$\left(2a-3ab\:^2\right)^3$
$\frac{dx}{dw}=3xw^2$
$\int\left(\frac{2-x}{x^2+4x}\right)dx$
$4\:sec^2-7\:tan^2x=3$
$-4+9\cdot\frac{12}{36}-5+3$
$\frac{dy}{dx}=\frac{3x}{5y+1}$
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