Derivative with respect to a vector

1. Oct 31, 2009

I've stumbled across something I've never seen before. I am taking a class outside of my major and the notation seems to be quite different from what I am used to, and I am completely baffled as to how to solve what I feel are some simple problems.

1. The problem statement, all variables and given/known data

Find $$\frac{d f(\vec{k})}{\vec{k}}$$ where $$f(\vec{k}) = sin(ak_x)-cos(bk_y)+cos(ck_z)$$. f itself is a scalar function that operates on the components of the vector $$\vec{k}$$.

3. The attempt at a solution

What does this notation mean? I have never seen a notation in which there is a derivative with respect to a vector. Is this the same thing as the gradient $$\nabla f$$?

2. Oct 31, 2009

HallsofIvy

Staff Emeritus
In general, if a function from $R^m$ to $R^n$ is "differentiable at $\vec{v_0}$" if and only if there exist a linear function, L, from $R^m$ to $R^n$ and a function $\epsilon$ from $R^m$ to $R^n$ such that $f(\vec{v})= f(\vec{v_0})+ L(\vec{v}- \vec{v_0})+ \epsilon(\vec{v})$ and $\lim_{\vec{v}\to \vec{0}} \epsilon(\vec{v}/|\vec{v}|$$= 0$. In this case, we say that L is the "derivative of f with respect to $\vec{v}$ at $vec{v_0}$".

Notice that L is a linear operator, not a number or even a vector. In the case f from $R^1$ to $R^1$, a real valued function of a single real variable as in Calculus I, a linear function from $R^m$ to $R^n$ is a multiple: $mx$ for some number m. In calculus I, we think of that number as being the derivative.

In the case of f from $R^m$ to $R^1$, a real valued function of several real variables, we can represent a linear function from $R^m$ to $R^1$ as a "dot product"- any linear function L(<x, y, z>) can be written as ax+ by+ cz= <a, b, c>$\cdot$<x, y, z>. In that case, we can identify the derivative with that vector- which is, in fact, the gradient.

That is the case in your example. f is a real valued function of a variable in $R^3$. Strictly speaking, its derivative is the linear transformation corresponding to taking the dot product of the position vector $x\vec{i}+ y\vec{j}+ z\vec{k}$ with the vector
$$\frac{\partial f}{\partial x}\vec{i}+ \frac{\partial f}{\partial y}\vec{j}+ \frac{\partial f}{\partial z}\vec{k}$$
which is, as you say, the gradient of f.

(Notice that we need "vector space" structure in the range space of f because we need to be able to subtract f(x+h)- f(x). We don't need to do "arithmetic" in the domain space because all we do there is take the limit as |h| goes to 0. That's why we tend to talk about "functions of several variables" rather than "functions of a vector variable".)

In the general case of f from $R^m$ to $R^n$ with neither m nor n equal to 1, a linear function from $R^m$ to $R^n$ can be written as a matrix multiplication and we can identify the derivative with that matrix. In the case of $\vec{f}(\vec{v})= f_x(x, y, z)\vec{i}+ f_y(x,y,z)\vec{j}+ f_z(x,y,z)\vec{k}$ (where \vec{v}= x\vec{i}+ y\vec{j}+ z\vec{k}[/itex]) that matrix is
$$\begin{bmatrix}\frac{\partial^2 f_x}{\partial x^2} & \frac{\partial^2 f_y}{\partial x\partial y} & \frac{\partial^2 f_z}{\partial x\partial z} \\ \frac{\partial^2 f_x}{\partial y\partial x} & \frac{\partial^2 f_y}{\partial y^2} & \frac{\partial^2 f_z}{\partial y\partial z}\\ \frac{\partial^2 f_z}{\partial z\partial x} & \frac{\partial^2 f_z}{\partial z\partial y} & \frac{\partial^2 f_z}{\partial z^2}\end{bmatrix}$$

Last edited: Oct 31, 2009
3. Oct 31, 2009

wow HallsOfIvy, thank you for the detailed reply, I understand now; I had never really stopped to think about what $$\nabla{F}$$ meant. I am a bit light on the theory side of calculus so I have to rely on my gut instinct a lot.