39.3. Найдите производную функции:

Решение:

1. \( y = (x + 2)(x^2 - 4x + 5) \) \(\implies y' = 1 \cdot (x^2 - 4x + 5) + (x + 2)(2x - 4) = x^2 - 4x + 5 + 2x^2 + 4x - 8 = 3x^2 - 3\).
2. \( y = (3x + 5)(2x^2 - 1) \) \(\implies y' = 3(2x^2 - 1) + (3x + 5)(4x) = 6x^2 - 3 + 12x^2 + 20x = 18x^2 + 20x - 3\).
3. \( y = x^2 \sin x \) \(\implies y' = 2x \sin x + x^2 \cos x\).
4. \( y = x \mathrm{ctg} x \) \(\implies y' = 1 \cdot \mathrm{ctg} x + x \cdot (-\frac{1}{\sin^2 x}) = \mathrm{ctg} x - \frac{x}{\sin^2 x}\).
5. \( y = (2x + 1)\sqrt{x} \) \(\implies y' = 2\sqrt{x} + (2x + 1)\frac{1}{2\sqrt{x}} = 2\sqrt{x} + \frac{2x + 1}{2\sqrt{x}} = \frac{4x + 2x + 1}{2\sqrt{x}} = \frac{6x + 1}{2\sqrt{x}}\).
6. \( y = \sqrt{x} \cos x \) \(\implies y' = \frac{1}{2\sqrt{x}} \cos x - \sqrt{x} \sin x\).

Ответ: 1) \( 3x^2 - 3 \) 2) \( 18x^2 + 20x - 3 \) 3) \( 2x \sin x + x^2 \cos x \) 4) \( \mathrm{ctg} x - \frac{x}{\sin^2 x} \) 5) \( \frac{6x + 1}{2\sqrt{x}} \) 6) \( \frac{\cos x}{2\sqrt{x}} - \sqrt{x} \sin x \).