Нужно решить все номера .вложение внутри

Question 1

Question 2

A1.
a) $(3x^5-2x^2)'=15x^4-4x$
б) $(2x^8-5)'=16x^7$
в) $(tgx-2cosx)'=\frac{1}{cos^2x}+2sinx$
г) $(sin(3x-\frac{\pi}{4}))'=cos(3x-\frac{\pi}{4})*(3x-\frac{\pi}{4})'=3cos(3x-\frac{\pi}{4})$

A2.
a) $((x^3+3)(x-x^3))'=(x^4-x^6+3x-3x^3)'=4x^3-6x^5+3-6x^2$
б) $(\frac{x^4-x^2}{x-1})'=\frac{(x^4-x^2)'(x-1)-(x^4-x^2)(x-1)'}{(x-1)^2}=\\ =\frac{(4x^3-2x)(x-1)-(x^4-x^2)}{(x-1)^2}=\frac{2x(2x^2-1)(x-1)-x^2(x-1)(x+1)}{(x-1)^2}=\\ =\frac{(x-1)(4x^3-2x-x^3-x^2)}{(x-1)^2}=\frac{3x^3-x^2-2x}{x-1}=\frac{x(3x^2-x-2)}{x-1}=\frac{3x(x-1)(x+\frac{2}{3})}{x-1}=\\ =x(3x+2)$

$f(x)=0.5cos(3x-\frac{\pi}{6}), x_0=\frac{\pi}{9}\\ f'(x)=-0.5*3sin(3x-\frac{\pi}{6})=-1,5sin(3x-\frac{\pi}{6})\\ f'(x_0)=-1.5*sin(\frac{\pi}{3}-\frac{\pi}{6})=-1.5sin\frac{\pi}{6}=-1.5*0.5=-0.75$

В1.
$f(x)=6x+x\sqrt{x}=6x+x^{3/2}\\ f'(x)=6+\frac{3}{2}x^{1/2}\ \textgreater \ 0|*2\\ 12+3\sqrt{x}\ \textgreater \ 0|:3\\ 4+\sqrt{x}\ \textgreater \ 0\\ \sqrt{x}\ \textgreater \ -4\\ x\geq 0$

В2.
$y=\frac{x+4}{\sqrt{x}}\\ y'=\frac{\sqrt{x}-(x+4)\frac{1}{2\sqrt{x}}}{x}=\frac{2x-x-4}{2x\sqrt{x}}=\frac{x-4}{2x\sqrt{x}}$

С1.
$y=(3-x)^4(2x+1)^3\\ y'=4(3-x)^3(-1)(2x+1)^3+(3-x)^43(2x+1)^22=\\= -4(3-x)^3(2x+1)^3+6(3-x)^4(2x+1)^2=2(3-x)^3(2x+1)^2*\\*(-2(2x+1)+3(3-x))=2(3-x)^3(2x+1)^2(-4x-2+9-3x)=\\= -2(x-3)^3(2x+1)^2(-7x+7)=14(x-3)^3(2x+1)^2(x-1)\\ y'\ \textless \ 0\\ (x-3)^3(2x+1)^2(x-1)\ \textless \ 0$
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