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The questions is

If $f(u,v,w)$ is differentiable and $u=x-y$, $v=y-z$, and $w=z-x$, show that $$\begin{equation} \frac{\partial{f}}{\partial{x}}+\frac{\partial{f}}{\partial{y}}+\frac{\partial{f}}{\partial{z}}=0 \end{equation}$$

The things that I could only get is $$\begin{equation} \frac{\partial{f}}{\partial{x}}+\frac{\partial{f}}{\partial{y}}+\frac{\partial{f}}{\partial{z}}=\frac{\partial{f}}{\partial{u}}+\frac{\partial{f}}{\partial{v}}+\frac{\partial{f}}{\partial{w}} \end{equation}$$

At that point, I even don`t know what $f(x)$ is....so how can I get the answer?

Alain Remillard
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Chatanee
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  • To start with, what did you obtain for $\partial_x f$, $\partial_y f$, and $\partial_z f$ in terms of the partials with respect to $u,v,w$? – bames Apr 13 '20 at 01:08
  • Oh, I see. I should find each of them and then sum these equations. Thanks!! – Chatanee Apr 13 '20 at 01:11

2 Answers2

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It doesn't matter what $f$ is. When you want partial derivative with respect to à variable, you need to take every path that leads to this variable. Since $u$ and $w$ depends on $x$, $$\frac{\partial f}{\partial x}=\frac{\partial f}{\partial u}\frac{\partial u}{\partial x}+\frac{\partial f}{\partial w}\frac{\partial w}{\partial x}=\frac{\partial f}{\partial u}-\frac{\partial f}{\partial w}$$ Same goes for $y$ and $z$ $$\frac{\partial f}{\partial y}=\frac{\partial f}{\partial u}\frac{\partial u}{\partial y}+\frac{\partial f}{\partial v}\frac{\partial v}{\partial y}=-\frac{\partial f}{\partial u}+\frac{\partial f}{\partial v}$$ $$\frac{\partial f}{\partial z}=\frac{\partial f}{\partial v}\frac{\partial u}{\partial z}+\frac{\partial f}{\partial w}\frac{\partial w}{\partial z}=-\frac{\partial f}{\partial v}+\frac{\partial f}{\partial w}$$ If you sum those three equations, you find the $0$ you are looking for.

Alain Remillard
  • 3,625
  • I didn't know that $f$ doesn't matter...The point was I had to do partial derivative with each of $u,v,w$. Thanks for amazing explanation! – Chatanee Apr 13 '20 at 01:14
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$$\frac{\partial{f}}{\partial{x}}=\frac{\partial{f}}{\partial{u}}\frac{\partial{u}}{\partial{x}}+\frac{\partial{f}}{\partial{v}}\frac{\partial{v}}{\partial{x}}+\frac{\partial{f}}{\partial{w}}\frac{\partial{w}}{\partial{x}} $$ $$\frac{\partial{f}}{\partial{y}}=\frac{\partial{f}}{\partial{u}}\frac{\partial{u}}{\partial{y}}+\frac{\partial{f}}{\partial{v}}\frac{\partial{v}}{\partial{y}}+\frac{\partial{f}}{\partial{w}}\frac{\partial{w}}{\partial{y}} $$ $$\frac{\partial{f}}{\partial{z}}=\frac{\partial{f}}{\partial{u}}\frac{\partial{u}}{\partial{z}}+\frac{\partial{f}}{\partial{v}}\frac{\partial{v}}{\partial{z}}+\frac{\partial{f}}{\partial{w}}\frac{\partial{w}}{\partial{z}}$$

Eeyore Ho
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