Question

Помогите пожалуйста если вам не сложно)

Answer 1

1) x' + 12' = 1

2) -12x' = - 12*1 = - 12

3)

${x}^{11} = {11x}^{11 - 1} = {11x}^{10}$

4)

${x}^{ - 7} = { - 7x}^{ - 8}$

5)

${x}^{ - \frac{2}{7} } = - \frac{2}{7} {x}^{ - 1 \frac{2}{7} }$

6)

7)

8)

${(1 - 2x)}^{ - 6} = { - 6(1 - 2x)}^{ - 7}$

9)

${(4x)}^{5} = {5(4x)}^{4}$

10)

11)

$\frac{11}{x} = \frac{ -11}{ {x}^{2} }$

12)

$\frac{x}{7} = \frac{7}{49} = \frac{1}{7}$

13) (-4x²) '+ 7x' + 8' = -8x +7

14)

15)

Это, правда, не всё

Answer 2

$(x+12)'=x'+12'=1+0=1\\\\(-12x)'=-12\cdot x'=-12\cdot 1=-12\\\\(x^{11})'=11x^{10}\\\\(x^{-7})'=-7\cdot x^{-7-1}=-7\cdot x^{-8}\\\\(x^{\frac{-2}{7}})'=-\frac{2}{7}\cdot x^{-\frac{2}{7}-1}=-\frac{2}{7}\cdot x^{-\frac{9}{7}}\\\\(\frac{1}{x^{12}})'=(x^{-12})'=-12\cdot x^{-13}=-\frac{12}{x^{13}}\\\\(\frac{1}{\sqrt[4]{x^3}})'=(x^{-\frac{3}{4}})'=-\frac{3}{4}\cdot x^{-\frac{7}{4}}=-\frac{3}{4\sqrt[4]{x^7}}\\\\(1-2x)^{-6}=-6\cdot (1-2x)^{-7}\cdot (-2)=12\cdot (1-2x)^{-7}\\\\((4x)^5)'=5\cdot (4x)^4\cdot 4=20(4x)^4=5120x^4$

$(\sqrt[4]{5x-1})'=\frac{1}{4}\cdot (5x-1)^{\frac{1}{4}-1}=\frac{1}{4}\cdot (5x-1)^{-\frac{3}{4}}=\frac{1}{4\sqrt[4]{(5x-1)^3}}\\\\(\frac{11}{x})'=11\cdot (x^{-1})'=11\cdot (-1)\cdot x^{-2}=-\frac{11}{x^2}\\\\(\frac{x}{7})'=(\frac{1}{7}\cdot x)'=\frac{1}{7}\\\\(-4x^2+7x+8)'=-8x+7\\\\(\frac{1}{x^5}+5x^2)'=-5\cdot x^{-6}+10x=-\frac{5}{x^6}+10x\\\\(-7\sqrt[7]{x}-14\sqrt[14]{x})'=-7\cdot \frac{1}{7}\cdot x^{-\frac{6}{7}}-14\cdot \frac{1}{14}\cdot x^{-\frac{13}{14}}=-\frac{1}{\sqrt[7]{x^6}}-\frac{1}{\sqrt[14]{x^{13}}}$