$\left(4x^2-5y\right)^2$
$\frac{dy}{dx}=-\frac{1}{4}4\left(y-27\right)$
$\int x\left(10x^3-2\sec^2x\right)dx$
$\frac{3x^4}{\left(1+2x\right)^2}$
$\frac{x^3-2x^2+31x+20}{x+5}$
$4a+6b\:5x^4y-20xy^3$
$\frac{dy}{dx}=\frac{2y}{x-100}$
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