$\frac{\sin^2\left(x\right)}{\cos\left(x\right)}$

$\left(xy^2-4x^2y^3+5x^3y+4x^2\right).\left(-2x^3y\right)$

$sin\left(2x\right)dx\:+\:4y\:cos^2\:\left(2x\right)dy\:=\:0$

$\frac{d}{dx}\left(4x^3y^3=10x-6y^2\right)$

$\frac{d}{dx}\:\left(2x-2\right)^4\left(x^2+x+1\right)^5$

$\frac{dy}{dx}=40\sin\left(60x\right)-20y$