$y'=\frac{y+1}{x-1}$
$\left(1-m^3\right)^2$
$3x^3-x^2\cdot\left(-2x\right)$
$15-1600$
$2\cdot11$
$\left(16p+20q\right).\left(16p-20q\right)$
$\frac{\tan^2\left(x\right)}{\sec\left(x\right)-1}$
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