$\frac{d}{dx}\sqrt[5]{3x^{3}+4x}$
$\sqrt[4]{-256}$
$cos2x-1=0$
$\frac{\left(\tan^2\left(17\right)+1\right)}{\sec^2\left(17\right)}$
$\left(9x+30\right)\left(9x-6\right)$
$4x+3+2x+1$
$\frac{d}{dx}4x^3-2x^2+8\sqrt{x}$
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