$x=r\cosφ, y=r\sinφ$, how to calculate $\dfrac{\partial φ}{\partial x}$?

Question

One approach I see is one approach

But what if I just put the expression $x=r\cosφ$ on the $\partial x$, it gives $\dfrac{-1}{r\sin φ}$, so which is wrong?

Answer 1

We have $$ \begin{align} \mathrm{d}x&=\cos(\varphi)\,\mathrm{d}r-r\sin(\varphi)\,\mathrm{d}\varphi\\ \mathrm{d}y&=\sin(\varphi)\,\mathrm{d}r+r\cos(\varphi)\,\mathrm{d}\varphi \end{align}\tag{1} $$ $(1)$ implies $$ \begin{align} \sin(\varphi)\,\mathrm{d}x&=\sin(\varphi)\cos(\varphi)\,\mathrm{d}r-r\sin^2(\varphi)\,\mathrm{d}\varphi\\ \cos(\varphi)\,\mathrm{d}y&=\sin(\varphi)\cos(\varphi)\,\mathrm{d}r+r\cos^2(\varphi)\,\mathrm{d}\varphi \end{align}\tag{2} $$ and subtracting the equations of $(2)$ gives $$ -\sin(\varphi)\,\mathrm{d}x+\cos(\varphi)\,\mathrm{d}y=r\,\mathrm{d}\varphi\tag{3} $$ Thus, $$ \begin{align} \frac{\partial\varphi}{\partial x} &=-\frac{\sin(\varphi)}r\\ &=-\frac y{x^2+y^2} \end{align}\tag{4} $$