Question

Помогите пжст , срочно

Answer 1

1)

$y'=(2x-1)'=(2x^1-1)'=2\cdot 1 \cdot x^{(1-1)}-0=2\cdot x^0=2 \cdot 1=2$

2)

$y'=(-8)'=0$

3)

$y'=(x^{-2.3})'=-2,3\cdot x^{(-2,3-1)}=-2,3\cdot x^{-3,3}=-\frac{2,3}{x^{3,3}}$

4)

$y'=(3x^7-6x^5-4x^2+17)'=3\cdot 7\cdot x^{(7-1)}-6\cdot 5 \cdot x^{(5-1)}-4\cdot 2 \cdot x^{(2-1)}+0= \\ \\ =21x^6-30x^4-8x=x\cdot (21x^5-30x^3-8)$

5)

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

6)

$y'=(\frac{2}{x^2}-\frac{3}{x^3})'=(2x^{-2}- 3x^{-3})'=2\cdot (-2)\cdot x^{(-2-1)}-3\cdot (-3)\cdot x^{(-3-1)}=\\ \\ = -4x^{-3}+9x^{-4}=-\frac{4}{x^3}+\frac{9}{x^4}=\frac{9}{x^4}-\frac{4}{x^3}=\frac{9-4x}{x^4}$

7)

$y'=((x^3-2)\cdot(x^2+1))'=(x^3-2)'\cdot (x^2+1)+(x^3-2)\cdot(x^2+1)'=\\ \\ =3x^2\cdot (x^2+1)+(x^3-2)\cdot 2x=x\cdot (3x\cdot (x^2+1)+2\cdot (x^3-2))= \\ \\ = x\cdot (3x^3+3x+2x^3-4)=x\cdot (5x^3+3x-4)$

8)

$y'=(3x\cdot tg \, x) ' = (3x)'\cdot tg \, x+3x\cdot (tg \, x)'=3tg \, x +3x\cdot \frac{1}{\cos^2{x}}=3\cdot (tg \, x +\frac{x}{\cos^2{x}})= \\ \\ =3\cdot (\frac{\sin{x}}{\cos{x}}+\frac{x}{\cos^2{x}})$

9)

$y'=(\frac{x^2-5x}{x-3})'=\frac{(x^2-5x)'\cdot (x-3)- (x^2-5x)\cdot (x-3)'}{(x-3)^2}=\frac{(2x-5)\cdot (x-3)-(x^2-5x)\cdot1}{(x-3)^2}=\frac{2x^2-6x-5x+15-x^2+5x}{(x-3)^2}=\\ \\ = \frac{x^2-6x+15}{(x-3)^2}=\frac{x^2-6x+9+6}{(x-3)^2}=\frac{(x-3)^2+6}{(x-3)^2}=1+\frac{6}{(x-3)^2}$

10) $x'(t)=v(t)=(3t^2-5t+8)'=6t-5 \\ \\ v(4)=6\cdot 4-5=24-5=19$