$-5\left(x-10\right)^2$
$\frac{1}{16}\:x\:^4\:.\:3$
$\tan\left(x\right)\cdot\cot\left(x\right)+\csc\left(x\right)$
$\int xe^{x^2}dt$
$\left(x+2y^3\right)^5$
$4.62\cdot2$
$y=\frac{\sin\left(x\right)\cos\left(x\right)\tan^5\left(x\right)}{\sqrt{x}}$
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