Laplacian, partial derivatives

[SpY]]

Homework Statement

Find the Laplacian of F = sin(k_x x)sin(k_y y)sin(k_z z)

Homework Equations

\nabla^2 f = \left( \frac{\partial}{\partial x} +\frac{\partial}{\partial y} + \frac{\partial}{\partial z} \right) \cdot \left( \frac{\partial}{\partial x} + \frac{\partial}{\partial y} + \frac{\partial}{\partial z} \right) \cdot F
Where F is a scalar function

The Attempt at a Solution

Biggest problem is with partial derivatives. I don't know how to approach taking a partial derivative of such a big multivariate product :( Just got the derivative of sin(x) is cos(x) and the second derivative -sin(x)

Just want to makes sure, but \frac{\partial}{\partial x}y = 0 but \frac{\partial}{\partial x}xy = y ?

 

[Dick]

Yes to everything you've written. The dot of the two vector operators is the sum of the second derivatives of F. There is no dot product with F. F is a scalar. What do you get if you add the three second derivatives of F with respect to x, y and z?

 

[SpY]]

It will just be the sum that is the Laplacian.. as you just said :O

Problem is how to do one partial derivative, for starters, on
<br /> sin(k_x x)sin(k_y y)sin(k_z z)<br />

 

[HallsofIvy]

What is the derivative, with respect to x, of AB sin(k_x x) with A and B constants?

What is the derivative, with respect to y, of AB sin(k_y y) with A and B constants?

What is the derivative, with respect to z, of AB sin(k_z z) with A and B constants?

 

[SpY]]

I think when you're taking the partial derivative of the first term, the k term gets pulled out by the chain rule, then just change it to cos. Taking the second derivative brings the k out again by the chain rule, leaving k^. So it will be -k^2(sin-sin-sin)