Упр.869 ГДЗ Алимов 10-11 класс (Алгебра)

869 1) 2×4-x3+3x+4;
2) -x5+2×3-3×2-1;
3) 6 (корень 3 степени x)+1/x2;
4) 2/x3 — 8 (корень 4 степени x);
5) (2x+3)8;
6) (4-3x)7;
7) корень 3 степени (3x-2);
8) 1/ корень (1-4x).

1. $$f(x)=2x^4-x^3+3x+4$$

   $$f'(x)=(2x^4)’- (x^3)’ + (3x)’ + 4′ = 8x^3-3x^2+3$$

2. $$f(x)=-x^5+2x^3-3x^2-1$$

   $$f'(x)=-(x^5)’+2(x^3)’-3(x^2)’-1’=-5x^4+6x^2-6x$$

3. $$f(x)=6\sqrt[3]{x}+\frac{1}{x^2}$$

   $$f'(x)=6\left(x^{\frac13}\right)’+(x^{-2})’=6\cdot\frac13x^{-\frac23}-2x^{-3}$$

   $$f'(x)=\frac{2}{\sqrt[3]{x^2}}-\frac{2}{x^3}$$

4. $$f(x)=\frac{2}{x^3}-8\sqrt[4]{x}$$

   $$f'(x)=2(x^{-3})’-8(x^{\frac14})’ = 2\cdot(-3)x^{-4}-8\cdot\frac14x^{-\frac34}$$

   $$f'(x)=-\frac{6}{x^4}-\frac{2}{\sqrt[4]{x^3}}$$

5. $$f(x)=(2x+3)^8$$

   $$f'(x)=8(2x+3)^7\cdot(2x+3)’=16(2x+3)^7$$

6. $$f(x)=(4-3x)^7$$

   $$f'(x)=7(4-3x)^6\cdot(4-3x)’=7(4-3x)^6\cdot(-3)=-21(4-3x)^6$$

7. $$f(x)=\sqrt[3]{3x-2}=(3x-2)^{\frac13}$$

   $$f'(x)=\frac13(3x-2)^{-\frac23}\cdot 3=(3x-2)^{-\frac23}$$

   $$f'(x)=\frac{1}{\sqrt[3]{(3x-2)^2}}$$

8. $$f(x)=\frac{1}{\sqrt{1-4x}}=(1-4x)^{-\frac12}$$

   $$f'(x)=-\frac12(1-4x)^{-\frac32}\cdot(-4)=2(1-4x)^{-\frac32}$$

   $$f'(x)=\frac{2}{(1-4x)\sqrt{1-4x}}$$

Ответ

1. $$8x^3-3x^2+3$$
2. $$-5x^4+6x^2-6x$$
3. $$\frac{2}{\sqrt[3]{x^2}}-\frac{2}{x^3}$$
4. $$-\frac{6}{x^4}-\frac{2}{\sqrt[4]{x^3}}$$
5. $$16(2x+3)^7$$
6. $$-21(4-3x)^6$$
7. $$\frac{1}{\sqrt[3]{(3x-2)^2}}$$
8. $$\frac{2}{(1-4x)\sqrt{1-4x}}$$