$\frac{d}{dx}x^2y^3=x^3y^2+4$
$-85+1-6$
$6x+3y=6$
$\frac{3}{7}x^2y^3\left(\frac{7}{3}xy-\frac{1}{3}x^2y+\frac{5}{7}\right)$
$\frac{5x^3y^4}{-25xy^2}$
$10u\:=\:1.500$
$-25x^2y^8-121$
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