1) найдите производную а) y = 7x - 3

1) Найдите производную a) $y = 7x - 3$ $y' = 7$ б) $y = \frac{1}{x} = x^{-1}$ $y' = -1 \cdot x^{-2} = -\frac{1}{x^2}$ в) $y = x^5 - 4\sqrt{x} = x^5 - 4x^{\frac{1}{2}}$ $y' = 5x^4 - 4 \cdot \frac{1}{2}x^{-\frac{1}{2}} = 5x^4 - \frac{2}{\sqrt{x}}$ 2) Найдите значение производной в точке $x_0$ a) $y = \frac{x^6}{3} + \frac{8}{x} = \frac{1}{3}x^6 + 8x^{-1}$, $x_0 = 2$ $y' = \frac{1}{3} \cdot 6x^5 + 8 \cdot (-1)x^{-2} = 2x^5 - \frac{8}{x^2}$ $y'(2) = 2 \cdot 2^5 - \frac{8}{2^2} = 2 \cdot 32 - \frac{8}{4} = 64 - 2 = 62$ б) $y = (4x^2 - 3)(2x^3 - 1)$, $x_0 = -1$ Воспользуемся формулой производной произведения $(uv)' = u'v + uv'$ $u = 4x^2 - 3 \Rightarrow u' = 8x$ $v = 2x^3 - 1 \Rightarrow v' = 6x^2$ $y' = 8x(2x^3 - 1) + (4x^2 - 3)(6x^2) = 16x^4 - 8x + 24x^4 - 18x^2 = 40x^4 - 18x^2 - 8x$ $y'(-1) = 40(-1)^4 - 18(-1)^2 - 8(-1) = 40(1) - 18(1) + 8 = 40 - 18 + 8 = 30$ 3) Найдите производную сложной функции a) $y = (8x + 2)^5$ $y' = 5(8x + 2)^4 \cdot (8x + 2)' = 5(8x + 2)^4 \cdot 8 = 40(8x + 2)^4$ б) $y = \sqrt{4x - 3} = (4x - 3)^{\frac{1}{2}}$ $y' = \frac{1}{2}(4x - 3)^{-\frac{1}{2}} \cdot (4x - 3)' = \frac{1}{2\sqrt{4x - 3}} \cdot 4 = \frac{2}{\sqrt{4x - 3}}$