$\frac{dy}{dx}-\frac{y}{x}=x\cos\left(x\right)$
$\frac{d}{dx}\:in\sqrt[5]{4-3x^5}$
$\frac{\tan^2x}{\sin^2x}=1\:+\:\tan^2x$
$7^2-2\cdot7+2^3$
$\frac{18.05}{19}$
$\left(4x+1\right)\cdot\left(2x^3+3x^2+5x\right)$
$-9-\left(+3+\right)\left(-5+10\right)$
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