Derivative and transformation in the derivation of the normal distribution I am trying to understand the normal distribution and its derivation through an available derivation, but I don't understand how this first equation is transformed into the differential equation at the bottom of this question.
$\theta$ is an angle, x and y are cartesian coordinates
$g(r) = p(x)p(y)$
differentiate with respect to $\theta$ :
$0 =  p(x) \frac{d~p(y)}{d~\theta } ~ p(y) \frac{d~p(x)}{d~\theta}$
Using $x = r\cos(\theta)$ and $y = r\sin(\theta)$ rewrite derivatives as:
$0 = p(x)~p(y)~(r\cos(\theta)) + p(y)~p(x)~(-r\sin(\theta))$
rewrite again as 
$0 = p(x)~p(y)~x - p(y)~p(x)~y$
giving differential equation
$\frac{p(x)}{x~p(x)}$ = $\frac{p(y)}{y~p(y)}$
how can the polar form of the x and y coordinates be used to rewrite the derivative? I am confused about how the differential equation was found from the beginning equation. What is the relationship between the polar coordinates $r\cos(\theta), r\sin(\theta)$ and the result?
Source: http://courses.ncssm.edu/math/Talks/PDFS/normal.pdf
 A: There seem to be some formatting issues with your pdf.   That should be:
$$\def\d{\mathop{\rm d}}\begin{align}g(r) &= p(x)~p(y) && \text{given}
\\[2ex] g(r) &= p(r\cos\theta)~p(r\sin\theta) & & x\gets r\cos\theta~,~ y\gets r\sin\theta
\\[2ex] \dfrac{\d g(r)}{\d \theta} & = \dfrac{\d p(r\cos\theta)~p(r\sin\theta)}{\d \theta}& &\text{derivative respective to }\theta
\\[2ex] 0 &= p(r\cos\theta)\dfrac{\d p(r\sin\theta)}{\d \theta}+p(r\sin\theta)\dfrac{\d p(r\cos\theta)}{\d \theta} & &\text{product rule}
\\[2ex] 0 &= p(r\cos\theta)~\dfrac{\d r\sin\theta}{\d \theta}\dfrac{\d p(r\sin\theta)}{\d r\sin\theta} + p(r\sin\theta)~\dfrac{\d r\cos\theta}{\d \theta}~\dfrac{\d p(r\cos\theta)}{\d r\cos\theta} & & \text{chain rule}
\\[2ex] 0 &= r\cos\theta~p(r\cos\theta)~p'(r\sin\theta) -r\sin\theta~p(r\sin\theta)~p'(r\cos\theta) & & \text{}
\\[2ex] 0 &= x~p(x)~p'(y)-y~p(y)~p'(x) & & r\cos\theta\gets x, r\sin\theta\gets y
\\[2ex] \dfrac{p'(y)}{y~p(y)} &= \dfrac{p'(x)}{x~p(x)} && \text{algebraic rearrangement}
\end{align}$$
