$\sqrt{1-3x^2}y'=x$
$\lim_{x\to2}\sqrt[3]{x+6}$
$\left(3x^2-10y^3\right)\left(3x^2+10y^3\right)$
$-40\:x\:-3$
$\tan\left(\arccos\frac{-4}{50}\right)$
$\cos x-\sin^2x+1=0$
$\left(2x^2y^3+3x^4\right)^2$
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