Question

Срочно нужна помощь в этом

Answer 1

$1)\; \; (2x^3-4x+8)'=6x^2-4\\\\(\frac{3}{x}-2\sqrt{x}+10)'=-\frac{3}{x^2}-2\cdot \frac{1}{2\sqrt{x}}=-\frac{3}{x^2}-\frac{1}{\sqrt{x}}\\\\\Big (3x\cdot (3x^2-4x)\Big )'=(9x^3-12x^2)'=27x^2-24x\\\\\Big ((2x-7)(4-3x)\Big )'=(29x-6x^2-28)'=29-12x\\\\(sin2x-2cos3x)'=2\, cos2x+6\, sin3x\\\\(cos(5x+4))'=-sin(5x+4)\cdot (5x+4)'=-5\cdot sin(5x+4)$

$2)\; \; f(x)=x^3-4x^2\; \; ,\; \; \; f'(x)=3x^2-8x\\\\f'(0)=0\; \; ,\; \; f'(1)=3-8=-5\\\\b)\; \; f(x)=\frac{1+4x}{1+2x}\\\\f'(x)=\frac{4(1+2x)-2(1+4x)}{(1+2x)^2}=\frac{2}{(1+2x)^2}\; \; ,\\\\f'(-1)=\frac{2}{1}=2\; \; ,\; \; f'(1)=\frac{2}{9}\\\\c)\; \; f(x)=sin2x\; \; ,\; \; \; f'(x)=-2\, cos2x\\\\f'(0)=-2\cdot cos0=-2\cdot 1=-2\\\\f'(\pi )=-2\cdot cos2\pi =-2\cdot 1-2$