$\frac{4}{\left(x-3\right)\left(x+1\ \right)}$
$\frac{dy}{dx}=1-cosx$
$\left(5x^3y^2+4x\cdot y^4\right)^3$
$h\left(t\right)=\left(2t^3+t^2\right)^7$
$\frac{dy}{dx}=\frac{4x^{\frac{1}{2}}}{y^{-\frac{1}{2}}}$
$3\left(x-2\right)>2x+3$
$2a-8ab$
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