$\frac{dy}{dx}=y\frac{1}{3}$
$\frac{d}{dx}4x^2+y^4=7x^5y^3$
$\frac{\left(-1492896\right)}{\left(-96\right)}$
$\left(4x^8-9y^7\right)^2$
$16b^2-48b+36$
$\lim_{x\to\infty}\left(\frac{x+1}{x-2}\right)^{\frac{x}{2}}$
$\left(8-5y\right)^2$
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