 2022-07-18

I want to solve $dy/dx$ for the following:
${x}^{2}+{y}^{2}={R}^{2}$ where $R$ is a constant.
I know to use implicit differentiation, though I have a question. When I derive ${R}^{2}$, do I obtain $2R$ or 0?
Additionally, deriving ${y}^{2}$ with respect to x yields $2y\left(dy/dx\right)$? This is different from a partial derivative?
Thanks! bulgarum87

Suppose $R$ is a function of $x$ and $y$; then
$\begin{array}{}\text{(1)}& {x}^{2}+{y}^{2}={R}^{2}\left(x,y\right);\end{array}$
if we define
$\begin{array}{}\text{(2)}& F\left(x,y\right)={x}^{2}+{y}^{2}-{R}^{2}\left(x,y\right),\end{array}$
we may also write (1) as
$\begin{array}{}\text{(3)}& F\left(x,y\right)={x}^{2}+{y}^{2}-{R}^{2}\left(x,y\right)=0;\end{array}$
by the implicit funtion theorem, this equation in fact may be seen as defining $y\left(x\right)$, a function of $x$, provided that
$\begin{array}{}\text{(4)}& \frac{\mathrm{\partial }F\left(x,y\right)}{\mathrm{\partial }y}\ne 0;\end{array}$
we have
$\begin{array}{}\text{(5)}& \frac{\mathrm{\partial }F\left(x,y\right)}{\mathrm{\partial }y}=2y-2R\left(x,y\right)\frac{\mathrm{\partial }R\left(x,y\right)}{\mathrm{\partial }y}\ne 0\end{array}$
provided
$\begin{array}{}\text{(6)}& y\ne R\left(x,y\right)\frac{\mathrm{\partial }R\left(x,y\right)}{\mathrm{\partial }y};\end{array}$
under such circumstances, we may affirm $y\left(x\right)$ is uniquely determined as a differentiable function of $x$ in some neighborhood of any point $\left(x,y\right)$; then we have
$\begin{array}{}\text{(7)}& F\left(x,y\right)={x}^{2}+{y}^{2}\left(x\right)-{R}^{2}\left(x,y\left(x\right)\right)=0;\end{array}$
we may take the total derivative with respect to x to obtain
$\begin{array}{}\text{(8)}& \frac{dF\left(x,y\right)}{dx}=2x+2y\frac{dy\left(x\right)}{dx}-2R\left(x,y\left(x\right)\right)\left(\frac{\mathrm{\partial }R\left(x,y\left(x\right)\right)}{\mathrm{\partial }x}+\frac{\mathrm{\partial }R\left(x,y\left(x\right)\right)}{\mathrm{\partial }y}\frac{dy\left(x\right)}{dx}\right)=0;\end{array}$
a little algebra allows us to isolate the terms containing $dy\left(x\right)/dx$:
$\begin{array}{}\text{(9)}& x+y\frac{dy\left(x\right)}{dx}-R\left(x,y\left(x\right)\right)\left(\frac{\mathrm{\partial }R\left(x,y\left(x\right)\right)}{\mathrm{\partial }x}+\frac{\mathrm{\partial }R\left(x,y\left(x\right)\right)}{\mathrm{\partial }y}\frac{dy\left(x\right)}{dx}\right)=0;\end{array}$
$\begin{array}{}\text{(10)}& y\frac{dy\left(x\right)}{dx}-R\left(x,y\left(x\right)\right)\frac{\mathrm{\partial }R\left(x,y\left(x\right)\right)}{\mathrm{\partial }y}\frac{dy\left(x\right)}{dx}=R\left(x,y\left(x\right)\right)\frac{\mathrm{\partial }R\left(x,y\left(x\right)\right)}{\mathrm{\partial }x}-x;\end{array}$
$\begin{array}{}\text{(11)}& \left(y-R\left(x,y\left(x\right)\right)\frac{\mathrm{\partial }R\left(x,y\left(x\right)}{\mathrm{\partial }y}\right)\frac{dy\left(x\right)}{dx}=R\left(x,y\left(x\right)\right)\frac{\mathrm{\partial }R\left(x,y\left(x\right)\right)}{\mathrm{\partial }x}-x;\end{array}$
for the sake of compactess and brevity, we introduce the subscript notation
$\begin{array}{}\text{(12)}& {R}_{x}=\frac{\mathrm{\partial }R}{\mathrm{\partial }x},\phantom{\rule{thickmathspace}{0ex}}\text{etc.},\end{array}$
and write (11) in the form
$\begin{array}{}\text{(13)}& {y}^{\prime }\left(x\right)=\frac{R{R}_{x}-x}{y-R{R}_{y}}=-\frac{x-R{R}_{x}}{y-R{R}_{y}},\end{array}$
which gives a general expression for ${y}^{\prime }\left(x\right)$; in the event that $R\left(x,y\right)$ is constant, we obtain
$\begin{array}{}\text{(14)}& {y}^{\prime }\left(x\right)=-\frac{x}{y},\end{array}$
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 $\left(x,y\right)$ where $y\ne 0$. Freddy Friedman

By the chaine rule you will get
$2x+2y\cdot {y}^{\prime }=0$

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