Question

Тема: Производная сложной функции.
Практическая работа. 1 Вариант нужен.

Answer 1

Ответ:

$1)f'(x)= \frac{1}{ \sin(x) } \times \cos(x) = ctg(x)$

$2)f'(x) = 4 {e}^{4x + 3}$

$3)f'(x) = 4 {( {x}^{2} + 2x - 1) }^{3} \times (2x + 2) \\ f'( - 1) = x {(1 - 2 - 1)}^{3} \times ( - 2 + 2) = 0$

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

$5)f'(x) = ( {( {x}^{2} + 1)}^{ \frac{3}{2} } )' = \frac{3}{2} {( {x}^{2} + 1) }^{ \frac{1}{2} } \times 2x = 3x \sqrt{ {x}^{2} + 1 } \\ f'( \sqrt{3} ) = 3 \sqrt{3} \sqrt{3 + 1} = 3 \sqrt{3} \times 2 = 6 \sqrt{3}$

$6)f'(x) = \frac{9 \sqrt{ {x}^{2} + 1} - \frac{1}{2} { ({x}^{2} + 1)}^{ - \frac{1}{2} } \times 2x \times 9x }{ {x}^{2} + 1 } = \frac{9 \sqrt{ {x}^{2} + 1 } - \frac{9 {x}^{2} }{ \sqrt{ {x}^{2} + 1} } }{ {x}^{2} + 1 } = \frac{9}{ \sqrt{ {x}^{2} + 1 } } - \frac{9 {x}^{2} }{ {( {x}^{2} + 1) }^{ \frac{3}{2} } }$

$f'(2 \sqrt{2} ) = \frac{9}{ \sqrt{8 + 1} } - \frac{9 \times 8}{ {(8 + 1)}^{ \frac{ 3}{2} } } = 3 - \frac{9 \times 8}{27} = 3 - \frac{8}{3} = \frac{1}{3}$