1

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!

  • If you are looking for $\frac{dy}{dx},$ then you need to derive the given equation in terms of $x.$ $R$ is a constant. Notice that $\frac{d}{dx} c = 0.$ – K. Jiang Jan 09 '19 at 19:47
  • You are differentiating with respect to $x$, and since $R$ is independent of $x$, its derivative is zero. For the derivative of $y^2=f(x)^2$, use the chain rule; its derivative is $\dfrac{d(y^2)}{df(x)}\cdot\dfrac{df(x)}x$ – ə̷̶̸͇̘̜́̍͗̂̄︣͟ Jan 09 '19 at 19:49

2 Answers2

1

By the chaine rule you will get $$2x+2y\cdot y'=0$$

Dr. Sonnhard Graubner
  • 95,283
0

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 $y \ne 0$.

Robert Lewis
  • 71,180