$\lim_{x\to0}\left(\left(2x+4\right)^2-16\right)$

$\frac{\left(4+x^6\right)dy}{dx}=\frac{x^5}{y}$

$c^2-16c-7$

$\frac{\cos\left(2b\right)}{2}=\cos\left(b\right)$

$8\left(\frac{2}{3}\right)^3-17\left(\frac{2}{3}\right)^2+12\left(\frac{2}{4}\right)+14$

$\frac{dy}{dx}3x^2\cdot y=x^2$

$2\left(3^2\right)-4$