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

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

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

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

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

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

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

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

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

$$f'(x)=2(x^{-3})’-8(x^{\frac14})’=2\cdot(-3)x^{-4}-8\cdot\frac14x^{-\frac34}=-\frac{6}{x^4}-\frac{2}{\sqrt[4]{x^3}}.$$

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

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

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

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

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

$$f'(x)=(3x-2)^{\frac13}{}’=\frac13(3x-2)^{-\frac23}\cdot 3=\frac{1}{\sqrt[3]{(3x-2)^2}}.$$

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

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