Partial derivatives for polar coordiantes I'm given that $\varphi = \arctan\left(\frac{y}{x}\right)$ and I'm asked to show that $$\frac{\partial x}{\partial \varphi}=-r\sin\varphi$$
I've tried to do this and I'm pretty sure this isn't true. 
I've said the following:
$$\tan\varphi=\frac{y}{x}$$
$$\sec^2(\varphi) \frac{\partial \varphi}{\partial x}=\frac{-y}{x^2}$$ $$ \frac{\partial \varphi}{\partial x}=-\frac{1}{r}\sin\varphi$$ $$\frac{\partial x}{\partial \varphi}=\frac{-r}{\sin\varphi}$$
Am I wrong or is it the question?
 A: This sort of error often arises because of our terrible notation for partial derivatives that fails to indicate what's being held constant. In the context of $x,y$ and $r,\phi$, it's often implicitly understood that these form two pairs of variables that belong together, and then it's assumed that if you form the partial derivative with respect to $r$ you're holding $\phi$ constant and vice versa unless otherwise indicated.
Written in unambiguous notation, what you did is:
$$
\tan\varphi=\frac yx\;,
$$
$$
\def\part#1#2#3{\left.\frac{\partial#1}{\partial#2}\right|_#3}\part{}xy\tan\varphi=\part{}xy\frac{y}{x}\;,
$$
$$
\sec^2(\varphi)\part\varphi xy=\frac{-y}{x^2}\;,
$$
$$
\part\varphi xy=-\frac1r\sin\varphi\;,
$$
$$
\part x\varphi y=\frac{-r}{\sin\varphi}\;.
$$
Note especially the last step. You can take reciprocals of partial derivatives just like reciprocals of total derivatives, but that doesn't change what's being held constant. The exercise is using the convention mentioned above and is asking for
$$
\part x\phi r\;.
$$
You can't get this as the reciprocal of a single partial derivative without mixing the variable pairs. You can write
$$
\part x\phi r=\left(\part\phi xr\right)^{-1}\;.
$$
But if you want to get it as a reciprocal in terms of the usual partial derivatives that don't mix the variable pairs (I don't know whether that's an efficient way to get it), it has to be as part of the matrix reciprocal
$$
\pmatrix{
\part x\phi r&\part xr\phi\\
\part y\phi r&\part yr\phi
}
=
\pmatrix{
\part\phi xy&\part\phi yx\\
\part rxy&\part ryx
}^{-1}\;,
$$
which you can verify by multiplying the matrices and using the chain rule.
Update: I figured it might be helpful to collect all the various partial derivatives we can form here.
We have
\begin{align}
x&=r\cos\varphi\;,\\
y&=r\sin\varphi\;,
\end{align}
and
\begin{align}
r&=\sqrt{x^2+y^2}\;,\\
\varphi&=\arctan\frac yx\;.
\end{align}
So the "normal" partial derivatives that don't mix the coordinate pairs are
\begin{align}
\part xr\varphi&=\cos\varphi\;,\\
\part x\varphi r&=-r\sin\varphi\;,\\
\part yr\varphi&=\sin\varphi\;,\\
\part y\varphi r&=r\cos\varphi\;,
\end{align}
and
\begin{align}
\part rxy&=\frac x{\sqrt{x^2+y^2}}\;,\\
\part ryx&=\frac y{\sqrt{x^2+y^2}}\;,\\
\part\varphi xy&=\frac{-y}{x^2+y^2}\;,\\
\part\varphi yx&=\frac x{x^2+y^2}\;.
\end{align}
From these we can obtain $8$ "mixed" partial derivatives by taking the reciprocals:
\begin{align}
\part rx\varphi&=\frac1{\cos\varphi}\;,\\
\part\varphi xr&=-\frac1{r\sin\varphi}\;,
\end{align}
and so on. There are $8$ more partial derivatives that can't be obtained in this way because one variable of a pair is differentiated with respect to the other, but they can be expressed as products of the other ones:
\begin{align}
\part xyr&=\part x\varphi r\part\varphi yr\\
&=\frac{-r\sin\varphi}{r\cos\varphi}\\
&=-\tan\varphi\;,\\\\
\part xy\varphi&=\part xr\varphi\part ry\varphi\\
&=\frac{\cos\varphi}{\sin\varphi}\\
&=\cot\varphi\;,\\\\
\part r\varphi x&=\part ryx\part y\varphi x\\
&=\frac y{\sqrt{x^2+y^2}}\frac{x^2+y^2}x\\
&=\frac yx\sqrt{x^2+y^2}\\
&=r\tan\varphi\;,\\\\
\part r\varphi y&=\part rxy\part x\varphi y\\
&=\frac x{\sqrt{x^2+y^2}}\frac{x^2+y^2}{-y}\\
&=-\frac xy\sqrt{x^2+y^2}\\
&=-r\cot\varphi\;,
\end{align}
and their four reciprocals. And then there are $24$ trivial partial derivatives of the forms
$$
\part uvu=0
$$
and
$$
\part uuv=1\;.
$$
In total, that makes $8+8+8+24=48$ partial derivatives, corresponding to all $4^3=64$ combinations of the variables except for the $16$ of the form
$$
\part uvv
$$
(including the case $u=v$), which make no sense.
A: Trigonometry tells us that the tangent is the ratio between two sides of the triangle, the side opposite to the angle and the side adjacent to the angle. Using this, $\tan(\varphi)=y/x$ tells us that $y$ can be seen as the opposite side, and $x$ the adjacent. That means $y=r\sin(\varphi)$ and $x=r\cos(\varphi)$ where the hypotenuse is $r$ and since it is an independent variable in polar coordinates we don't need to write it as a function of $x$ and $y$. Now doing the derivative on $x$ gives you the expected answer:
$$\frac{\partial x}{\partial\varphi}=\frac{\partial (r\cos(\varphi))}{\partial\varphi}=r\frac{\partial\cos(\varphi)}{\partial\varphi}=-r\sin(\varphi)$$
When we change coordinate systems, we consider the variables of one system as dependent and the variables of the other system as independent like this: $x(r,\varphi)$ and $y(r,\varphi)$, or $r(x,y)$ and $\varphi(x,y)$. This is only a convention, since you could use any pair of variables from $x$, $y$, $r$ and $\varphi$ as the dependent and independent variables. Now with that in mind, we see that the partial derivative you are being asked to calculate uses $r$ and $\varphi$ as independent variables, since you are using polar coordinates, and the derivative is taken with respect to $\varphi$. This shows that in that particular partial derivative it is $r$ which is being held constant. But we are using $x(r,\varphi)$ and $y(r,\varphi)$ as independent variables, so consequently, holding $r$ constant and $\varphi$ variable, makes $y=y(\varphi)$, showing that $y$, as a dependent variable, cannot be considered constant (like you did) when we take that derivative holding $r$ constant, since $y$ is now a function of $\varphi$.
A: When performing partial derivatives with multiple variables, you should always keep track of the quantity you are holding fixed. The problem seems to want $(\dfrac{\partial x}{\partial \phi})_r$, while you've computed $(\dfrac{\partial x}{\partial \phi})_{y=r\sin\phi}$. The subscript indicates what quantity you're holding fixed. Though I must say it seems to be an error on the problem's part.
A: It is : $ \tan(φ)=\frac{y}{x} \Leftrightarrow x = \frac{y}{\tan(φ)} \Leftrightarrow \frac{\partial x}{\partial φ} = \frac{\frac{\partial \tan(φ) y}{\partial φ}}{\tan^2(φ)} \Leftrightarrow \frac{\partial x}{\partial φ} =1 + \tan^2(φ)y $ $....$
A: I prefer to do calculation with differentials, because unlike usual partial derivative notation, differentials say what they mean. Furthermore, their calculation better parallels single-variable calculus; in particular, recall implicit differentiation.
The question asks for $\frac{\partial x}{\partial \varphi}$. In terms of differentials, this usually means "Compute $\mathrm{d}x$, and substitute $\mathrm{d}\varphi \to 1$ and $\mathrm{d}r \to 0$".
Following your work idea, the part where you go off the rails is easy to get right with differentials:
$$ \tan(\varphi) = \frac{y}{x} $$
$$ \mathrm{d}  \tan(\varphi) = \mathrm{d} \left( \frac{y}{x} \right) $$
$$ \sec^2(\varphi) \mathrm{d} \varphi
= -\frac{y}{x^2} \mathrm{d}x + \frac{1}{x} \mathrm{d}y $$
In terms of differentials, your error is that you proceeded with $\mathrm{d}y = 0$, but that's not the intended question. The intended question asks for $\mathrm{d}r = 0$.
To figure out what effect $\mathrm{d}r = 0$ has, we need to use some of the other identities. For example,
$$ \mathrm{d}y = \mathrm{d}(r \sin(\varphi)) = \sin(\varphi) \mathrm{d} r + r \cos(\varphi) \mathrm{d}\varphi$$
Since we're setting $\mathrm{d}r = 0$ and $\mathrm{d}\varphi = 1$, we can compute $\mathrm{d}y = r \cos(\varphi)$. Substituting into the above simplifies to
$$ \sec^2(\varphi) = -\frac{y}{x^2} \mathrm{d}x + \frac{r}{x} \cos(\varphi) $$
However, taking a cue from this answer, a shorter path through the algebra is to start with a different identity:
$$ \mathrm{d}x = \mathrm{d}(r \cos(\varphi)) = \cos(\varphi) \mathrm{d}r
- r \sin(\varphi) \mathrm{d} \varphi $$
Now, substituting $\mathrm{d}r \to 0$ and $\mathrm{d}\varphi \to 1$ gives us $-r \sin(\varphi)$, as desired.
