Вариант №2 1. Найдите производную функции: 1) y = x⁷; 2) y = 5; 3) y = -6/x; 4) y = 4x+5; 5) y = sinx+0,5√x; 6) y = x·cosx; 7) y = lgx/x; 8) y = (3x-4)⁶; 9) y = x·ctgx.

Решение:

1. \( y' = (x^7)' = 7x^6 \)
2. \( y' = (5)' = 0 \)
3. \( y' = (-6x^{-1})' = -6(-1)x^{-2} = 6x^{-2} = \frac{6}{x^2} \)
4. \( y' = (4x+5)' = 4 \)
5. \( y' = (sinx+0,5\sqrt{x})' = cosx + 0,5 \cdot \frac{1}{2\sqrt{x}} = cosx + \frac{1}{4\sqrt{x}} \)
6. \( y' = (x \cdot cosx)' = 1 \cdot cosx + x \cdot (-sinx) = cosx - xsinx \)
7. \( y' = \left(\frac{lg x}{x}\right)' = \frac{(lg x)' \cdot x - lg x \cdot x'}{x^2} = \frac{\frac{1}{x \ln 10} \cdot x - lg x \cdot 1}{x^2} = \frac{\frac{1}{\ln 10} - lg x}{x^2} \)
8. \( y' = ((3x-4)^6)' = 6(3x-4)^5 \cdot (3x-4)' = 6(3x-4)^5 \cdot 3 = 18(3x-4)^5 \)
9. \( y' = (x \cdot ctgx)' = 1 \cdot ctgx + x \cdot (ctgx)' = ctgx + x \cdot \left(-\frac{1}{sin^2 x}\right) = ctgx - \frac{x}{sin^2 x} \)

Ответ: 1) \( 7x^6 \); 2) 0; 3) \( \frac{6}{x^2} \); 4) 4; 5) \( cosx + \frac{1}{4\sqrt{x}} \); 6) \( cosx - xsinx \); 7) \( \frac{\frac{1}{\ln 10} - lg x}{x^2} \); 8) \( 18(3x-4)^5 \); 9) \( ctgx - \frac{x}{sin^2 x} \).