$\left(x^2-4\right)\frac{dy}{dx}=\frac{1}{y}$
$x^2-6x+13$
$\left(4x^4+3x^3+2x^2-x-1\right)+\left(5x^2-2x+3\right)$
$\lim_{x\to\frac{1}{2}}\left(\frac{6x^2-5x-4}{4x^2+16x-9}\right)$
$x+2-z=-5$
$\frac{dy}{dx}=x^2y^{-5}$
$4tan^2\left(x\right)-4sec^2\left(x\right)$
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