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I am having some trouble proving the following:

Prove that if $f$ is a differentiable function of $3$ variables and $g(x,y,z)=f(x-y,y-z,z-x)$, then $g_x(x,y,z)+g_y(x,y,z)+g_z(x,y,z)=0$

I tried setting $u=x-y,v=y-z,w=z-x$, so that $g(x,y,z)=f(u(x,y,z),v(x,y,z),w(x,y,z)).$ If this is right, can someone help me finish this?

sasha
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  • $g_x = f_u u_x +f_vv_x + f_ww_x= f_u -f_w$ Does that help? –  May 21 '15 at 02:27
  • @Bye_World Kind of, but what about the others? – sasha May 21 '15 at 02:31
  • $g_y = f_u u_y + f_vv_y + f_ww_y = ?$ ($\leftarrow$ you fill in the '$?$')This is just the definition of the chain rule for functions of more than one variable. –  May 21 '15 at 02:33

1 Answers1

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You have the right idea, you just have to use the multivariate chain rule now.

$$\frac{\partial g}{\partial x} = \frac{\partial f}{\partial u}\frac{\partial u}{\partial x} + \frac{\partial f}{\partial v}\frac{\partial v}{\partial x} + \frac{\partial f}{\partial w}\frac{\partial w}{\partial x} = f_u + 0 - f_w.$$

Can you complete the others?

Cameron Williams
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  • If you can, can you go on. I'm not completely firm on this concept. – sasha May 21 '15 at 02:28
  • I don't want to do the entire problem for you since this is a very good learning exercise. Try repeating what I did above with derivatives with respect to $y$ and $z$. (It is very similar.) – Cameron Williams May 21 '15 at 02:30
  • Can you do y, so I get the idea. I'll do z by myself – sasha May 21 '15 at 02:30
  • I sure can :) The way to think of chain rule (informally) is like fractions. Notice above that the $\partial u$, $\partial v$ and $\partial w$ sort of cancel, leaving you with derivatives of $f$ with respect to $x$. (This isn't a perfect analogy in more than one variable but it gives you the idea.) Using this idea, the chain rule for $y$ gives us $$\frac{\partial g}{\partial y} = \frac{\partial f}{\partial u} \frac{\partial u}{\partial y} + \frac{\partial f}{\partial v} \frac{\partial v}{\partial y} + \frac{\partial f}{\partial w} \frac{\partial w}{\partial y} = -f_u + f_v + 0.$$ – Cameron Williams May 21 '15 at 02:33
  • So would z be: $$\frac{\partial g}{\partial z} = \frac{\partial f}{\partial u}\frac{\partial u}{\partial z} + \frac{\partial f}{\partial v}\frac{\partial v}{\partial z} + \frac{\partial f}{\partial w}\frac{\partial w}{\partial z} = 0 - f_v + f_w$$? – sasha May 21 '15 at 02:36
  • Yep! Exactly right. What happens if you add all of those up? – Cameron Williams May 21 '15 at 02:38
  • equals 0, and we're done – sasha May 21 '15 at 02:38
  • Precisely. Glad you got it! – Cameron Williams May 21 '15 at 02:39