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

1.
1) Фиксируем x. y(x) = x? — 6x + 4
2) Даём приращение аргументу. y(x +

?x) = (x + ?x)? — 6(x + ?x) + 4 = x? + 2x?x + ?x? — 6x — 6?x + 4
3)

Находим приращение функции. ?y = y(x +

?x) — y(x) = x? + 2x?x + ?x? — 6x — 6?x + 4 — x? + 6x — 4 = 2x?x + ?x? — 6?x
4) Составляем отношение ?y/?x.

?y/?x = (2x?x + ?x? — 6?x)/?x = 2x + ?x — 6
5) Находим предел:

$\lim_{\Delta x \to 0} (2x + \Delta x - 6) = 2x-6$

2.
f(x) = 2x? + x? — 3x + 3
f'(x) = 6x? + 2x — 3
f'(-2) = 6 * (-2)? + 2 * (-2) — 3 = 24 — 4 — 3 = 17

3.
f(x) = 3? * log?x
f'(x) = 3? * ln3 * log?x + 3? * 1/xln3 = 3?(ln3*log?x + 1/xln3) = 3?(lnx + 1/xln3)
f'(1) = 3(ln1 + 1/ln3) = 3(0 + 1/ln3) = 3/ln3

4.
f(x) = (x? — 2)/(x? — 4)
f'(x) = (2x(x? — 4) — 2x(x? — 2)) / (x? — 4)? = (2x? — 8x — 2x? + 4x) / (x? — 4)? = — 4x/(x?-4)?
f'(3) = — 12/(9-4)? = -12/25 = -0,48

5.
f(x) = x^(1/10)
f'(x) = 1/10 * x^(1/10-1) = 1/10 * x^(-9/10)

$f'(5) = \frac{1}{10} *5^ {-\frac{9}{10} } = \frac{1}{10} * \frac{1}{5} ^ {\frac{9}{10} } = \frac{1}{10} \sqrt[10]{ (\frac{1}{5}) ^ 9}$