$y'\:=\:2xy2$
$\cos^2\left(x\right)\tan\left(-x\right)$
$\lim_{x\to2}\left(\frac{\left(2^x-4\right)}{2-x}\right)$
$\int\left(4x^3\cdot\cos\left(2x^4\right)\right)dx$
$-8d+10d$
$16x\:+\:11=18x-9$
$\frac{d}{dx}2x=4y^2$
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