$10x\:-\:8y\:-\:11y\:-7x$
$y'=\frac{\left(x-5\right)}{y^2}$
$\frac{\left(6\cdot x\right)}{\left(5\cdot x^2-4x-1\right)}$
$tan^4\left(1+xtan^2x\right)$
$3u\:x\:2w^9\:x\:8u^{8\:}x\:w^4$
$3y^2dx=6x^4dy$
$\int_0^{\sqrt{2}}\frac{\left(2x-1\right)}{\left(y\cdot\sqrt{y^2-9}\right)}dy$
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