Найти производные следующих функций:

Привет! Давай решим эти примеры на нахождение производных. Будем использовать основные правила: $(x^n)' = n \cdot x^{n-1}$ и $(c \cdot f(x))' = c \cdot f'(x)$. 1) $y = 3x^4 \implies y' = 3 \cdot 4x^3 = 12x^3$ 2) $y = 2x^{-5} \implies y' = 2 \cdot (-5)x^{-6} = -10x^{-6} = -\frac{10}{x^6}$ 3) $y = 4x^{\frac{1}{3}} \implies y' = 4 \cdot \frac{1}{3}x^{-\frac{2}{3}} = \frac{4}{3x^{\frac{2}{3}}} = \frac{4}{3\sqrt[3]{x^2}}$ 4) $y = 5x^{-\frac{2}{5}} \implies y' = 5 \cdot (-\frac{2}{5})x^{-\frac{7}{5}} = -2x^{-\frac{7}{5}} = -\frac{2}{x \sqrt[5]{x^2}}$ 5) $y = 5\sqrt[5]{x^3} = 5x^{\frac{3}{5}} \implies y' = 5 \cdot \frac{3}{5}x^{-\frac{2}{5}} = 3x^{-\frac{2}{5}} = \frac{3}{\sqrt[5]{x^2}}$ 6) $y = \frac{1}{2x^3} = \frac{1}{2}x^{-3} \implies y' = \frac{1}{2} \cdot (-3)x^{-4} = -\frac{3}{2x^4}$ 7) $y = 3x^2 \cdot x^{\frac{1}{3}} = 3x^{\frac{7}{3}} \implies y' = 3 \cdot \frac{7}{3}x^{\frac{4}{3}} = 7x\sqrt[3]{x}$ 8) $y = \frac{2x^2}{\sqrt[3]{x}} = 2x^2 \cdot x^{-\frac{1}{3}} = 2x^{\frac{5}{3}} \implies y' = 2 \cdot \frac{5}{3}x^{\frac{2}{3}} = \frac{10}{3}\sqrt[3]{x^2}$ 9) $y = \frac{1}{x^4} = x^{-4} \implies y' = -4x^{-5} = -\frac{4}{x^5}$. Тогда: $y'(-1) = -\frac{4}{(-1)^5} = 4$ $y'(2) = -\frac{4}{2^5} = -\frac{4}{32} = -\frac{1}{8} = -0,125$ 10) $y = 4x^3 + 2x^2 + x - 5 \implies y' = 12x^2 + 4x + 1$ 11) $y = 2x^{\frac{1}{2}} + 3 \cdot \frac{1}{2}x^{-\frac{2}{3}} - 2x^{-\frac{1}{2}} - x^{-1} + 1$ $y' = 2 \cdot \frac{1}{2}x^{-\frac{1}{2}} + \frac{3}{2} \cdot (-\frac{2}{3})x^{-\frac{5}{3}} - 2 \cdot (-\frac{1}{2})x^{-\frac{3}{2}} - (-1)x^{-2} + 0$ $y' = \frac{1}{\sqrt{x}} - \frac{1}{x\sqrt[3]{x^2}} + \frac{1}{x\sqrt{x}} + \frac{1}{x^2}$ 12) $y = (x^3 - 1)(x^2 + x + 1) = x^5 + x^4 + x^3 - x^2 - x - 1$ $y' = 5x^4 + 4x^3 + 3x^2 - 2x - 1$ 13) $y = (3x^2 + 1)(2x^2 + 3) = 6x^4 + 9x^2 + 2x^2 + 3 = 6x^4 + 11x^2 + 3$ $y' = 24x^3 + 22x$ 14) $y = \frac{x^2+1}{x^2-1}$. Используем $(u/v)' = \frac{u'v - uv'}{v^2}$: $y' = \frac{2x(x^2-1) - (x^2+1)(2x)}{(x^2-1)^2} = \frac{2x^3 - 2x - 2x^3 - 2x}{(x^2-1)^2} = -\frac{4x}{(x^2-1)^2}$ 15) $y = \frac{x^2-x+1}{x^2+x+1}$. Аналогично: $u' = 2x-1, v' = 2x+1$ $y' = \frac{(2x-1)(x^2+x+1) - (x^2-x+1)(2x+1)}{(x^2+x+1)^2} = \frac{(2x^3+2x^2+2x-x^2-x-1) - (2x^3+x^2-2x^2-x+2x+1)}{(x^2+x+1)^2} = \frac{(2x^3+x^2+x-1) - (2x^3-x^2+x+1)}{(x^2+x+1)^2} = \frac{2x^2 - 2}{(x^2+x+1)^2}$