$\frac{d}{dx}\:\frac{10}{\sqrt[5]{x}}$
$1.8^4\cdot8^8$
$0x+3=1x+2$
$10-10.\left[-6+5.\left(-4+7-3\right)\right]$
$\frac{d}{dx}\left(2y^2+xy+x^2=16\right)$
$3\sin\left(4x\right)=-6\sin\left(2x\right)$
$\frac{2-\sec^2\left(x\right)}{\tan^2\left(x\right)}+1$
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