Multivariable derivatives problem?

[Jormungandr]

Homework Statement

Let f(x,y,z)=u(t), where t=xyz. Show that f_{xyz} = F(t) and find F(t).

The Attempt at a Solution

I'm a little confused about the presentation of the variables in this problem. What does F(t) refer to? This isn't a chain rule question, because it's presented before chain rule is introduced. I'm just not sure how to go about finding each partial derivative since u(t) isn't explicitly given... Some advice would be appreciated!

 

[HallsofIvy]

It is a chain rule problem, but uses the chain rule for a function of a single variable that you saw back in first semester Calculus. Saying "f(x,y,z)=u(t), where t=xyz" means that f is really as single function of the product xyz rather than a more general function of x, y, and z. For example, f(x, y, z) might be "xyz+ 1" or it might be "(xyz)^3" or "sin(xyz)" but cannot be "x+ 3y+ 2z" or "x^2+yz".

If f(x, y, z)= u(t) with t= xyz, then f_x= u'(xyz)(yz), f_y= u'(xyz)(xz), and f_z= u'(xyz)(xy).

with f_x= u'(xyz)yz, then f_{xy}= u''(xyz)(xz)(yz)+ u'(xyz)z= u''(xyz)xyz^2+ u'(xyz)z and f_{xyz)= u'''(xyz)(xy)(xyz^2)+ 2u''(xyz)xyz+ u''(xyz)xyz+ u'(xyz)=
u'''(xyz)x^2y^2z^2+ 3u''(xyz)xyz)+ u'(xyz).

Now, what do you get if you replace every "xyz" in that with t?