$x-y+xy'=0$
$-7x^{-\frac{2}{5}}y^6z^{-4}$
$\frac{dy}{dx}=y=-7$
$\sqrt[3]{\frac{128a^5}{8b^3}}$
$\frac{2\cos\left(2x\right)}{\sin\left(2x\right)}=\frac{1}{\tan\left(x\right)}-\tan\left(x\right)$
$\frac{8x^3+y^3}{2x-y}$
$-5+2\cdot2$
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