Deriviative of a multivariable function with respect to a constant?

[Fractal20]

Homework Statement

So this is arising in my applied math course in solving the wave equation in n dimensions. So we have a function u(\vec{x}+r\vec{z},t) and where x and z are n dimensional vectors and r is a scalar (also, u is a scalar function). Then when we take the partial derivative with respect to r we get:

∇u(\vec{x}+r\vec{z},t)\bullet\vec{z}
(sorry that huge dot is suppose to be a dot product)

I am just use to take derivatives of multivariable functions with respect to variables and then there is the old rule that the result is the gradient of the function dotted with the direction vector. But this is not case. I am okay with excepting this as a rule. It is the same as the single variable analog if ∇ was replaced with the partial with respect to r. But I would still like some sense of why it should be this way. Doe this result somehow follow from the limit as h approaches zero of

(1/h)*(u(\vec{x}+(r+h)\vec{z},t)-u(\vec{x}+r\vec{z},t))<br /> <br /> ? Can anybody offer some insight? Thanks<br /> <br /> <h2>Homework Equations</h2><br /> <br /> <br /> <br /> <h2>The Attempt at a Solution</h2><br />

 

[SteamKing]

I think you must expand u into its vector components and differentiate term by term.
It might be possible also to use the chain rule.

Without knowing the nature of u or the constituent vectors, that is all the advice I can offer.

 

[LCKurtz]

Homework Statement

So this is arising in my applied math course in solving the wave equation in n dimensions. So we have a function u(\vec{x}+r\vec{z},t) and where x and z are n dimensional vectors and r is a scalar (also, u is a scalar function). Then when we take the partial derivative with respect to r we get:

∇u(\vec{x}+r\vec{z},t)\bullet\vec{z}
(sorry that huge dot is suppose to be a dot product)

Don't be sorry, use Tex: $\nabla u(\vec x+r \vec z,t)\cdot \vec z$
Here's what I typed to get that, it's actually easier that what you did:

##\nabla u(\vec x+r \vec z,t)\cdot \vec z##