Предмет: Математика, автор: evgenia98745688

Производная функция

Приложения:

Ответы

Автор ответа: Miroslava227

Ответ:

$f'(x) = 2 \times 7 {x}^{6} = 14 {x}^{6}$

$f'(x) = \frac{1}{10} \times ( - 10) {x}^{ - 11} = - \frac{1}{ {x}^{11} } \\$

$f'(x) = ( - 8 {x}^{ - 1} )' = - 8 \times ( - 1) {x}^{ - 2} = \frac{8}{ {x}^{2} } \\$

$f'(x) = (11 {x}^{ \frac{1}{2} } )' = 11 \times \frac{1}{2} {x}^{ - \frac{1}{2} } = \frac{11}{2 \sqrt{x} } \\$

$f'(x) = 0 \\$

$f'(x) = 13 {x}^{12} - 33 {x}^{10} + 8 {x}^{7} + 16x - 1 + 0 \\$

$f'(x) = - \frac{1}{ { \sin}^{2}x } + 3 \sin(x) + \cos(x) \\$

$f'(x) = \frac{1}{ { \cos }^{2} x} + 12 {x}^{11} - \frac{1}{2 \sqrt{x } } \\$

$f'(x) = - \sin(x) - 10 \times ( - 1) {x}^{ - 2} - 2 = \\ = - \sin(x) + \frac{10}{ {x}^{2} } - 2$

$f'(x) = ((5x + 14) \times {x}^{9} ) '= (5 {x}^{10} + 14 {x}^{9} ) '= \\ = 50 {x}^{9} + 126 {x}^{8}$

$f'(x) = (4x - 18)'(5x + 7) + (5x + 7)'(4x - 18) = \\ = 4(5x + 7) + 5(4x - 18) = \\ = 20x + 28 + 20x - 80 = 40x - 52$

$f'(x) = (10x)' \times (x - tgx) + (x - tgx)' \times 10x = \\ = 10(x - tgx) + (1 - \frac{1}{ { \cos }^{2}x } ) \times 10x$

$f'(x) = \frac{(6x - 3)'(2 + 3x) - (2 + 3x)'(6x - 3)}{ {(3x + 2)}^{2} } = \\ = \frac{6(3x + 2) - 3(6x - 3)}{ {(3x + 2)}^{2} } = \\ = \frac{18x + 12 - 18x + 9}{ {(3x + 2)}^{2} } = \frac{21}{ {(3x + 2)}^{2} }$

$f'(x) = (\frac{5 {x}^{5} + {x}^{4} + x}{x} ) '= (5 {x}^{4} + {x}^{3} + 1) ' = \\ = 20 {x}^{3} + 3 {x}^{2}$

$f'(x) = \frac{(0.5 {x}^{6} - 1)'2 \sqrt{x} - (2 \sqrt{x} )'(0.5 {x}^{6} - 1)}{4x} = \\ = \frac{3 {x}^{5} \times 2 \sqrt{x} - 2 \times \frac{1}{2} {x}^{ - \frac{1}{2} } \times (0.5 {x}^{6} - 1) }{ 4x } = \\ = \frac{6 {x}^{5} \sqrt{x} - \frac{0.5 {x}^{6} - 1 }{ \sqrt{x} } }{4x} = \frac{3}{2} {x}^{4} \sqrt{x} - \frac{0.5 {x}^{6} - 1}{4x \sqrt{x} }$