$\frac{8x^8y^8}{2x^3y^2}$
$-58\cdot\left(-11\right)$
$\:\left(\sqrt{5}-y\right)^2$
$\left(-12x^2y^3\right)\left(-8x^5y^6\right)\left(16xy\right)$
$4\sqrt{3\left(1+2\right)}$
$8+4-13$
$\frac{dy}{dx}=\frac{1}{2}xy$
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