$\left(x^2-8x\right)\left(+16\right)$

$3\:-\:12\:\cdot\:-4\:+\:7$

$x^2\frac{dy}{dx}=4x^3+3x+2$

$\lim_{x\to1}\left(\left\{x+1\right\}\left\{x-1\right\}\right)$

$xdy-ydx=ydy$

$\tan\left(x\right)\left(\cos\left(x\right)\right)\frac{1}{\sin\left(x\right)}=1$

$tan\left(x\right)^2=\frac{sin\left(x\right)^2}{cos\left(x\right)^2}$