$\frac{d}{dx}y=\tan\frac{1}{x}$

$\frac{d}{dx}\left(y^2-3y=x^3+8\right)$

$\:\left(x+2\right)^2\:+\left(x+4\right)^2\:-\:2\left(x+3\right)^2$

$x^{18}-\frac{1}{8}n^{12}c^{18}$

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

$\left(4a-3\right)^3$

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