$\frac{2x+5}{-4x+8}$

$\frac{dy}{dx}=-2xy+1$

$\lim_{x\to0}\left(\frac{e^x}{4\sin\left(x\right)}\right)$

$-\left(\frac{6}{-3}\right)$

$\frac{1-\cos\left(x\right)}{1+\cos\left(x\right)}=\tan^2\left(\frac{x}{2}\right)$

$3\left(-2\right)^2-3\left(-2\right)+1$

$\frac{d^2}{dx^2}e^{2x}$