$\frac{d}{dx}\left(3x^2+4y^3=ln\left(xy\right)\right)$
$-232\:-\:345$
$4-12\cdot\left(-2\right)$
$\left(\:6x^2-3xt\right)^3$
$-8t^2+7t-9$
$-5^2+8\left|-1\right|+\left(-3\right)$
$4xy+20xy+4y^2$
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