Multivariable Chain Rule Question

[Master1022]

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 $f(x,y) = \frac{x}{y}$ and we had used the substitutions $x = r \cos\theta$ and $y = r \sin\theta$. In the previous parts of the question, we have shown the following result:
$$\frac{\partial f}{\partial x} = \cos\theta \Big(\frac{\partial f}{\partial r}\Big) - \frac{\sin\theta}{r} \Big(\frac{\partial f}{\partial \theta}\Big)$$

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:
$$\frac{\partial f}{\partial x} = \Big(\frac{\partial f}{\partial r}\Big) \Big(\frac{\partial r}{\partial x}\Big) + \Big(\frac{\partial f}{\partial \theta}\Big) \Big(\frac{\partial \theta}{\partial x}\Big)$$

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 $r^2 = x^2 + y^2$ and $\tan\theta = \frac{y}{x}$ and have gone on to the answer. I, however, just used the substitution above $x = r\cos\theta$, but this doesn't seem to lead us in the right direction. Showing my method, we get $\frac{\partial r}{\partial x} = \frac{1}{cos\theta}$ and $\frac{\partial \theta}{\partial x} = \frac{-1}{r\sin\theta}$. Substituting into the above chain rule yields:

$$\frac{\partial f}{\partial x} = \Big(\frac{\partial f}{\partial r}\Big) \Big(\frac{1}{cos\theta} \Big) + \Big(\frac{\partial f}{\partial \theta}\Big) \Big(\frac{-1}{r\sin\theta}\Big)$$. 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 $r$ in terms of $x$ and $y$ 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.

 

[Orodruin]

Please show your work in taking the derivatives.

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

 

[Master1022]

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:
$$r = \frac{x}{cos\theta} -------> \frac{\partial r}{\partial x} = \frac{1}{\cos\theta}$$
and for the theta/r relationship:
$$\cos\theta = \frac{x}{r} -------> \frac{\partial}{\partial x} (\cos\theta) = \frac{\partial}{\partial x} \Big(\frac{x}{r}\Big)$$
$$\frac{\partial}{\partial \theta} (\cos\theta) \Big(\frac{\partial \theta}{\partial x}\Big) = \frac{1}{r}\$$
$$\frac{\partial \theta}{\partial x} = \frac{-1}{r\sin\theta}$$

 

[Orodruin]

Thank you for your fast response. So my working was as follows:
$$r = \frac{x}{cos\theta} -------> \frac{\partial r}{\partial x} = \frac{1}{\cos\theta}$$

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.

 

[Master1022]

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!