$5x^3+4x^2+9x$

$\left(\frac{1}{5}z-10z^3\right)^3$

$\frac{d}{dx}\left(\frac{e^{-y}}{x^2}\right)$

$\left(x^7y\frac{1}{2}-4a^4b\right)^4$

$\frac{1-1}{2}$

$\left(x-12\right)\left(x-7\right)$

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