Question

Решите вот это вот пожалуйста!!!!!!!!

Answer 1

Ответ:

1.

$f'(x) = (4 {x}^{ - 2} + 3 \cos(x)) '= 4 \times ( - 2) {x}^{ - 3} - 3 \sin(x) = \\ = - \frac{8}{ {x}^{3} } - 3 \sin(x)$

2.

$f'(x) = ( {x}^{ - 4} - 2 \sin(x))' = - 4 {x}^{ - 5} - 2 \cos(x) = \\ = - \frac{4}{ {x}^{5} } - 2\cos(x)$

3.

$f'(x) = 3 \times 3 {x}^{2} - 2 \times 2x + 4 = 9 {x}^{2} - 4x + 4$

4.

$f'(x) = (7( {x - 7)}^{ - \frac{1}{2} } ) '= 7 \times ( - \frac{1}{2} ) {(x - 7)}^{ - \frac{3}{2} } = \\ = - \frac{7}{2 \sqrt{ {(x - 7)}^{3} } }$

5.

$f'(x) = 3 \cos(x + \frac{3\pi}{4} ) \times (x + \frac{3\pi}{4} ) '= 3 \sin(x + \frac{3\pi}{4} )$

6.

$f'(x) = 2 \cos( \frac{\pi}{3} - x ) \times ( \frac{\pi}{3} - x)' = - 2 \cos( \frac{\pi}{3} - x)$

7.

$f'(x) = 0 + ( \sin {}^{ - 2} (4x)) ' = - 2 \sin {}^{ - 3} (4x) \times ( \sin(4x)) ' \times (4x) '= \\ = - \frac{2}{ \sin {}^{3} (4x) } \times \cos(4x) \times 4 = - \frac{8 \cos(4x) }{ \sin {}^{3} (4x) }$