1.
1) Фиксируем x. y(x) = 5x - 9
2) Даём приращение аргументу. y(x + Δx) = 5(x + Δx) - 9 = 5x + 5Δx - 9
3) Находим приращение функции. Δy = y(x +
Δx) - y(x) = 5x + 5Δx - 9 - 5x + 9 = 5Δx
4) Составляем отношение Δy/Δx.
Δy/Δx = 5Δx/Δx = 5
5) Находим предел:
f'(x) = 5
2.
f(x) = x³ + 2x² - 5x + 1
f'(x) = 3x² + 4x - 5
f'(-1) = 3 * (-1)² - 4 - 5 = 3 - 9 = -6
3.
f(x) = eˣ * sinx
f'(x) = eˣ * sinx + eˣ * cosx = eˣ (sinx + cosx)
f'(0) = e⁰(sin0 + cos0) = 1 * (0 + 1) = 1
4.
f(x) = (x²+3) / (x-1)
f'(x) = (2x(x-1) - (x²+3)) / (x-1)² = (2x² - 2x - x² - 3) / (x - 1)² = (x² - 2x - 3)/(x - 1)²
f'(0) = (0 - 0 - 3)/(0 - 1)² = -3/1 = -3
5.
f(x)
= x^(1/6)
f'(x) = 1/6 * x^(1/6-1) = 1/6 * x^(-5/6)
f'(64) = 1/6 * 64^(-5/6) = 1/6 * 2^(-5) = 1/6 * 1/32 = 1/192