$\sqrt[5]{x^3}y^5z^2$
$9^xdx=e^ydy$
$x^2-1=1$
$\left(3.59\cdot10^{-2}\right)-\left(1.23\cdot10^{-2}\right)$
$\lim_{x\to\infty}\left(\frac{4x^2}{2x^2+1}\right)$
$\left(\frac{1}{2+x}\right)\frac{dy}{dx}=\left(3-4y\right)$
$\frac{d^2}{dx^2}\left(2y-3x^2\right)$
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