$\left(y^2+1\right)=\left(1+xy\right)\left(\frac{dy}{dx}\right)$
$\frac{d}{dx}y\:=\:t^{\sqrt{3t}}$
$\int\frac{10-x}{x^3-10x^2-16x}dx$
$-4x^2-16y$
$-8x-6x+3x-5x+4-x$
$y'-4xy^2=0$
$\left(sec^2y\right)y'\:-3tany\:+1\:=\:0$
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