$y^3dx+3xy^2dy$
$\left(2x^3y+x^5y^2\right)^2$
$\left(\sec^2\left(x\right)-1\right)\cdot\cos^2\left(x\right)=\sin^2\left(x\right)$
$\frac{dy}{dx}=\sqrt{\left(2x+y-1\right)}$
$9,6-8,071$
$-19^0$
$\tan^2t-\sec^2t$
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