Partial derivatives using polar coordinates I was given the following problems as practice, and I've sold all but one. However, I am not sure that my answers are correct.

Calculate $(\partial r/\partial x)_y$ , $(\partial r/\partial y)_x$ , $(\partial θ/\partial x)_y$ ,$(\partial y/\partial x)_r$, $(\partial r/\partial \theta)_x$
$x=r\cos(\theta)$, $y=r\sin(\theta)$

I have that

$(\partial r/\partial x)_y=\cos(\theta)$
$(\partial r/\partial y)_x=\sin(\theta)$
$(\partial y/\partial x)_r=-\cot(\theta)$
$(\partial \theta/\partial x)y=-\sin(\theta)/r$

Would $(\partial r/\partial \theta)_x$ then be equal to $r\tan\theta$?
 A: Note that $r=\sqrt{x^2+y^2}$.  Holding $y$ fixed we have
$$\begin{align}
\frac{\partial r}{\partial x}&=\frac{x}{\sqrt{x^2+y^2}}\\\\
&=\frac xr\\\\
&=\cos(\theta)
\end{align}$$

Similarly, holding $x$ fixed we have
$$\begin{align}
\frac{\partial r}{\partial y}&=\frac{y}{\sqrt{x^2+y^2}}\\\\
&=\frac yr\\\\
&=\sin(\theta)
\end{align}$$

Then, holding $r$ fixed we see that $\sqrt{x^2+y^2}$ is constant.  Hence, 
$$\begin{align}
\frac{\partial r}{\partial x}&=0\\\\
&=\frac{x}{\sqrt{x^2+y^2}}+\frac{y}{\sqrt{x^2+y^2}}\frac{\partial y}{\partial x}
\end{align}$$
whereupon solving for $\frac{\partial y}{\partial x}$ we find that 
$$\begin{align}
\frac{\partial y}{\partial x}&=-\frac xy\\\\
&=-\cot(\theta)
\end{align}$$

Finally, holding $y$ fixed, we have $x=y\cot(\theta)$ so that 
$$\begin{align}
\frac{\partial x}{\partial x}&=1\\\\
&=y\left(-\csc^2(\theta)\frac{\partial \theta}{\partial x}\right)
\end{align}$$
whereupon solving for $\frac{\partial \theta}{\partial x}$ we find that
$$\begin{align}
\frac{\partial \theta}{\partial x}&=-\sin^2(\theta)/y\\\\
&=-\sin^2(\theta)/(r\sin(\theta))\\\\
&=-\sin(\theta)/r
\end{align}$$
