$\frac{dy}{dx}=\frac{\left(x-1\right)}{\left(\ln\left(y\right)\right)\left(x^2+2x+3\right)^2}$

$16.21-87.23$

$y\left(4+3x\right)^4$

$1,39\cdot2$

$\frac{3x^3}{x^2+x-2}$

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

$\frac{20x^3-18x^2+4}{4x-2}$