Question

Решить только 3-ий вариант ">

Answer 1

$1)\ \ f(x)=\dfrac{x^3}{3}-\dfrac{x^2}{2}-2x+3\\\\f'(x)=x^2-x-2\\\\x_0=3:\ f'(3)=9-3-2=4\\\\\\2)\ \ a)\ \ y=4x^2+\dfrac{4}{x^2}-\dfrac{1}{x}\\\\y'=4\cdot 2x+\dfrac{-4\cdot 2x}{x^4}-\dfrac{-1}{x^2}=8x-\dfrac{8}{x^3}+\dfrac{1}{x^2}\\\\b)\ \ y=(3x+1)\cdot sinx\\\\y'=3\cdot sinx+(3x+1)\cdot cosx\\\\c)\ \ y=\dfrac{x^2+1}{x^2-1}\\\\y'=\dfrac{2x(x^2-1)-(x^2+1)\cdot 2x}{(x^2-1)^2}=\dfrac{-4x}{(x^2-1)^2}$