$\left(2\right)\frac{3}{4}n\:\left(1\right)$
$\frac{3x^3-5x^2+10x-3}{3x+1}$
$\left(4+y\right)^3$
$\left(4c-7d\right)^2$
$\left(y^2+xy^2\right)\cdot y'$
$ab^2c^2\cdot a^4bc^3$
$\frac{d}{dx}sen\:\frac{5x+4}{5x-4}$
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