Question

СРОЧНО НАДО, ДАЮ 50 БАЛЛОВ СДЕЛАТЬ ВСЕ 4 ПРИМЕРА С РЕШЕНИЕМ

Answer 1

Ответ:

1.

$f(x) = 4 \cos( \frac{5x}{2} ) - 7x + 3$

$f'(x) = - 4 \sin( \frac{5x}{2} ) \times ( \frac{5x}{2} ) '- 7 + 0 = \\ = - 4 \sin( \frac{5x}{2} ) \times \frac{5}{2} - 7 = - 10 \sin( \frac{5x}{2} ) - 7$

$f'( - \frac{\pi}{3} ) = - 10 \sin( \frac{5}{2} \times ( - \frac{\pi}{3} ) ) - 7 = \\ = 10 \sin( \frac{5\pi}{6} ) - 7 = \frac{10}{2} - 7 = 5 - 7 = - 2$

2.

$f(x) = {(2x + 1)}^{3}$

$f'(x) = 3 {(2x + 1)}^{2} \times (2x + 1)' = \\ = 3 {(2x + 1)}^{2} \times 2 = 6 {(2x + 1)}^{2}$

$f'(3) = 6(2 \times 3 + 1) {}^{2} = 6 \times 49 = 294$

3.

$( \sin(3 {x}^{2} ) ) '= \cos(3 {x}^{2} ) \times (3 {x}^{2} )' = 6x \cos(3 {x}^{2} )$

4.

$( \sqrt{6 {x}^{4} + 1 } ) '= ({(6x {}^{4} + 1)}^{ \frac{1}{2} } ) '= \\ = \frac{1}{2} {(6 {x}^{4} + 1) }^{ - \frac{1}{2} } \times (6 {x}^{4} + 1) '= \\ = \frac{24 {x}^{3} }{2 \sqrt{6 {x}^{4} + 1} } = \frac{12 {x}^{3} }{ \sqrt{6 {x}^{4} + 1} }$