Предмет: Математика, автор: renogfavi42

Вычислить производные следующих функций:

Приложения:

Ответы

Автор ответа: Miroslava227

1.

а)

$y' = 6 {x}^{5}$

б)

$y '= 0$

в)

$y '= 3 {x}^{2} - 3$

г)

$y' = 4 \times 9 {x}^{8} - ( - \frac{5}{ \sin {}^{2} (x) } ) + \frac{3}{ \cos {}^{2} (x) } - 16 \times \frac{1}{2} {x}^{ - \frac{1}{2} } + 0 = \\ = 36 {x}^{8} + \frac{5}{ \sin {}^{2} (x) } + \frac{3}{ \cos {}^{2} (x) } - \frac{8}{ \sqrt{x} }$

д)

$y' = \frac{(5 {x}^{2} - 7)'( {x}^{5} - x) - ( {x}^{5} - x)'(5 {x}^{2} - 7) }{ {( {x}^{5} - x)}^{2} } = \\ = \frac{10x( {x}^{5} - x) - (5 {x}^{4} - 1)(5 {x}^{2} - 7) }{ {( {x}^{5} - x)}^{2} } = \\ = \frac{10 {x}^{6} - 10 {x}^{2} - 25 {x}^{6} + 35 {x}^{4} + 5 {x}^{2} - 7}{ {( {x}^{5} - x)}^{2} } = \\ = \frac{ - 15 {x}^{6} - 5 {x}^{2} + 35 {x}^{4} - 7 }{ {x}^{2} ( {x}^{4} - 1) {}^{2} }$

е)

$y '= (5 {x}^{2} - 6)'(3x - 12) + (3x - 12)'(5 {x}^{2} - 6) = \\ = 10x(3 {x}^{} - 12) + 3(5 {x}^{2} - 6) = 30 {x}^{2} - 120x + 15 {x}^{2} - 18 = \\ = 45 {x}^{2} - 120x - 18$

ж)

$y '= 5 {(5x - 6)}^{4} \times (5x - 6) '= 5 {(5x - 6)}^{4} \times 5 = \\ = 25 {(5x - 6)}^{4}$

з)

$y '= \cos(10 - 4x) \times (10 - 4x)' = - 4 \cos(10 - 4x) \\$

2.

$v = y'(x_0)$

$y' = 2x - 2 \\ v = y'( - 0.1) = 2 \times ( - 0.1) - 2 = \\ = - 0.2 - 2 = - 2.2$

3.

$k = y'(x_0)$

$y' = \frac{1}{10} \times 10 {x}^{9} - \frac{1 }{7} \times 7 {x}^{6} + \sqrt{3} - 0= \\ = {x}^{9} - {x}^{6} + \sqrt{3} \\ k = y'(1) = 1 - 1 + \sqrt{3} = \sqrt{3}$