Упр.793 ГДЗ Алимов 10-11 класс (Алгебра)
1) f(x) = x6, x0=1/2;
2) f(x) = x^-2, x0=3;
3) f(x) = корень x, x0=4;
4) f(x) = корень 3 степени x, x0=8;
5) f(x) = корень (5-4x), x0=1;
6) f(x) = 1/(корень (3x+1), x0=1.
$$f(x)=x^6,\quad x_0=\frac12$$
$$f'(x)=(x^6)’=6x^5$$
$$f’\!\left(\frac12\right)=6\left(\frac12\right)^5=\frac{6}{32}=\frac{3}{16}$$
$$f(x)=x^{-2},\quad x_0=3$$
$$f'(x)=(x^{-2})’=-2x^{-3}=-\frac{2}{x^3}$$
$$f'(3)=-\frac{2}{3^3}=-\frac{2}{27}$$
$$f(x)=\sqrt{x},\quad x_0=4$$
$$f'(x)=(\sqrt{x})’=\left(x^{\frac12}\right)’=\frac12x^{-\frac12}=\frac{1}{2\sqrt{x}}$$
$$f'(4)=\frac{1}{2\sqrt{4}}=\frac{1}{2\cdot 2}=\frac14$$
$$f(x)=\sqrt[3]{x},\quad x_0=8$$
$$f'(x)=\left(x^{\frac13}\right)’=\frac13x^{-\frac23}=\frac{1}{3\sqrt[3]{x^2}}$$
$$f'(8)=\frac{1}{3\sqrt[3]{8^2}}=\frac{1}{3\sqrt[3]{64}}=\frac{1}{3\cdot 4}=\frac{1}{12}$$
$$f(x)=\sqrt{5-4x},\quad x_0=1$$
$$f'(x)=\left((5-4x)^{\frac12}\right)’=\frac12(5-4x)^{-\frac12}\cdot(-4)=-\frac{2}{\sqrt{5-4x}}$$
$$f'(1)=-\frac{2}{\sqrt{5-4\cdot 1}}=-\frac{2}{\sqrt1}=-2$$
$$f(x)=\frac{1}{\sqrt{3x+1}},\quad x_0=1$$
$$f(x)=(3x+1)^{-\frac12}$$
$$f'(x)=-\frac12(3x+1)^{-\frac32}\cdot 3=-\frac{3}{2(3x+1)^{\frac32}}$$
$$f'(1)=-\frac{3}{2(3\cdot 1+1)^{\frac32}}=-\frac{3}{2\cdot 4^{\frac32}}=-\frac{3}{16}$$
Ответ
$$\frac{3}{16};\ -\frac{2}{27};\ \frac14;\ \frac{1}{12};\ -2;\ -\frac{3}{16}$$
