# Chain Rule in Partial Derivative

• Mar 31st 2010, 06:18 AM
Lafexlos
Chain Rule in Partial Derivative
$f\left(u,v,w\right)$ is differentiable and $u=\;x-y,\; v=\;y-z,\;w=\;z-x$, show that $\frac{\partial{f}}{\partial{x}}\;+\;\frac{\partial {f}}{\partial{y}}\;+\;\frac{\partial{f}}{\partial{ z}}\;=\;0$.

I think i'll use chain rule and that equation will be $\frac{\partial{f}}{\partial{u}}\frac{\partial{u}}{ \partial{x}}\;+\;\frac{\partial{f}}{\partial{v}}\f rac{\partial{v}}{\partial{y}}\;+\;\frac{\partial{f }}{\partial{w}}\frac{\partial{w}}{\partial{z}}\;=\ ;0$
From now on i can not find any value for $\frac{\partial{f}}{\partial{u}}\;$ ,, $\frac{\partial{f}}{\partial{v}}\;$ and $\;\frac{\partial{f}}{\partial{x}}$. (Headbang) So how can i find these values?

If that's not the way of solution, i need some hints.
• Mar 31st 2010, 06:31 AM
Ruun
This is only intuition

$f\left(u,v,w\right) = f\left(x-y,y-z,z-x\right)$

$\frac{\partial f}{\partial u} = \frac{\partial f}{\partial (x - y)} = \frac{\partial f}{\partial x} - \frac{\partial f}{\partial y}$.
• Mar 31st 2010, 06:36 AM
Lafexlos
wow.! I didn't know, i can split derivative like that.
Thanks. =)
• Mar 31st 2010, 06:39 AM
Ruun
Neither do I lol, it`s just the only thing I can think of, may be wrong. Wait for a serious mathematician or something..
• Mar 31st 2010, 01:12 PM
ICanFly
Hint
Hint: $\frac{{\partial f}}{{\partial x}} = \frac{{\partial f}}{{\partial u}} \times \frac{{\partial u}}{{\partial x}} + \frac{{\partial f}}{{\partial w}} \times \frac{{\partial w}}{{\partial x}}
$
• Mar 31st 2010, 01:31 PM
Lafexlos
Quote:

Originally Posted by Lafexlos
I think i'll use chain rule and that equation will be $\frac{\partial{f}}{\partial{u}}\frac{\partial{u}}{ \partial{x}}\;+\;\frac{\partial{f}}{\partial{v}}\f rac{\partial{v}}{\partial{y}}\;+\;\frac{\partial{f }}{\partial{w}}\frac{\partial{w}}{\partial{z}}\;=\ ;0$
From now on i can not find any value for $\frac{\partial{f}}{\partial{u}}\;$ ,, $\frac{\partial{f}}{\partial{v}}\;$ and $\;\frac{\partial{f}}{\partial{x}}$. (Headbang) So how can i find these values?

If that's not the way of solution, i need some hints.

.
• Mar 31st 2010, 02:45 PM
ICanFly
$\frac{{\partial f}}{{\partial x}} = \frac{{\partial f}}{{\partial u}} \times \frac{{\partial u}}{{\partial x}} + \frac{{\partial f}}{{\partial w}} \times \frac{{\partial w}}{{\partial x}} = \frac{{\partial f}}{{\partial u}} - \frac{{\partial f}}{{\partial w}}$

$\frac{{\partial f}}{{\partial y}} = \frac{{\partial f}}{{\partial v}} \times \frac{{\partial v}}{{\partial y}} + \frac{{\partial f}}{{\partial u}} \times \frac{{\partial u}}{{\partial y}} = \frac{{\partial f}}{{\partial v}} - \frac{{\partial f}}{{\partial u}}$

$\frac{{\partial f}}{{\partial z}} = \frac{{\partial f}}{{\partial v}} \times \frac{{\partial v}}{{\partial x}} + \frac{{\partial f}}{{\partial w}} \times \frac{{\partial w}}{{\partial z}} = - \frac{{\partial f}}{{\partial v}} + \frac{{\partial f}}{{\partial w}}$