Vector Derivative (not del)

What does it mean to operate with $\frac{d}{\vec{v}}$ where v is a vector?

Say you have another vector q, how do you do $\frac{d\vec{v}} {d\vec{q}}[\itex]? What about [itex]\frac{d\vec{v}} {d\vec{v}}$?
(Can't remember how to do the proper font sorry)

Last edited by a moderator:

HallsofIvy
Homework Helper
First you will have to say what kind of vector space $\vec{v}$ is in and what kind of function of v you are talking about. If f: Rn to Rm, that is if the variable, $\vec{v}$ is an n dimensional vector variable and f maps it to an m dimensional vector, the $d\vec{f}/d\vec{v}$, at $\vec{v}_0$ is "the linear transformation from Rn to Rm that best approximates $\vec{f}$ in some region around $\vec{v}_0$".

More precisely, a function,$\vec{f}$, from Rn to Rm, is said to be differentiable at $\vec{v}_0$ if and only if there exist a linear transformation, L, from Rn to Rm, and a function $\epsilon(\vec{v})$, from Rn to Rm, such that
1) $f(\vec{v})= f(\vec{v}_0)+ L(\vec{v}- \vec{v_0})+ \epsilon(\vec{v})$
2) $\lim_{\vec{v}\rightarrow \vec{0}}\epsilon(\vec{v})/||\vec{v}-\vec{v}_0||= 0$

It can be shown that the linear transformation, L, in (1), is unique and we say that L is the derivative of f at $\vec{v}_0$.

Notice that, if we reduce this to R1 to R1, we are saying that the derivative is NOT the slope of the tangent line y= mx+ b but, rather, the linear function y= mx.

If f is from R1 to R3, a "vector valued function of a single real variable", then the L above is linear transformation from R1 to R3 given by
$$Lt= \left<\frac{df_x}{dt},\frac{df_y}{dt},\frac{df_z}{dt}\right>t$$
which we can think of as being "represented" by the usual derivative vector,
$$\left<\frac{df_x}{dt},\frac{df_y}{dt}, \frac{df_z}{dt}\right>$$

If f is from R3 to R, a "real valued function of 3 real variables, x, y, z", then the derivative, in the sense above, is the linear transformation from R3 to R given by the dot product
$$\left<\frac{\partial f}{\partial x},\frac{\partial f}{\partial y}, \frac{\partial f}{\partial z}\left>\cdot \vec{v}[/itex] which we can think of as represented by the gradient vector, [tex]\left<\frac{\partial f}{\partial x},\frac{\partial f}{\partial y}, \frac{\partial f}{\partial z}\left>[/itex] More generally, if f is from Rn to Rm, it derivative, at any "point", is the linear transformation which can be represented, in some basis, as the m by n matrix having the partial derivatives of the components of f as elements. In any case, of course, $\frac{d\vec{v}}{d\vec{v}}$, where $\vec{v}$ is a vector function from Rn to itself (NOT just a single vector- derivatives are only defined for functions) is the identity transformation on Rn which can be reprsented by the n by n identity matrix. Thanks mate. I didn't expect such a long reply and one quite so theoretical. It'll take me a bit to mull over so cheers! HallsofIvy More generally, if f is from Rn to Rm, it derivative, at any "point", is the linear transformation which can be represented, in some basis, as the m by n matrix having the partial derivatives of the components of f as elements. I want to understand exactly how do I derivate vector by vector. My problem: I try to solve Euler equations with some method... In this method I use a linearization to the first vector [tex]\hat{Q}$$. I need to write the $$\partial$$F/$$\partial$$Q.
Where Q and F are vectors [1,4].
Can you help me to understand how to build the matrix of linear transformation?

Thank you, I have found the reference: http://www.met.rdg.ac.uk/~ross/Documents/VectorDeriv.html

Last edited:
HallsofIvy
Homework Helper
HallsofIvy

I want to understand exactly how do I derivate vector by vector.
My problem: I try to solve Euler equations with some method... In this method I use a linearization to the first vector $$\hat{Q}$$. I need to write the $$\partial$$F/$$\partial$$Q.
Where Q and F are vectors [1,4].
I presume that you mean they are vectors in R2, not just that specific vector!

Can you help me to understand how to build the matrix of linear transformation?
Thank you.

Say F= <f(x,y), g(x,y)> and Q= <p(x,y), q(x,y)>. Let r= <x,y>. Then, by the chain rule,
$$\frac{dF}{dQ}= \frac{dF}{dr}/\frac{dQ}{dr}[/itex]. And those two derivatives are the matrices having the partial derivatives as entries. (Strictly speaking, they are the linear transformations represented by these matrices in the <1, 0>, <0, 1> basis for R2.) [tex]\frac{dF}{dr}= \left[\begin{array}{cc}\frac{\partial f}{\partial x} & \frac{\partial f}{\partial y} \\ \frac{\partial g}{\partial x} & \frac{\partial g}{\partial y}\end{array}\right]$$

Similarly,
$$\frac{dQ}{dr}= \left[\begin{array}{cc}\frac{\partial p}{\partial x} & \frac{\partial p}{\partial y} \\ \frac{\partial q}{\partial x} & \frac{\partial q}{\partial y}\end{array}\right]$$

So that
$$\frac{dF}{dW}= \left[\begin{array}{cc}\frac{\partial f}{\partial x} & \frac{\partial f}{\partial y} \\ \frac{\partial g}{\partial x} & \frac{\partial g}{\partial y}\end{array}\right]\left[\begin{array}{cc}\frac{\partial p}{\partial x} & \frac{\partial p}{\partial y} \\ \frac{\partial q}{\partial x} & \frac{\partial q}{\partial y}\end{array}\right]^{-1}$$

The existance of that derivative depends not only on the existance of these partial derivatives but also on dQ/dr being invertible.