$\frac{d}{dx}\left(\sin x-\cos y-3=0\right)$
$\frac{x^5+x^4+2x+3}{x-1}$
$-9\:y-18$
$m^3+3m^2-13m-15$
$x^2-4x+1=\:0$
$\lim_{h\to0}\left(\frac{2x^3x-3}{x^3+2x^2-x+1}\right)$
$\frac{dy}{dx}=\frac{\left(y-2x\right)}{\left(4x+y\right)}$
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