Question

с математикой.10 класс. 44.2(2,4) и 44.3(2,3). От

Answer 1

44.2

2.

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

4.

$f'(x) = (2 {x}^{ - 1} + 2 \cos(x) + 4 \sin(x) ) '= \\ = - 2 {x}^{ - 2} - 2 \sin(x) + 4 \cos(x) = \\ = - \frac{2}{ {x}^{2} } - 2 \sin(x) + 4 \cos(x) \\ f'( \frac{2\pi}{3} ) = - 2 \times {( \frac{3}{2\pi}) }^{2} - 2 \sin( \frac{2\pi}{3} ) + 4 \cos( \frac{2\pi}{3} ) = \\ = - 2 \times \frac{9}{4 {\pi}^{2} } - 2 \times \frac{ \sqrt{3} }{2} + 4 \times ( - \frac{1}{2} ) = \\ = - \frac{9}{2 {\pi}^{2} } - \sqrt{3} - 2$

44.3

2.

$f'(x) = \cos(x) - \frac{1}{ { \sin }^{2}(x) } \\$

3.

$f'(x) = - \sin(x) - \frac{5}{ { \cos}^{2}(x) } - 3 {x}^{ - 4} = \\ = - \sin(x) - \frac{5}{ { \cos}^{2} (x)} - \frac{3}{ {x}^{4} }$