$4\cdot12\cdot\left(-2\right)$
$y=\tan\left(3x^2+2x\right)$
$\frac{d}{dx}\sqrt[5]{x^2+3x-\frac{3}{x}}$
$\lim_{x\to-\infty}\left(2x^3+5x^2+3\right)$
$4+-5+-7+8+-10+-9+12+11+-3$
$4x^2+4$
$3\:\left(4x\:+\:\frac{2}{3}\right)\:-\:7x\:+\:\frac{3}{8}$
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