$\lim_{x\to-1}\left(\sqrt{16-4x}\right)$
$\frac{dy}{dx}=-4x^2y$
$x^2+12x+8$
$0\:x\:-2\:+-2$
$3x^3+x^2-12x-4$
$8y+\left(-4x\right)-5x+1-6y+2$
$\frac{d^2y}{dx^2}=x^2$
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