1. Найдите производные: 1) y = x^4; 2) y = 4; 3) y = -3/x; 4) y = 3x + 2; 5) y = 2cos x - 4√x; 6) y = x · sin x; 7) y = ctgx / x; 8) y = (2x - 3)^8; 9) y = x · tgx.

Решение:

Найдём производные для каждого выражения:

1. \( y = x^4 \) \( \implies y' = 4x^3 \)
2. \( y = 4 \) \( \implies y' = 0 \)
3. \( y = -\frac{3}{x} = -3x^{-1} \) \( \implies y' = -3(-1)x^{-2} = \frac{3}{x^2} \)
4. \( y = 3x + 2 \) \( \implies y' = 3 \)
5. \( y = 2 \cos x - 4\sqrt{x} = 2\cos x - 4x^{1/2} \) \( \implies y' = -2\sin x - 4(\frac{1}{2}x^{-1/2}) = -2\sin x - \frac{2}{\sqrt{x}} \)
6. \( y = x \cdot \sin x \) \( \implies y' = (x)'\sin x + x(\sin x)' = 1 \cdot \sin x + x \cdot \cos x = \sin x + x\cos x \)
7. \( y = \frac{\text{ctg}x}{x} \) \( \implies y' = \frac{(\text{ctg}x)' \cdot x - \text{ctg}x \cdot (x)'}{x^2} = \frac{-\frac{1}{\sin^2 x} \cdot x - \text{ctg}x \cdot 1}{x^2} = \frac{-\frac{x}{\sin^2 x} - \frac{\cos x}{\sin x}}{x^2} = \frac{-x - \sin x \cos x}{x^2 \sin^2 x} \)
8. \( y = (2x - 3)^8 \) \( \implies y' = 8(2x - 3)^7 \cdot (2x - 3)' = 8(2x - 3)^7 \cdot 2 = 16(2x - 3)^7 \)
9. \( y = x \cdot \text{tg}x \) \( \implies y' = (x)'\text{tg}x + x(\text{tg}x)' = 1 \cdot \text{tg}x + x \cdot \frac{1}{\cos^2 x} = \text{tg}x + \frac{x}{\cos^2 x} \)

Ответ: 1) 4x³; 2) 0; 3) 3/x²; 4) 3; 5) -2sin x - 2/√x; 6) sin x + x cos x; 7) (-x - sin x cos x) / (x² sin² x); 8) 16(2x - 3)⁷; 9) tg x + x/cos² x.