$0,00478\:x\:1,5$

$\int\left(\tan\left(5u\right)\right)\sec\left(5u\right)du$

$\frac{d}{dx}\left(\left(3x+1\right)^5\right)\left(5x+4\right)^{-2}$

$6x,\:18xy,\:24x^2\:y\:26x^3$

$3x^3+x+3+18-6x^3-8x+3x^3+7x-2x^2$

$-4x+2x=-18$

$\frac{dy}{dx}=\frac{\left(x-2\right)\left(y+1\right)}{xy-2y+2x-4}$