Помогите, пожалуйста, срочно надо!!!!!!!!!!

1.
1) Фиксируем x. y(x) = 5x — 9
2) Даём приращение аргументу. y(x + ?x) = 5(x + ?x) — 9 = 5x + 5?x — 9
3) Находим приращение функции. ?y = y(x +

?x) — y(x) = 5x + 5?x — 9 — 5x + 9 = 5?x
4) Составляем отношение ?y/?x.

?y/?x = 5?x/?x = 5
5) Находим предел:
$\lim_{\Delta x \to0} (5) = 5$

f'(x) = 5

2.
f(x) = x? + 2x? — 5x + 1
f'(x) = 3x? + 4x — 5
f'(-1) = 3 * (-1)? — 4 — 5 = 3 — 9 = -6

3.
f(x) = e? * sinx
f'(x) = e? * sinx + e? * cosx = e? (sinx + cosx)
f'(0) = e?(sin0 + cos0) = 1 * (0 + 1) = 1

4.
f(x) = (x?+3) / (x-1)
f'(x) = (2x(x-1) — (x?+3)) / (x-1)? = (2x? — 2x — x? — 3) / (x — 1)? = (x? — 2x — 3)/(x — 1)?
f'(0) = (0 — 0 — 3)/(0 — 1)? = -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

Оцени ответ

1) f ' (x) = 5

2) f ' (x) = 3x^2+4x-5
f ' (-1) = 3 * (-1)^2 + 4 * (-1) - 5 = -6

3) $f'(x) = e^x * (sin(x) + cos(x))$
f ' (0) = 1 * (1 + 0) = 1

4) $f'(x) = \frac{x^2 - 2x -3}{(x-1)^2}$
f ' (0) = -3

5) $f'(x) =\frac{1}{6 * \sqrt[6]{ x^{5} } }$
f ' (64) = $\frac{1}{6 * 32} = \frac{1}{192}$