Multivariable Chain Rule Question

In summary, when using the chain rule for partial derivatives, it is important to take into account the relationships between the variables involved. In the given equation, the substitution of x and y for r and theta respectively led to incorrect partial derivatives due to not considering the constant values. To arrive at the correct result, it is necessary to use the expressions r^2 = x^2 + y^2 and \tan\theta = \frac{y}{x}.
  • #1
Master1022
611
117
Homework Statement
Show that the result obtained can be done directly via the chain rule for partial derivatives.
Relevant Equations
Chain rule for partial derivatives
For context, we have an equation [itex] f(x,y) = \frac{x}{y} [/itex] and we had used the substitutions [itex] x = r \cos\theta [/itex] and [itex] y = r \sin\theta [/itex]. In the previous parts of the question, we have shown the following result:
[tex] \frac{\partial f}{\partial x} = \cos\theta \Big(\frac{\partial f}{\partial r}\Big) - \frac{\sin\theta}{r} \Big(\frac{\partial f}{\partial \theta}\Big) [/tex]

Now we have been asked to get to this result directly using the chain rule for partial derivatives.
My attempt: the chain rule states that:
[tex] \frac{\partial f}{\partial x} = \Big(\frac{\partial f}{\partial r}\Big) \Big(\frac{\partial r}{\partial x}\Big) + \Big(\frac{\partial f}{\partial \theta}\Big) \Big(\frac{\partial \theta}{\partial x}\Big) [/tex]

This is where I seem to have gone wrong, and I would appreciate any insight as to why that may have been the case. The answers have used [itex] r^2 = x^2 + y^2 [/itex] and [itex] \tan\theta = \frac{y}{x} [/itex] and have gone on to the answer. I, however, just used the substitution above [itex] x = r\cos\theta [/itex], but this doesn't seem to lead us in the right direction. Showing my method, we get [itex] \frac{\partial r}{\partial x} = \frac{1}{cos\theta} [/itex] and [itex] \frac{\partial \theta}{\partial x} = \frac{-1}{r\sin\theta} [/itex]. Substituting into the above chain rule yields:

[tex] \frac{\partial f}{\partial x} = \Big(\frac{\partial f}{\partial r}\Big) \Big(\frac{1}{cos\theta} \Big) + \Big(\frac{\partial f}{\partial \theta}\Big) \Big(\frac{-1}{r\sin\theta}\Big) [/tex]. I am not sure I can see any clever way to get us to the right answer from here.

Looking back, I think the fact that I had two expressions for [itex] r [/itex] in terms of [itex] x [/itex] and [itex] y[/itex] could have been part of the problem. However, I cannot seem to put an exact reason as to why that would be the case.

Thanks in advance.
 
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  • #2
Please show your work in taking the derivatives.

Note that generally
$$
\frac{\partial f}{\partial g} \neq \frac{1}{\frac{\partial g}{\partial f}}.
$$
 
  • #3
Orodruin said:
Please show your work in taking the derivatives.

Note that generally
$$
\frac{\partial f}{\partial g} \neq \frac{1}{\frac{\partial g}{\partial f}}.
$$

Thank you for your fast response. So my working was as follows:
[tex] r = \frac{x}{cos\theta} -------> \frac{\partial r}{\partial x} = \frac{1}{\cos\theta} [/tex]
and for the theta/r relationship:
[tex] \cos\theta = \frac{x}{r} -------> \frac{\partial}{\partial x} (\cos\theta) = \frac{\partial}{\partial x} \Big(\frac{x}{r}\Big) [/tex]
[tex] \frac{\partial}{\partial \theta} (\cos\theta) \Big(\frac{\partial \theta}{\partial x}\Big) = \frac{1}{r}\ [/tex]
[tex] \frac{\partial \theta}{\partial x} = \frac{-1}{r\sin\theta} [/tex]
 
  • #4
Master1022 said:
Thank you for your fast response. So my working was as follows:
[tex] r = \frac{x}{cos\theta} -------> \frac{\partial r}{\partial x} = \frac{1}{\cos\theta} [/tex]
This is incorrect. When you are taking the partial derivative with respect to x, you are keeping y constant, not ##\theta##. Similar arguments for the other partial derivative.
 
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  • #5
Orodruin said:
This is incorrect. When you are taking the partial derivative with respect to x, you are keeping y constant, not ##\theta##. Similar arguments for the other partial derivative.
Ahh yes, that makes sense. Many thanks for helping me out!
 

Related to Multivariable Chain Rule Question

1. What is the multivariable chain rule?

The multivariable chain rule is a formula used in calculus to find the derivative of a function with multiple variables. It is an extension of the chain rule for functions with a single variable.

2. How is the multivariable chain rule applied?

The multivariable chain rule is applied by taking the partial derivatives of the outer function and the inner function, and then multiplying them together. This is done for each variable in the function.

3. What is the purpose of the multivariable chain rule?

The purpose of the multivariable chain rule is to find the rate of change of a function with multiple variables. It is commonly used in physics, engineering, and other fields to solve problems involving multiple variables.

4. Are there any limitations to the multivariable chain rule?

Yes, the multivariable chain rule can only be applied to functions with multiple variables that are differentiable. Additionally, it can only be used for functions with a finite number of variables.

5. Can the multivariable chain rule be used for higher order derivatives?

Yes, the multivariable chain rule can be extended to find higher order derivatives of functions with multiple variables. This involves taking multiple partial derivatives and multiplying them together.

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