$\tan\alpha\sec\alpha=\frac{1}{1-\sen\alpha}-\frac{1}{1+\sen\alpha}$
$\left(4x^3+y^2\right)\left(4x^2\right)$
$7x+5=3x-15$
$y'+19\frac{y}{x}=0$
$74.\left(-85\right).3$
$\lim_{x\to0}\left(\frac{x^2-x^4}{x^6-x^2}\right)$
$-5=\frac{7}{3}x+16$
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