Can I write ##\frac{\partial x}{\partial x}=1## from Problems, ML Boas

[agnimusayoti]

Homework Statement

given $u(x,y)$ and $y(x,z)$: show that
$$(\frac {\partial u}{\partial x})_z = (\frac {\partial u}{\partial x})_y + (\frac {\partial u}{\partial y})_x (\frac {\partial y}{\partial x})_z$$

Relevant Equations

Chain rule

First, I tempted to use this kind of chain rule:
$$(\frac {\partial u}{\partial x})_z = (\frac {\partial u}{\partial x})_y (\frac {\partial x}{\partial x})_z + (\frac {\partial u}{\partial y})_x (\frac {\partial y}{\partial x})_zz$$
Then, I think that $(\frac {\partial x}{\partial x})_z = 1$
i got the result.

But, I doubt about my answer. So I try to find different answer. I did in this way (i think below is the right answer).
For $u(x,y) \Rightarrow du = (\frac {\partial u}{\partial x})_y dx + (\frac {\partial u}{\partial y})_x dy$.
For $y(x,z) \Rightarrow dy = (\frac {\partial y}{\partial x})_z dx + (\frac {\partial y}{\partial z})_x dz$.
Then, I substitute dy from 2nd equation to the 1st one and get the same result.

Now, the question is can I use the first method, where I define $(\frac {\partial x}{\partial x})_z = 1$?
Thanks.

 

[Delta2]

I think 1st method is correct, I don't see anything wrong, you essentially use the 2-variable chain rule.

 

[agnimusayoti]

I think 1st method is correct, I don't see anything wrong, you essentially use the 2-variable chain rule.

But, there is no x function as x and z independent variable. This was confusing me. Could you explain? Thanks

 

[Mark44]

But, there is no x function as x and z independent variable. This was confusing me. Could you explain? Thanks

x is certainly a function of itself, so you can write $\frac{dx}{dx} = 1$ or $\frac{\partial x}{\partial x} = 1$

 

[Delta2]

But, there is no x function as x and z independent variable. This was confusing me. Could you explain? Thanks

I think I understand your confusion. We need to introduce a new variable/function $w(x,z)=x$. We will have the composite function $u(w(x,z),y(x,z))$ in which you can now clearly apply the 2-variable chain rule. Because essentially w=x you ll get the desired result.