$36-18x^3+9x^6$
$\sin3x=-05$
$\int t\sin^{-1}\left(x\right)dx$
$3x^4y^3\left(-5x^2y^2\right)$
$\frac{dy}{dx}=\frac{xsecy}{ye^{t^2}}$
$\frac{3}{4}\:=\:\frac{1}{4}\:x$
$tan^2\left(x\right)-\tan\left(x\right)$
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