$\frac{d}{dx}\:\frac{y^2}{x^3}=1+y^2$
$\frac{x^2+7}{x-5}$
$\frac{\sqrt[4]{8a^7b^{32}}}{5xy}$
$\left(-4\right)^2-5\cdot\left(-4\right)-4$
$\frac{sinxcosx}{sinx+cosx}=\:\frac{sinx}{1+cosx}$
$1-8v+1$
$a^4\:+\:4\:a^2\:b\:+\:4\:b^2$
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