$-9+7+3-5-9+14+16-17-2-5$
$y'=y^2$
$\tan\left(x\right)\left(1-\sin^2\left(x\right)\right)=\frac{1}{2}\sin\left(2x\right)$
$\frac{4x^3+5x+2}{x-1}$
$14x^2y-21xy2$
$f\left(x\right)=x^2sin\left(x\right)+x\cos\left(x\right)$
$-2-18-\left(-22\right)+\left(-4\right)$
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