Answer
$ \frac{dy}{dx} =-\frac{x}{y}$
OR
$\frac{dy}{dx}= \mp \frac{x}{ \sqrt {100-x^2 }}$
Work Step by Step
$x^2+y^2=100$
$ \frac{d}{dx}(x^2+y^2)=\frac{d}{dx}(100) $
$ \frac{d}{dx}(x^2) + \frac{d}{dx}(y^2) =0$
$ 2x + \frac{d}{dy}(y^2)\frac{dy}{dx} =0$
$ 2x + 2y \frac{dy}{dx} =0$
$ 2y \frac{dy}{dx} =-2x$
$ y \frac{dy}{dx} =-x$
$ \frac{dy}{dx} =-\frac{x}{y}$
Since
$x^2+y^2=100$
$y^2=100-x^2$
$\sqrt{y^2}= \sqrt {100-x^2}$
$y=\pm \sqrt {100-x^2}$
Putting in
$ \frac{dy}{dx} =-\frac{x}{y}$
$\frac{dy}{dx}= \mp \frac{x}{ \sqrt {100-x^2 }}$
