$\frac{1}{x^4+5x^2+4}$
$\frac{d}{dx}10\ln x$
$6x^4+6x$
$\frac{dy}{dx}=\frac{x+3x^2}{y^2}$
$\sqrt[2]{\frac{27}{3}}$
$\left(3z-4a\right)^3$
$x-5>3x$
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