$\frac{1}{16x^3+4x^2}$
$\frac{1-\sin^2x}{\cos x\sin x}=\cot x$
$y'=2xy+3y-4x-6$
$\lim_{x\to\infty}\left(\frac{1}{3}x\right)$
$2.9=y-6.3$
$-32\cdot10$
$2xy^3-x^2y=y^{2-3x\ln\tan\left(\sin\left(2xy\right)\right)}$
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