$\frac{sin^2x}{tan^2x}=\frac{1+cos\:2x}{2}$
$\frac{dy}{dx}=\sqrt{\frac{sen^2x}{\left(1+cos\:x\right)}}$
$f\left(x\right)=\frac{x^4-5x^3+\sqrt{x}}{x^2}$
$\frac{20x\:-5}{\left(4y\:-\:1\right)^2}dy\:-\:3dx\:=\:0$
$\left|68+9\right|+\left(-62\right)$
$2x^2-6x+2=0$
$\frac{2}{x}+\frac{x}{x-1}$
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