Linearizing f(x,y,z) at (1,1,1) for Quick Calculations

[Mindscrape]

I am linearizing the function $$f(x,y,z) = tan^{-1}(xyz)$$ at the point (1,1,1).

Since $$f(x_0,y_0,z_0)= \frac{\pi}{4} + \pi*n$$ should I just take the first value or do I have to carry all the solutions through the linearization process?

Um, anybody remember this? I can put up some work if it helps.

$$L(x,y,z) = \Delta f = f_x \Delta x + f_y \Delta y + f_z \Delta z$$

So

$$L(x,y,z) = f(x_0,y_0,z_0) + f_x(x-x_0) + f_y (y-y_0) + f_z (z-z_0)$$

where $$f_x = \frac{\partial f}{\partial x}$$ (and f_y and f_z)

So does $$L(x,y,z) = \frac{\pi}{4} +\pi n + \frac{1}{2}[(x-1)+(y-1)+(z-1)]$$
?

 

[Mindscrape]

Surely somebody must know. I have doubts that I should include the general solutions, because that would be a strange linearization. I'm not really sure though, since the general solutions would be just as valid, unless I am overlooking something.