Implicit differentiation questions 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 $2y (dy/dx)$? This is different from a partial derivative? 
Thanks!
 A: By the chaine rule you will get $$2x+2y\cdot y'=0$$
A: Suppose $R$ is a function of $x$ and $y$; then
$x^2 + y^2 = R^2(x, y); \tag 1$
if we define
$F(x, y) = x^2 + y^2 - R^2(x, y), \tag 2$
we may also write (1) as
$F(x, y) = x^2 + y^2 - R^2(x, y) = 0; \tag 3$
by the implicit funtion theorem, this equation in fact may be seen as defining $y(x)$, a function of $x$, provided that 
$\dfrac{\partial F(x, y)}{\partial y} \ne 0; \tag 4$
we have
$\dfrac{\partial F(x, y)}{\partial y} = 2y - 2R(x, y) \dfrac{\partial R(x, y)}{\partial y} \ne 0 \tag 5$
provided
$y \ne R(x, y) \dfrac{\partial R(x, y)}{\partial y};\tag 6$
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
$F(x, y) = x^2 + y^2(x) - R^2(x, y(x)) = 0; \tag 7$
we may take the total derivative with respect to $x$ to obtain
$\dfrac{dF(x, y)}{dx} = 2x + 2y\dfrac{dy(x)}{dx} - 2R(x, y(x)) \left ( \dfrac{\partial R(x, y(x))}{\partial x} + \dfrac{\partial R(x, y(x))}{\partial y} \dfrac{dy(x)}{dx} \right ) = 0; \tag 8$
a little algebra allows us to isolate the terms containing $dy(x)/dx$:
$x + y\dfrac{dy(x)}{dx} - R(x, y(x)) \left ( \dfrac{\partial R(x, y(x))}{\partial x} + \dfrac{\partial R(x, y(x))}{\partial y} \dfrac{dy(x)}{dx} \right ) = 0; \tag 9$
$y\dfrac{dy(x)}{dx} - R(x, y(x)) \dfrac{\partial R(x, y(x))}{\partial y} \dfrac{dy(x)}{dx} = R(x, y(x))\dfrac{\partial R(x, y(x))}{\partial x} - x; \tag{10}$
$\left ( y -  R(x, y(x)) \dfrac{\partial R(x, y(x)}{\partial y} \right )\dfrac{dy(x)}{dx} = R(x, y(x))\dfrac{\partial R(x, y(x))}{\partial x} - x; \tag{11}$
for the sake of compactess and brevity, we introduce the subscript notation
$R_x = \dfrac{\partial R}{\partial x}, \; \text{etc.}, \tag{12}$
and write (11) in the form
$y'(x) = \dfrac{RR_x - x}{y - RR_y} = -\dfrac{x - RR_x}{y - RR_y}, \tag{13}$
which gives a general expression for $y'(x)$; in the event that $R(x, y)$ is constant, we obtain
$y'(x) = -\dfrac{x}{y}, \tag{14}$
which the reader may recognize as the slope of the circle
$x^2 + y^2 = R^2 \tag{15}$
at any point $(x, y)$ where $y \ne 0$.
