$\frac{d^2}{dx^2}\left(16x^4-4x^2\right)$
$\frac{d}{dx}\sqrt[3]{120}\:$
$y'=t^2+1,\:y\left(1\right)=4$
$\left(16xb^2-5xy^3\right)^2$
$8\:+\:x\:=\:14$
$8x^4-2x^2$
$u^6+8t^3$
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