Question

Нужна помощь в решении.

Answer 1

$1)\; \; y=5x^2+4\sqrt[3]{x^5}+3\\\\y'=5\cdot 2x+4\cdot \frac{5}{3}\cdot x^{\frac{2}{3}}+0=10x+\frac{20}{3}\cdot \sqrt[3]{x^2}\\\\2)\; \; y=\frac{2x^2}{1-3x}\\\\y'=\frac{4x(1-3x)-2x^2\cdot (-3)}{(1-3x)^2}=\frac{4x-12x^2+6x^2}{(1-3x)^2}=\frac{4x-6x^2}{(1-3x)^2}=\frac{2x(2-3x)}{(1-3x)^2}$