Вариант №1 1. Найдите производную функции: 1) y = x⁴; 2) y = 4; 3) y = -3/x; 4) y = 3x+2; 5) y = 2cosx-4√x; 6) y = x·sinx; 7) y = ctgx/x; 8) y = (2x-3)⁸; 9) y = x·tgx.

Решение:

1. \( y' = (x^4)' = 4x^3 \)
2. \( y' = (4)' = 0 \)
3. \( y' = (-3x^{-1})' = -3(-1)x^{-2} = 3x^{-2} = \frac{3}{x^2} \)
4. \( y' = (3x+2)' = 3 \)
5. \( y' = (2cosx-4\sqrt{x})' = -2sinx - 4 \cdot \frac{1}{2\sqrt{x}} = -2sinx - \frac{2}{\sqrt{x}} \)
6. \( y' = (x \cdot sinx)' = 1 \cdot sinx + x \cdot cosx = sinx + xcosx \)
7. \( y' = \left(\frac{ctg x}{x}\right)' = \frac{(ctg x)' \cdot x - ctg x \cdot x'}{x^2} = \frac{-\frac{1}{sin^2 x} \cdot x - ctg x \cdot 1}{x^2} = \frac{-x - ctg x \cdot sin^2 x}{x^2 \cdot sin^2 x} = \frac{-x - \frac{cosx}{sinx} \cdot sin^2 x}{x^2 sin^2 x} = \frac{-x - cosx sinx}{x^2 sin^2 x} \)
8. \( y' = ((2x-3)^8)' = 8(2x-3)^7 \cdot (2x-3)' = 8(2x-3)^7 \cdot 2 = 16(2x-3)^7 \)
9. \( y' = (x \cdot tgx)' = 1 \cdot tgx + x \cdot (tgx)' = tgx + x \cdot \frac{1}{cos^2 x} = tgx + \frac{x}{cos^2 x} \)

Ответ: 1) \( 4x^3 \); 2) 0; 3) \( \frac{3}{x^2} \); 4) 3; 5) \( -2sinx - \frac{2}{\sqrt{x}} \); 6) \( sinx + xcosx \); 7) \( \frac{-x - cosx sinx}{x^2 sin^2 x} \); 8) \( 16(2x-3)^7 \); 9) \( tgx + \frac{x}{cos^2 x} \).