3})$ as a function $V:\mathbb{R}^{3n}\rightarrow \mathbb{R}$?. If yes, is it then correct to say $\frac{\partial}{\partial \vec{x_k}}V=\frac{\partial}{\partial (x_{k1},x_{k2},x_{k3})^T}V=(\frac{\partial}{\partial x_{k1}}V,\frac{\partial}{\partial x_{k2}}V,\frac{\partial}{\partial x_{k3}}V)^T$ ? I think here is where my main problem lies. Commented Apr 1, 2021 at 10:42
• I edited my answer, I hope it helps. Commented Apr 1, 2021 at 11:59
• Your example was exactly what I needed. Thank you! Commented Apr 1, 2021 at 12:37

Your expression appears to be referring to the scalar potential, V, (of an object) the value of which depends on its position in a multi-space specified by coordinates: x, y, z, and perhaps more. Then the force on the object depends on the gradient of the potential which is found by summing the partial derivatives with respect to each dimension. Though I have not seen a formal definition of the operation of dividing a scalar by a vector, this expression seems to imply that would be done in terms of components.