$-\:-5+6-3+6+3-+5-2+1$
$\frac{dy}{dx}=4xy=8$
$5x^2+6x-4=0$
$\int\frac{3x+6}{\left(2x^2+8x+3\right)^2}\:dx$
$\left(\frac{3y-1}{4x}\right)\left(\frac{dy}{dx}\right)=\frac{y^2}{x^2+9}$
$y=\frac{1}{x}-ln\:x$
$\frac{dy}{dx}\left(\frac{x^2+3}{x+4}\right)$
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