Suppose $R$ is a function of $x$ and $y$; then

$\begin{array}{}\text{(1)}& {x}^{2}+{y}^{2}={R}^{2}(x,y);\end{array}$

if we define

$\begin{array}{}\text{(2)}& F(x,y)={x}^{2}+{y}^{2}-{R}^{2}(x,y),\end{array}$

we may also write (1) as

$\begin{array}{}\text{(3)}& F(x,y)={x}^{2}+{y}^{2}-{R}^{2}(x,y)=0;\end{array}$

by the implicit funtion theorem, this equation in fact may be seen as defining $y(x)$, a function of $x$, provided that

$\begin{array}{}\text{(4)}& {\displaystyle \frac{\mathrm{\partial}F(x,y)}{\mathrm{\partial}y}}\ne 0;\end{array}$

we have

$\begin{array}{}\text{(5)}& {\displaystyle \frac{\mathrm{\partial}F(x,y)}{\mathrm{\partial}y}}=2y-2R(x,y){\displaystyle \frac{\mathrm{\partial}R(x,y)}{\mathrm{\partial}y}}\ne 0\end{array}$

provided

$\begin{array}{}\text{(6)}& y\ne R(x,y){\displaystyle \frac{\mathrm{\partial}R(x,y)}{\mathrm{\partial}y}};\end{array}$

under such circumstances, we may affirm $y(x)$ is uniquely determined as a differentiable function of $x$ in some neighborhood of any point $(x,y)$; then we have

$\begin{array}{}\text{(7)}& F(x,y)={x}^{2}+{y}^{2}(x)-{R}^{2}(x,y(x))=0;\end{array}$

we may take the total derivative with respect to x to obtain

$\begin{array}{}\text{(8)}& {\displaystyle \frac{dF(x,y)}{dx}}=2x+2y{\displaystyle \frac{dy(x)}{dx}}-2R(x,y(x))({\displaystyle \frac{\mathrm{\partial}R(x,y(x))}{\mathrm{\partial}x}}+{\displaystyle \frac{\mathrm{\partial}R(x,y(x))}{\mathrm{\partial}y}}{\displaystyle \frac{dy(x)}{dx}})=0;\end{array}$

a little algebra allows us to isolate the terms containing $dy(x)/dx$:

$\begin{array}{}\text{(9)}& x+y{\displaystyle \frac{dy(x)}{dx}}-R(x,y(x))({\displaystyle \frac{\mathrm{\partial}R(x,y(x))}{\mathrm{\partial}x}}+{\displaystyle \frac{\mathrm{\partial}R(x,y(x))}{\mathrm{\partial}y}}{\displaystyle \frac{dy(x)}{dx}})=0;\end{array}$

$\begin{array}{}\text{(10)}& y{\displaystyle \frac{dy(x)}{dx}}-R(x,y(x)){\displaystyle \frac{\mathrm{\partial}R(x,y(x))}{\mathrm{\partial}y}}{\displaystyle \frac{dy(x)}{dx}}=R(x,y(x)){\displaystyle \frac{\mathrm{\partial}R(x,y(x))}{\mathrm{\partial}x}}-x;\end{array}$

$\begin{array}{}\text{(11)}& (y-R(x,y(x)){\displaystyle \frac{\mathrm{\partial}R(x,y(x)}{\mathrm{\partial}y}}){\displaystyle \frac{dy(x)}{dx}}=R(x,y(x)){\displaystyle \frac{\mathrm{\partial}R(x,y(x))}{\mathrm{\partial}x}}-x;\end{array}$

for the sake of compactess and brevity, we introduce the subscript notation

$\begin{array}{}\text{(12)}& {R}_{x}={\displaystyle \frac{\mathrm{\partial}R}{\mathrm{\partial}x}},\phantom{\rule{thickmathspace}{0ex}}\text{etc.},\end{array}$

and write (11) in the form

$\begin{array}{}\text{(13)}& {y}^{\prime}(x)={\displaystyle \frac{R{R}_{x}-x}{y-R{R}_{y}}}=-{\displaystyle \frac{x-R{R}_{x}}{y-R{R}_{y}}},\end{array}$

which gives a general expression for ${y}^{\prime}(x)$; in the event that $R(x,y)$ is constant, we obtain

$\begin{array}{}\text{(14)}& {y}^{\prime}(x)=-{\displaystyle \frac{x}{y}},\end{array}$

which the reader may recognize as the slope of the circle

$\begin{array}{}\text{(15)}& {x}^{2}+{y}^{2}={R}^{2}\end{array}$

at any point $(x,y)$ where $y\ne 0$.

###### Not exactly what you’re looking for?

Ask My Question