Упр.181 ГДЗ Колягин Ткачёва 11 класс (Алгебра)
1) f(x) = 1/x3 + 1/x2;
2) f(x) = корень x + 1/x+1;
3) f(x) = 3/ корень 3 степени x — 2/x3;
4) f(x) = x3/2 — x^-3/2;
5) f(x) = (x-1)2(x-3);
6) f(x) = (x-3)3 (x-1);
7) f(x) = (x2-1)(x+3);
8) f(x) = (x2-9)(x+1).
$$f(x)=\frac{1}{x^3}+\frac{1}{x^2}=x^{-3}+x^{-2}$$
$$f'(x)=-3x^{-4}-2x^{-3}$$
$$f'(3)=-\frac{3}{3^4}-\frac{2}{3^3}=-\frac{3}{81}-\frac{2}{27}=-\frac{1}{27}-\frac{2}{27}=-\frac{1}{9}$$
$$f'(1)=-\frac{3}{1^4}-\frac{2}{1^3}=-3-2=-5$$
$$f(x)=\sqrt{x}+\frac{1}{x}+1=x^{1/2}+x^{-1}+1$$
$$f'(x)=\frac{1}{2}x^{-1/2}-x^{-2}$$
$$f'(3)=\frac{1}{2\sqrt{3}}-\frac{1}{9}=\frac{\sqrt{3}}{6}-\frac{1}{9}$$
$$f'(1)=\frac{1}{2}-1=-\frac{1}{2}$$
$$f(x)=\frac{3}{\sqrt[3]{x}}-\frac{2}{x^3}=3x^{-1/3}-2x^{-3}$$
$$f'(x)=3\cdot\left(-\frac{1}{3}\right)x^{-4/3}-2\cdot(-3)x^{-4}=-x^{-4/3}+6x^{-4}$$
$$f'(3)=-\frac{1}{3\sqrt[3]{3^4}}+\frac{6}{3^4}=-\frac{1}{3\sqrt[3]{81}}+\frac{2}{27}=\frac{2}{27}-\frac{1}{3\sqrt[3]{81}}$$
$$f'(1)=-1+6=5$$
$$f(x)=x^{3/2}-x^{-3/2}$$
$$f'(x)=\frac{3}{2}x^{1/2}+\frac{3}{2}x^{-5/2}$$
$$f'(3)=\frac{3\sqrt{3}}{2}+\frac{3}{2\cdot 3^2\sqrt{3}}=\frac{14\sqrt{3}}{9}$$
$$f'(1)=\frac{3}{2}+\frac{3}{2}=3$$
$$f(x)=(x-1)^2(x-3)$$
$$f'(x)=2(x-1)(x-3)+(x-1)^2$$
$$f'(3)=2\cdot 2\cdot 0+2^2=4$$
$$f'(1)=2\cdot 0\cdot(-2)+0^2=0$$
$$f(x)=(x-3)^3(x-1)$$
$$f'(x)=3(x-3)^2(x-1)+(x-3)^3$$
$$f'(3)=0$$
$$f'(1)=3\cdot(-2)^2\cdot 0+(-2)^3=-8$$
$$f(x)=(x^2-1)(x+3)$$
$$f'(x)=2x(x+3)+(x^2-1)$$
$$f'(3)=2\cdot 3\cdot 6+8=44$$
$$f'(1)=2\cdot 1\cdot 4+0=8$$
$$f(x)=(x^2-9)(x+1)$$
$$f'(x)=2x(x+1)+(x^2-9)$$
$$f'(3)=2\cdot 3\cdot 4+0=24$$
$$f'(1)=2\cdot 1\cdot 2-8=-4$$
Ответ
1) $$f'(3)=-\frac{1}{9},\ f'(1)=-5$$; 2) $$f'(3)=\frac{\sqrt{3}}{6}-\frac{1}{9},\ f'(1)=-\frac{1}{2}$$; 3) $$f'(3)=\frac{2}{27}-\frac{1}{3\sqrt[3]{81}},\ f'(1)=5$$; 4) $$f'(3)=\frac{14\sqrt{3}}{9},\ f'(1)=3$$; 5) $$f'(3)=4,\ f'(1)=0$$; 6) $$f'(3)=0,\ f'(1)=-8$$; 7) $$f'(3)=44,\ f'(1)=8$$; 8) $$f'(3)=24,\ f'(1)=-4$$.
