Вопрос:

Find the derivative of the following functions: 1. f(x) = 5x^4 + x^3/3 - 2x^2 + 6x 2. f(x) = (4x-8) * sqrt(x) 3. f(x) = (x^2 + 4x) / (x - 2) 4. f(x) = 2/x^3 - 6/x^4 5. f(x) = sqrt(2x-7), x0 = 1 6. f(x) = sin^3(x)

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Ответ:

Решение:

Найдем производные для каждой функции:

  1. \( f'(x) = \frac{d}{dx}(5x^4 + \frac{x^3}{3} - 2x^2 + 6x) \)
    \( f'(x) = 20x^3 + \frac{3x^2}{3} - 4x + 6 \)
    \( f'(x) = 20x^3 + x^2 - 4x + 6 \)
  2. \( f(x) = (4x-8)\sqrt{x} = 4x^{3/2} - 8x^{1/2} \)
    \( f'(x) = \frac{d}{dx}(4x^{3/2} - 8x^{1/2}) \)
    \( f'(x) = 4 \cdot \frac{3}{2}x^{1/2} - 8 \cdot \frac{1}{2}x^{-1/2} \)
    \( f'(x) = 6x^{1/2} - 4x^{-1/2} = 6\sqrt{x} - \frac{4}{\sqrt{x}} \)
  3. \( f(x) = \frac{x^2 + 4x}{x-2} \)
    Используем правило дифференцирования частного: \( (\frac{u}{v})' = \frac{u'v - uv'}{v^2} \)
    \( u = x^2 + 4x \), \( u' = 2x + 4 \)
    \( v = x - 2 \), \( v' = 1 \)
    \( f'(x) = \frac{(2x+4)(x-2) - (x^2+4x)(1)}{(x-2)^2} \)
    \( f'(x) = \frac{2x^2 - 4x + 4x - 8 - x^2 - 4x}{(x-2)^2} \)
    \( f'(x) = \frac{x^2 - 4x - 8}{(x-2)^2} \)
  4. \( f(x) = 2x^{-3} - 6x^{-4} \)
    \( f'(x) = \frac{d}{dx}(2x^{-3} - 6x^{-4}) \)
    \( f'(x) = 2(-3)x^{-4} - 6(-4)x^{-5} \)
    \( f'(x) = -6x^{-4} + 24x^{-5} = \frac{24}{x^5} - \frac{6}{x^4} \)
  5. \( f(x) = \sqrt{2x-7} = (2x-7)^{1/2} \)
    \( f'(x) = \frac{1}{2}(2x-7)^{-1/2} \cdot \frac{d}{dx}(2x-7) \)
    \( f'(x) = \frac{1}{2}(2x-7)^{-1/2} \cdot 2 \)
    \( f'(x) = (2x-7)^{-1/2} = \frac{1}{\sqrt{2x-7}} \)
  6. \( f(x) = \sin^3 x = (\sin x)^3 \)
    \( f'(x) = 3(\sin x)^2 \cdot \frac{d}{dx}(\sin x) \)
    \( f'(x) = 3\sin^2 x \cdot \cos x \)

Ответ:

1. \( f'(x) = 20x^3 + x^2 - 4x + 6 \)

2. \( f'(x) = 6\sqrt{x} - \frac{4}{\sqrt{x}} \)

3. \( f'(x) = \frac{x^2 - 4x - 8}{(x-2)^2} \)

4. \( f'(x) = \frac{24}{x^5} - \frac{6}{x^4} \)

5. \( f'(x) = \frac{1}{\sqrt{2x-7}} \)

6. \( f'(x) = 3\sin^2 x \cos x \)

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