$-yx^2z^3\left(4y^0z\right)^2$

$\frac{\left(y+z\right)^3}{x^3+y^3}$

$-y\:dy=\frac{\left(-xy\:dx\right)}{x^2}$

$\left(8y-1\right)^2$

$\tan^{-1}\left(-44\right)$

$\frac{dy}{dx}\left(y=\left(4x+1\right)^5\left(x^4-6\right)^6\right)$

$5x+3\left(x-6\right)$