Упр.36.7 ГДЗ Мерзляк 10 класс Базовый уровень (Алгебра)

Задача

1) f(x)=8/x+5x-2, x_0=2; 4) f(x)=(1+3x)vx, x_0=9;
2) f(x)=(2-3x)/(x+2), x_0=-3; 5) f(x)=3(x^(1/3))-10(x^(1/5)), x_0=1;
3) f(x)=(x^2+2)/(x-2)-2sin(x), x_0=0; 6) f(x)=x sin(x), x_0=0?

Подробный ответ

$$f(x)=\frac{8}{x}+5x-2,\quad x_0=2.$$
$$f'(x)=8\left(\frac{1}{x}\right)’+(5x-2)’=8\cdot\left(-\frac{1}{x^2}\right)+5=5-\frac{8}{x^2}.$$
$$f'(2)=5-\frac{8}{2^2}=5-\frac{8}{4}=5-2=3.$$
$$f(x)=\frac{2-3x}{x+2},\quad x_0=-3.$$
$$f'(x)=\frac{(2-3x)'(x+2)-(2-3x)(x+2)’}{(x+2)^2}.$$
$$f'(x)=\frac{-3(x+2)-(2-3x)\cdot 1}{(x+2)^2}=\frac{-3x-6-2+3x}{(x+2)^2}=-\frac{8}{(x+2)^2}.$$
$$f'(-3)=-\frac{8}{(-3+2)^2}=-\frac{8}{(-1)^2}=-8.$$
$$f(x)=\frac{x^2+2}{x-2}-2\sin x,\quad x_0=0.$$
$$f'(x)=\frac{(x^2+2)'(x-2)-(x^2+2)(x-2)’}{(x-2)^2}-2(\sin x)’.$$
$$f'(x)=\frac{2x(x-2)-(x^2+2)\cdot 1}{(x-2)^2}-2\cos x =\frac{2x^2-4x-x^2-2}{(x-2)^2}-2\cos x =\frac{x^2-4x-2}{(x-2)^2}-2\cos x.$$
$$f'(0)=\frac{0^2-4\cdot 0-2}{(0-2)^2}-2\cos 0=\frac{-2}{4}-2\cdot 1=-\frac12-2=-\frac52.$$
$$f(x)=(1+3x)\sqrt{x},\quad x_0=9.$$
$$f'(x)=(1+3x)’\sqrt{x}+(1+3x)(\sqrt{x})’$$
$$f'(x)=3\sqrt{x}+(1+3x)\cdot\frac{1}{2\sqrt{x}} =\frac{6x+1+3x}{2\sqrt{x}} =\frac{9x+1}{2\sqrt{x}}.$$
$$f'(9)=\frac{9\cdot 9+1}{2\sqrt{9}}=\frac{81+1}{2\cdot 3}=\frac{82}{6}=\frac{41}{3}=13\frac{2}{3}.$$
$$f(x)=3\sqrt[3]{x}-10\sqrt[5]{x},\quad x_0=1.$$
$$f'(x)=3(\sqrt[3]{x})’-10(\sqrt[5]{x})’$$
$$f'(x)=3\cdot\frac{1}{3\sqrt[3]{x^2}}-10\cdot\frac{1}{5\sqrt[5]{x^4}} =\frac{1}{\sqrt[3]{x^2}}-\frac{2}{\sqrt[5]{x^4}}.$$
$$f'(1)=\frac{1}{\sqrt[3]{1^2}}-\frac{2}{\sqrt[5]{1^4}}=1-2=-1.$$
$$f(x)=x\sin x,\quad x_0=0.$$
$$f'(x)=(x)’\sin x+x(\sin x)’=1\cdot\sin x+x\cos x.$$
$$f'(0)=1\cdot\sin 0+0\cdot\cos 0=0.$$