$\frac{\left(cos^2\left(x\right)-sin^2\left(x\right)\right)}{1-tan^2x}$
$\frac{d}{dx}\left(6y+3x=9-4x^2y^3\right)$
$\frac{\frac{1}{\sqrt{x}}}{x+3}$
$\left(sin\:5x\right)^2.cos5x$
$2\left(xy^2-x^3\right)+\:y^2-x^2$
$-962+-114$
$4x^2-3x\frac{5x}{4x^2-3x}-6=0$
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