$\frac{dy}{dx}=\frac{3-\sin\left(x\right)}{y}$

$\frac{6}{x+1}=\frac{5}{x-1}$

$8x^2y^3\cdot x^4$

$\int_0^1\left(\frac{6}{x^3}\right)dx$

$\lim_{x\to8}\left(\frac{\sqrt{7+\sqrt[3]{x-3}}}{x-8}\right)$

$\frac{d}{dx}\left(x^x\right)\left(x+7\right)^4\left(8x-3\right)^3$