Question

ПОЖАЛУЙСТА ПОМОГИТЕ! 40 БАЛЛОВ С МЕНЯ!

Answer 1

$1.\;f(x)=7,5x^2+\sqrt x+0,5x+2\pi\\f'(x)=15x+\frac1{2\sqrt x}+0,5\\2.\;g(x)=\frac{x^3}3-x^2+\pi\\g'(x)=x^2-2x\\g'(x_0)=g'(4)=16-8=8\\3.\;f(x)=\frac{x^3}3+2,5x^2+6x\\f'(x)=x^2+5x+6\\x^2+5x+6=0\\D=25-4\cdot6=1\\x_{1,2}=\frac{-5\pm1}2\\x_1=-3,\;x_2=-2\\4.\;a)\;g(x)=(x+1)\left(\frac{x^4}4-\frac1x\right)\\g'(x)=\left(\frac{x^4}4-\frac1x\right)+(x+1)(x^3+\frac1{x^2})$
$b)\;f(x)=\frac{x^2-x}{3x+2}\\f'(x)=\frac{(2x-1)(3x+2)-(x^2-x)\cdot3}{(3x+2)^2}=\frac{6x^2+x-2-3x^2+3x}{(3x+2)^2}=\frac{3x^2+4x-2}{(3x+2)^2}\\c)\;h(x)=\sqrt{3-5x}\\h'(x)=-\frac5{2\sqrt{3-5x}}\\d)\;f(x)=3\cos x-\frac12\sin2x\\f'(x)=-2\sin x-\cos2x$

Задание 2:
$1.\;5x^4+\frac{17}{x^3}+28\\2.\;\frac{2x(4x-7)-4x^2}{(4x-7)^2}=\frac{8x^2-14x-4x^2}{(4x-7)^2}=\frac{4x^2-14x}{(4x-7)^2}\\3.\;\frac{\sin3x}{\sqrt x}+3\sqrt x\cos x\\4.\;\frac1x\\5.\;7x^2+5x$