Question

CALC BC HELPPP!!!??? 100PTS!!!

Answer 1

the assumption being that "x" is a plain variable whilst "y" is a function, that matters because the chain rule would be needed for a function, not so for a plain variable.

$4x^2+4x+xy=5\implies 8x+4+\stackrel{\textit{product rule}}{\left( 1\cdot y+x\cdot \cfrac{dy}{dx} \right)}=0 \\\\\\ x\cfrac{dy}{dx}=-8x-4-y\implies \cfrac{dy}{dx}=\cfrac{-8x-4-y}{x}$

now, we know that y(5) = -23, which is another way of saying that when x = 5, y = -23, but we already knew that, we can get that by simply plugging it into the equation hmmm y'(5), well

$\left. \cfrac{dy}{dx}=\cfrac{-8x-4-y}{x} \right|_{\stackrel{x=5~}{\textit{\tiny y=-23}}}\implies \cfrac{-8(5)-4-(-23)}{5}\implies \cfrac{-21}{5}$