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

• agnimusayoti
In summary, In this conversation, the first method to solve the equation using the 2-variable chain rule was used.
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.

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

etotheipi
Delta2 said:
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

agnimusayoti said:
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##

agnimusayoti
agnimusayoti said:
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.

agnimusayoti

## 1. How do I solve for ##\frac{\partial x}{\partial x}## in a problem?

The partial derivative of a variable with respect to itself is always equal to 1. Therefore, you can write ##\frac{\partial x}{\partial x}=1## in any problem.

## 2. Is it always true that ##\frac{\partial x}{\partial x}=1##?

Yes, the partial derivative of a variable with respect to itself is always equal to 1. This is a fundamental rule in calculus.

## 3. Can I use ##\frac{\partial x}{\partial x}=1## in any type of problem?

Yes, you can use this equation in any problem that involves finding the partial derivative of a variable with respect to itself. It is a basic rule that applies to all types of problems.

## 4. How do I know when to use ##\frac{\partial x}{\partial x}=1## in a problem?

You should use this equation whenever you need to find the partial derivative of a variable with respect to itself. This is typically used in problems involving multivariable calculus or in physics and engineering problems.

## 5. Can I use ##\frac{\partial x}{\partial x}=1## in higher dimensions?

Yes, this equation can be used in any number of dimensions. It is a fundamental rule in calculus that applies to all dimensions.

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