Answer
a. equation: $y=±\sqrt{\frac{25}{3}-\frac{25}{36}x^2}$
b. derivatives: $y'=-\frac{25x}{36y}$
Work Step by Step
1. Rewrite the equation in terms of x:
$36y^2=300-25x^2$
$y^2=\frac{300}{36}-\frac{25}{36}x^2$ = $y^2= \frac{25}{3}-\frac{25}{36}x^2$
$\sqrt {y^2}= \sqrt {\frac{25}{3}-\frac{25}{36}x^2}$
a. $y= ± \sqrt{\frac{25}{3}-\frac{26}{36}x^2}$
2. Find the derivative of $y= ± \sqrt{\frac{25}{3}-\frac{26}{36}x^2}$
$y'= \frac {d}{dx} (- \sqrt{\frac{25}{3}-\frac{26}{36}x^2})$
$y'= \frac {d}{dx} (-\sqrt {\frac{300-25x^2}{36}})$
$y'=\frac{d}{dx}(-\frac{\sqrt{300-25x^2}}{6})$
$y'= -\frac{1}{6}\times\frac{d}{dx}(\sqrt{300-25x^2})$
$y' = -\frac{1}{6}\times(\frac{1}{2\sqrt{300-25x^2}})(-25)(2x) $
$y'= \frac{5x}{6\sqrt{12-x^2}}$
$y'= \frac {d}{dx} ( \sqrt{\frac{25}{3}-\frac{26}{36}x^2})$
$y'= \frac {d}{dx} (\sqrt {\frac{300-25x^2}{36}})$
$y'=\frac{d}{dx}(\frac{\sqrt{300-25x^2}}{6})$
$y'= \frac{1}{6}\times\frac{d}{dx}(\sqrt{300-25x^2})$
$y' = \frac{1}{6}\times(\frac{1}{2\sqrt{300-25x^2}})(-25)(2x) $
$y'= -\frac{5x}{6\sqrt{12-x^2}}$
3. Find the derivative of: $25x^2+36y^2=300$
$\frac{d}{dx} (25x^2) + \frac{d}{dx}(36y^2)= \frac{d}{dx}(300)$
$50x +72y (\frac{dy}{dx}) = 0$
b. $\frac{dy}{dx}= \frac{-50x}{72y} = -\frac{25x}{36y}$
4. Graph:
