$4x\left(3\right)-7x+3$
$x'=\frac{\left(1-x\right)}{\left(1+t\right)}$
$2x+3\:\left(6x-2\right)\left(4x-1\right)$
$-2.\left(-4x+1\right)$
$\int\frac{x^2+8}{x^3+4x}dx$
$2cos\left(-6x\right)$
$\frac{d}{dx}\ln y=x^{e^{2x}}$
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