Answer
equation: $y=±\sqrt{64-x^{2}}$
derivatives: $y'= -\frac{x}{y}$
Work Step by Step
1. Rewrite the equation in terms of x: $x^{2} + y^{2}=64$
$y^{2} = 64-x^{2}$
$y=±\sqrt{64-x^{2}}$
2. Find the derivative of y=-$\sqrt{64-x^{2}}$ and y=$\sqrt{64-x^{2}}$
$y'= \frac{d}{dx}(-\sqrt{64-x^{2}})$
$y'= \frac{x}{\sqrt{64-x^{2}}}$
$y'= \frac{d}{dx}(-\sqrt{64-x^{2}})$
$y'= -\frac{x}{\sqrt{64-x^{2}}}$
3. Find the derivative of: $x^{2}+y^{2}=64$
$\frac{d}{dx}(x^2)+\frac{d}{dx}(y^2)=\frac{d}{dx}(64)$
$2x+2y(\frac{dy}{dx})=0$
$2y(\frac{dy}{dx})=-2x$
$\frac{dy}{dx}=-\frac{x}{y}$
4. The derivatives are equivalent
5. Graph:
