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!