Задача 69949 Найти f'(x0), если:...
Условие
Решение
(x^n)' = n*x^(n-1)
1) f(x) = x^6; x0 = 1/2
f'(x) = 6x^5
f'(x0) = 6*(1/2)^5 = 6/2^5 = 6/32 = 3/16
2) f(x) = x^(-2); x0 = 3
f'(x) = -2x^(-3) = -2/x^3
f'(x0) = -2/3^3 = -2/27
3) f(x) = sqrt(x) = x^(1/2); x0 = 4
[m]f'(x) = \frac{1}{2}x^{-1/2} = \frac{1}{2\sqrt{x}}[/m]
[m]f'(x0) = \frac{1}{2\sqrt{4}} = \frac{1}{2 \cdot 2} = \frac{1}{4}[/m]
4) [m]f(x) = \sqrt[3]{x} = x^{1/3}[/m]; x0 = 8
[m]f'(x) = \frac{1}{3}x^{-2/3} = \frac{1}{3\sqrt[3]{x^2}}[/m]
[m]f'(x0) = \frac{1}{3\sqrt[3]{8^2}} = \frac{1}{3 \cdot 2^2} = \frac{1}{12}[/m]
5) f(x) = sqrt(5 - 4x); x0 = 1
[m]f'(x) = \frac{1}{2}(5-4x)^{-1/2}(-4) = \frac{-4}{2\sqrt{5-4x}} = -\frac{2}{\sqrt{5-4x}}[/m]
[m]f'(x0) = -\frac{2}{\sqrt{5-4 \cdot 1}} = -\frac{2}{\sqrt{1}} = -2[/m]
6) [m]f(x) = \frac{1}{\sqrt{3x+1}} = (3x+1)^{-1/2}[/m]; x0 = 1
[m]f'(x) = -\frac{1}{2}(3x+1)^{-3/2} \cdot 3 = -\frac{3}{2(3x+1)^{3/2}} = -\frac{3}{2\sqrt{(3x+1)^3}}[/m]
[m]f'(x0) = -\frac{3}{2\sqrt{(3 \cdot 1+1)^3}} = -\frac{3}{2\sqrt{4^3}} = -\frac{3}{2 \cdot 2^3} = \frac{3}{16}[/m]