$\frac{d}{dx}\left(\frac{x^3+2x^2+3x}{x^2-3x+1}\right)$

$\lim_{x\to0}\left(3x^2-4x+1\right)$

$\frac{5}{x}=\frac{6}{x^3-2x^2}$

$144x^2-1$

$\frac{dy}{dx}=xy\left(4-y\right)$

$\frac{x^2-4}{x+0}$

$\lim_{x\to\frac{\pi}{2}}\left(\frac{\ln\left(\sin\left(x\right)\right)}{\ln\left(\cos\left(4x\right)\right)}\right)$