$\frac{d}{dx}\frac{x\left(2x+1\right)^{\frac{5}{2}}}{\left(3x-4\right)^{\frac{2}{3}}}$
$\lim_{n\to\infty}ln\left(\frac{n}{n-2}\right)$
$\sqrt[3]{\left(x^3-y^3\right)}$
$6x^2+5\:x+1$
$\frac{dr}{d\theta\:\:}+r\cdot\:\:\:tan\theta\:\:=sec\theta\:$
$-4x^2+32=0$
$x-x^{-1}=4$
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