Найдите производную функции.

Решите задачу:

$A.1.\quad (3x^5-2x^2)'=15x^4-4x\\\\(2x^5-5)'=10x^4\\\\( \frac{2}{x^3} -x^2)'=-\frac{2\cdot 3x^2}{x^6}-2x=- \frac{6}{x^4} -2x\\\\(32\sqrt[3]{x^2})'=32\cdot \frac{2}{3}\cdot x^{-\frac{1}{3}}= \frac{64}{3\sqrt[3]{x}} \\\\A.2.\quad a)\; \; \; y=(x^3+3)(x-x^3)\\\\y'=3x^2(x-x^3)+(x^3+3)(1-3x^2)\\\\b)\; \; \; y= \frac{x^4-x^2}{x-1} \\\\y'= \frac{(4x^3-2x)(x-1)-(x^4-x^2)\cdot 1}{(x-1)^2} = \frac{4x^4-4x^3-2x^2+2x-x^4+x^2}{(x-1)^2} =\\\\= \frac{3x^4-4x^3-x^2+2x}{(x-1)^2} \\\\B.2.\quad y= \frac{x}{\sqrt{x-4}}$

$y'= \frac{1\cdot \sqrt{x-4}-x\cdot \frac{1}{2\sqrt{x-4}}}{(\sqrt{x-4})^2} = \frac{2(x-4)-x}{2\sqrt{(x-4)^3}}= \frac{x-8}{2\sqrt{(x-4)^3}}$