$\left(x-2\right)y'$

$0=-\frac{6x}{\left(x^2+3\right)^2}$

$\left(y+4\right)^2=-12\left(x-3\right)$

$\frac{d}{dx}\left(\left(3x+2\right)^4\left(8x-7\right)^2\right)$

$\lim_{x\to\infty}\left(\sqrt[3]{\frac{3+64x^3}{x^3+8}}\right)$

$4\left(4g+4\right)-5g$

$cosx-1=-cosx$