Правило 2 (u + v)' = u' + v' 1 y=x^6+ x^2 2 y=x^-2 + 1/x 3 y= -4x+x^4 4 y= x^4 + 2x^3 5 y=x^-1+1 6 y=6x-3x^5 7 y=4+3x 8 y=x + x^-5-2 9 y=x^8+3x-2 10 y=x 3+ π

Решение:

Используем правило дифференцирования суммы: \( (u+v)' = u' + v' \).

1. \( y = x^6 + x^2 \)
   \( y' = (x^6)' + (x^2)' = 6x^5 + 2x \)
2. \( y = x^{-2} + \frac{1}{x} \)
   \( y = x^{-2} + x^{-1} \)
   \( y' = (x^{-2})' + (x^{-1})' = -2x^{-3} - x^{-2} = -\frac{2}{x^3} - \frac{1}{x^2} \)
3. \( y = -4x + x^4 \)
   \( y' = (-4x)' + (x^4)' = -4 + 4x^3 \)
4. \( y = x^4 + 2x^3 \)
   \( y' = (x^4)' + (2x^3)' = 4x^3 + 6x^2 \)
5. \( y = x^{-1} + 1 \)
   \( y' = (x^{-1})' + (1)' = -x^{-2} + 0 = -\frac{1}{x^2} \)
6. \( y = 6x - 3x^5 \)
   \( y' = (6x)' - (3x^5)' = 6 - 15x^4 \)
7. \( y = 4 + 3x \)
   \( y' = (4)' + (3x)' = 0 + 3 = 3 \)
8. \( y = x + x^{-5} - 2 \)
   \( y' = (x)' + (x^{-5})' - (2)' = 1 - 5x^{-6} - 0 = 1 - \frac{5}{x^6} \)
9. \( y = x^8 + 3x - 2 \)
   \( y' = (x^8)' + (3x)' - (2)' = 8x^7 + 3 - 0 = 8x^7 + 3 \)
10. \( y = x · 3 + \pi \)
    \( y = 3x + \pi \)
    \( y' = (3x)' + (\pi)' = 3 + 0 = 3 \)

Ответ:
1. \( y' = 6x^5 + 2x \)
2. \( y' = -\frac{2}{x^3} - \frac{1}{x^2} \)
3. \( y' = -4 + 4x^3 \)
4. \( y' = 4x^3 + 6x^2 \)
5. \( y' = -\frac{1}{x^2} \)
6. \( y' = 6 - 15x^4 \)
7. \( y' = 3 \)
8. \( y' = 1 - \frac{5}{x^6} \)
9. \( y' = 8x^7 + 3 \)
10. \( y' = 3