$\frac{1}{secx}=secx\left(1-sin^2x\right)$
$\:7\:-\:12x\:-\:2x^2$
$2a^2+4y$
$24\cdot12$
$\left(2xy^4+x\right)dx+\left(4x^2y^3\right)dy=0$
$-4x^3+x^3+x^3-6y^3+y^3+4y^3-3$
$\frac{\left(x^3-15x\right)}{x^2-16}$
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