# Derivative of multivariable function

• rojan1918
In summary, the problem is that the order of differentiation doesn't matter for the second derivative of a multivariable function. All of the mentioned conditions must be met for the order of differentiation to change.
rojan1918
So this is a problem for microeconomics, but should follow under general calculus:

The point is x=(x1(p1,p2,u),x2(p1,p2,u)) where u is a constant on the function u(x1,x2). p1 is the price of x1 and p2 for x2. I'm supposed to show that (dx1/dp2)=(dx2/dp1). I've been given the info that for the second derivative of a multivariable function the order of differentation is irrelevant (df(x1,x2)/dx1dx2)=(df(x1,x2)/dx2dx1). All of the mentioned should also be the case for functions of more than two variables.

I've tried to set second derivatives equal to each other, but obviously ended up with the exact same on both sides. I don't know where to go from here. Hints and tips would be much appreciated.

rojan1918 said:
So this is a problem for microeconomics, but should follow under general calculus:

The point is x=(x1(p1,p2,u),x2(p1,p2,u)) where u is a constant on the function u(x1,x2). p1 is the price of x1 and p2 for x2. I'm supposed to show that (dx1/dp2)=(dx2/dp1). I've been given the info that for the second derivative of a multivariable function the order of differentation is irrelevant (df(x1,x2)/dx1dx2)=(df(x1,x2)/dx2dx1). All of the mentioned should also be the case for functions of more than two variables.

I've tried to set second derivatives equal to each other, but obviously ended up with the exact same on both sides. I don't know where to go from here. Hints and tips would be much appreciated.

Please stop using boldface typing; it looks like you are yelling at us.

Anyway, I do not think your question can be correct as written, because I can very easily devise two functions ##x_1(p_1,p_2,u)## and ##x_2(p_1,p_2,u)## with ##\partial x_1/\partial p_2 \neq \partial x_2/ \partial p_1##. Are you actually given some specific functions ##x_1## and ##x_2## as functions of ##(p_1,p_2)##? If you are given the right kind of functions then the result could be true.

Furthermore, if ##u## is really a function of ##(x_1,x_2)## as you state, then you are dealing with functions of the form ##x_1(p_1,p_2,u(x_1,x_2))##, in which the function ##x_1## is also a function of itself. That really makes no sense.

It is not always relevant. The necessary and sufficient condition for the change of order of differentiation is that all the functions and partial derivatives must be continuous.

Have you applied the product rule and chain rule appropriately for both derivatives?

The problem I'm running into is I think you need to assume that x is linear in x1 and x2. Or be given that info. Is there any additional information?

Alpharup
Does this equation talk about rate of substitution?
I assume whatever you have given.
x is a function of x1 and x2
write x =k(x1)m(x2) where k and m are functions of x1 and x2 alone, respectively.
x1 can similarly be written as V1(p1)U1(p2)
y1 can similarly be written as V2(p1)U2(p2)
You have also given function u.
let u be h(x1)s(x2)at this particular point 'x',
k(x1)m(x2) is constant.
I think it should be fine now for differentiating and using chain rule.

Last edited:
So assuming all the necessary conditions are met (continuity)
##\frac{\partial x}{\partial p_1} = \frac{\partial x}{\partial x_1}\frac{\partial x_1}{\partial p_1} + \frac{\partial x}{\partial x_2}\frac{\partial x_2}{\partial p_1}##
##\frac{\partial x}{\partial p_2} = \frac{\partial x}{\partial x_1}\frac{\partial x_1}{\partial p_2} + \frac{\partial x}{\partial x_2}\frac{\partial x_2}{\partial p_2}##
##\frac{\partial}{\partial p_2}\frac{\partial x}{\partial p_1} = \frac{\partial x}{\partial x_1}\frac{\partial^2 x_1}{\partial p_1 \partial p_2} + \frac{\partial x_1}{\partial p_1}\frac{\partial^2 x}{\partial x_1 \partial p_2} + \frac{\partial x}{\partial x_2}\frac{\partial^2 x_2}{\partial p_1 \partial p_2} + \frac{\partial x_2}{\partial p_1}\frac{\partial^2 x}{\partial x_2 \partial p_2} ##
##\frac{\partial}{\partial p_1}\frac{\partial x}{\partial p_2} = \frac{\partial x}{\partial x_1}\frac{\partial^2 x_1}{\partial p_2 \partial p_1} + \frac{\partial x_1}{\partial p_2}\frac{\partial^2 x}{\partial x_1 \partial p_1} + \frac{\partial x}{\partial x_2}\frac{\partial^2 x_2}{\partial p_2 \partial p_1} + \frac{\partial x_2}{\partial p_2}\frac{\partial^2 x}{\partial x_2 \partial p_1}##
setting them equal and cancelling:
## \frac{\partial x}{\partial x_1}\frac{\partial^2 x_1}{\partial p_2 \partial p_1} + \frac{\partial x_1}{\partial p_1}\frac{\partial^2 x}{\partial x_1 \partial p_1} + \frac{\partial x}{\partial x_2}\frac{\partial^2 x_2}{\partial p_2 \partial p_1} + \frac{\partial x_2}{\partial p_2}\frac{\partial^2 x}{\partial x_2 \partial p_1} = \frac{\partial x}{\partial x_1}\frac{\partial^2 x_1}{\partial p_1 \partial p_2} + \frac{\partial x_1}{\partial p_1}\frac{\partial^2 x}{\partial x_1 \partial p_2} + \frac{\partial x}{\partial x_2}\frac{\partial^2 x_2}{\partial p_1 \partial p_2} + \frac{\partial x_2}{\partial p_1}\frac{\partial^2 x}{\partial x_2 \partial p_2} ##
Should end up with
##\frac{\partial x_1}{\partial p_1}\frac{\partial^2 x}{\partial x_1 \partial p_1} + \frac{\partial x_2}{\partial p_1}\frac{\partial^2 x}{\partial x_2 \partial p_2} = \frac{\partial x_2}{\partial p_2}\frac{\partial^2 x}{\partial x_2 \partial p_1} + \frac{\partial x_1}{\partial p_2}\frac{\partial^2 x}{\partial x_1 \partial p_1}##

Essentially, you need ##\frac{\partial}{\partial p_2} \frac{\partial x}{\partial x_1} = 0## and ##\frac{\partial}{\partial p_1} \frac{\partial x}{\partial x_2} = 0##
I don't see how you can get here with this information.

P.S.
Ctrl+v & Ctrl+c made this post possible...

## 1. What is the definition of a derivative of a multivariable function?

The derivative of a multivariable function is a measure of how the function changes with respect to each of its input variables. It is a vector that contains the partial derivatives of the function with respect to each input variable.

## 2. How is the derivative of a multivariable function calculated?

The derivative of a multivariable function is calculated using the concept of partial derivatives. Each partial derivative is calculated by holding all other input variables constant and taking the derivative of the function with respect to the variable in question.

## 3. What is the significance of the derivative of a multivariable function?

The derivative of a multivariable function is used to find the slope of the function at a specific point. This allows us to determine the direction and magnitude of change in the function at that point, which is useful in optimization problems and understanding the behavior of the function.

## 4. Can the derivative of a multivariable function have more than one value at a single point?

Yes, the derivative of a multivariable function can have more than one value at a single point. This occurs when the function has more than one input variable and the derivatives with respect to each variable are not constant or proportional to each other.

## 5. How is the derivative of a multivariable function used in real-world applications?

The derivative of a multivariable function is used in many fields, including physics, economics, and engineering. It is used to model and analyze complex systems, optimize processes, and make predictions about how a system will behave under different conditions.

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