Question

здравствуйте, помогите пожалуйста решить задание​

Answer 1

Ответ:

a

$f'(x) = \frac{1}{4} \times 4 {x}^{3} - 6x + 0 = { x }^{3} - 6$

$f'( - 3) = - 27 +18 = - 9$

б

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

$f(1) = \frac{2 - 2 + 2}{1} = 2$

в

$f'(x) = (2 {x}^{2} + 1)'(4 + x) + (4 + x)'(2 {x}^{2} + 1) = \\ = 4x(4 + x) + 1 \times (2 {x}^{2} + 1) = \\ = 16x + 4 {x}^{2} + 2 {x}^{2} + 1 = 6 {x}^{2} + 16x + 1$

$f'(1) = 6 + 16 + 1 = 23$

г

$f'(x) = (2x)' \sin(x) + ( \sin(x) ) '\times 2x = \\ = 2 \sin(x) + 2x \cos(x)$

$f'( \frac{\pi}{4} ) = 2 \times \frac{ \sqrt{2} }{2} + 2 \times \frac{ \pi }{4} \times \frac{ \sqrt{2} }{2} = \\ = \sqrt{2} (1 + \frac{\pi}{4} )$