$x^2-4=3x\:$
$\frac{dy}{dx}=9x^2\left(4+y^2\right)^{\frac{3}{2}}$
$\frac{\frac{4}{u^2}}{\frac{4}{u}}$
$\left(-10\right)^2-5\left(-10\right)+10$
$x^2+8x+9$
$xy'+3y=x^6$
$-x^3+7x^2+12x-54$
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