- #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.