$h ( t ) = \sqrt { t } ( 1 - t ^ { 2 } )$

$\frac{13x^3+8x^2}{2x^2}$

$\left(\frac{\frac{1}{2}xy^2z}{\frac{3}{4}a^4m^2z^{10}}\right)^3$

$\tan\left(\arccos\left(-0.08\right)\right)$

$\left(6a^3b^2+4b^5c^4\right)\left(6a^3b^2-4b^5c^4\right)$

$x^2+16x+40$

$y=\frac{-2x^2-5}{cos\:2x^3}$