Answer
$\dfrac{dy}{dx}=-\dfrac{2xy}{x^2+4}.$
At $(2, 1)\rightarrow\dfrac{dy}{dx}=-\dfrac{2(2)(1)}{2^2+4}=-\dfrac{1}{2}$
Work Step by Step
$\dfrac{d}{dx}((x^2+4)y)=\dfrac{d}{dx}(8)\rightarrow$
$y(\dfrac{d}{dx}(x^2+4))+(x^2+4)(\dfrac{d}{dx}(y))=0\rightarrow$
$2xy+\dfrac{dy}{dx}(x^2+4)=0\rightarrow$
$\dfrac{dy}{dx}=-\dfrac{2xy}{x^2+4}.$
At $(2, 1)\rightarrow\dfrac{dy}{dx}=-\dfrac{2(2)(1)}{2^2+4}=-\dfrac{1}{2}$
