- #1
agnimusayoti
- 240
- 23
- 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.
$$(\frac {\partial u}{\partial x})_z = (\frac {\partial u}{\partial x})_y (\frac {\partial x}{\partial x})_z + (\frac {\partial u}{\partial y})_x (\frac {\partial y}{\partial x})_zz$$
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.