What is derivative of a vector respect to another vector?

**yungman** · September 17th 2010, 08:34 PM

I am confused. I never seen derivative of a vector respect to another vector. When I go on the web, the article just show divergence, curl, gradient etc. But not derivative of a vector respect to another vector?

For example what is $\frac{d(\vec{x}-\vec{x_0})^2}{d \vec{x}} ?$ For $\vec{x_0}$ is a constant vector.

The book seems to imply:

$\frac{d[(\vec{x}-\vec{x_0})^2]}{d \vec{x}} = 2(\vec{x}-\vec{x_0}) \frac{d \vec{x}}{d \vec{x}} = 2(\vec{x}-\vec{x_0})$

I guess I don't know how to do a derivative like this. Can anyone help?

**Opalg** · September 18th 2010, 03:48 AM

Originally Posted by yungman

I am confused. I never seen derivative of a vector respect to another vector. When I go on the web, the article just show divergence, curl, gradient etc. But not derivative of a vector respect to another vector?

For example what is $\frac{d(\vec{x}-\vec{x_0})^2}{d \vec{x}} ?$ For $\vec{x_0}$ is a constant vector.

The book seems to imply:

$\frac{d[(\vec{x}-\vec{x_0})^2]}{d \vec{x}} = 2(\vec{x}-\vec{x_0}) \frac{d \vec{x}}{d \vec{x}} = 2(\vec{x}-\vec{x_0})$

I guess I don't know how to do a derivative like this. Can anyone help?

I think you should start by asking what is meant by $(\vec{x}-\vec{x_0})^2$ . The square of a vector is not normally defined at all. Can you post a more detailed account of what "the book" actually says when it "seems to imply" that this (undefined?) product has an (undefined?) derivative?

**HallsofIvy** · September 18th 2010, 04:44 AM

Assuming that we are talking about vectors in [tex]R^[tex] (more specifically $R^3$ ) and not abstract vector spaces, my interpretation of $(\vec{x}- \vec{x}_0)^2$ would be a "scalar product"- the square of its length. In that case, this is a function from a vector space to its underlying field (probably the real numbers). Assuming this is a 3 dimensional vector space, we can think of that as a function of three variables, the x, y, z, components of the vector and then the "deriviative of $f:R^3 \rightarrow R$ " is the "gradient" vector, $\left<\begin{array}{ccc}\frac{\partial f}{\partial x} & \frac{\partial f}{\partial y} & \frac{\partial f}{\partial z}\end{array}\right>$ . If we have, specifically, $f= \left<\begin{array}{ccc}x- x_0 & y- y_0 & z- z_0 \end{array}\right>^2= (x- x_0)^2+ (y- y_0)^2+ (y- y_0)^2$ then the derivative, or gradient, is $\left<\begin{array}{ccc}2(x- x_0) & 2(y- y_0) & 2(z- z_0) \end{array}\right> = 2\left(\left<\begin{array}{ccc}x & y & z\end{array}\right>- \left<\begin{array}{ccc}x_0 & y_0 & z_0\right>\end{array}$ as claimed.

(Strictly speaking, the derivative, at a given $\right<\begin{array}{ccc}x_0 & y_0 & z_0\end{array}\right>$ , is the linear function that maps $\right<\begin{array}{ccc}x & y & z\end{array}\right>$ to $\frac{\partial f(x_0, y_0, z_0)}{\partial x}x+ \frac{\partial f(x_0, y_0, z_0)}{\partial y}y+ \frac{\partial f(x_0, y_0, z_0)}{\partial z}z$ but we can think of that as the dot product of $\left<\begin{array}{ccc}\frac{\partial f}{\partial x} & \frac{\partial f}{\partial y} & \frac{\partial f}{\partial z}\end{array}\right>$ with $\left<\begin{array}{ccc}x & y & z\end{array}\right>$ just as we think of the derivative of f(x) at $x= x_0$ to be the number $f'(x_0)$ when, more deeply, it is the linear function $y= f'(x_0)x$ .

If you have an actual "vector" function from $R^3$ to $R^3$ , say, $\left<\begin{array}{ccc}f(x,y,z), g(x,y,z), h(x,y,z)\end{array}\right>$ , then we can think of its derivative as the matrix
$\begin{bmatrix}\frac{\partial f}{\partial x} & \frac{\partial f}{\partial y} & \frac{\partial f}{\partial z} \\ \frac{\partial g}{\partial x} & \frac{\partial g}{\partial y} & \frac{\partial g}{\partial z} \\ \frac{\partial h}{\partial x} & \frac{\partial h}{\partial y} & \frac{\partial h}{\partial z}\end{bmatrix}$ .

More correctly, it is the linear transformation defined by that matrix.

**yungman** · September 18th 2010, 02:31 PM

Sorry guys and thanks for the help.

I type wrong, it was $|\vec{x}-\vec{x_0}|^2$ , sorry.

The book is PDE by Strauss p194 to p195. It is part of the derivation of the Green's Function for sphere. The part is about normal derivative of G. It talked about derivation respect to $\vec{x}$ and some very funcky statement I still don't understand. But the later part just went back to the ordinary definition of normal derivative:

$\frac{\partial G}{\partial n} = \nabla G \cdot \hat{n}$

and derive the equation accordinary as if nothing happened!!!! So it is a non question at this point. Strauss is not a good book in any stretch. I just cannot find any PDE book that cover the Green's Function and the EM book that I ordered is still in shipment!!!

Thanks

Alan

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