Question

Производные сложных функций.

Answer 1

Пошаговое объяснение:

1.

$y' = \cos(2x - 1) \times (2x - 1)' = 2 \cos(2x - 1)$

2.

$y '= 4 {(5x + 2)}^{3} \times (5x + 2)' = 4 {(5x + 2)}^{3} \times 5 = \\ = 20 {(5x + 2)}^{3}$

3.

$y' = \frac{1}{ \sqrt{1 - \frac{ {x}^{2} }{4} } } \times ( \frac{x}{2} ) '= \frac{1}{2 \sqrt{1 - \frac{ {x}^{2} }{4} } } = \\ = \frac{1}{ \sqrt{4(1 - \frac{ {x}^{2} }{4} )} } = \frac{1}{ \sqrt{4 - {x}^{2} } }$

4.

$y' = - \sin( {x}^{2} ) \times ( {x}^{2}) '= - 2x \sin( {x}^{2} )$

5.

$y' = \frac{1}{1 + {(3 - {x}^{2}) }^{2} } \times (3 - {x}^{2} ) '= \frac{1}{1 + 9 - 6 {x}^{2} + {x}^{4} } \times ( - 2x) = \\ = \frac{ - 2x}{ {x}^{4} - 6 {x}^{2} + 10}$

6.

$y' =( 2 {( \cos(5x)) }^{ - 1} )' = \\ = - 2 {( \cos(5x)) }^{ - 2} \times ( \cos(5x)) ' \times (5x) '= \\ = - \frac{2}{ { \cos}^{2}(5x) } \times ( - \sin(5x)) \times 5 = \\ = \frac{10 \sin(5x) }{ { \cos }^{2} (5x)}$

7.

$y' = {e}^{ {2}^{x} } \times ( {2}^{x}) '= {e}^{ {2}^{x} } \times {2}^{x} \times ln(2)$

8.

$y' = \frac{( \sin( {x}^{2} ))' \times ( {x}^{2})' \times x - x' \sin( {x}^{2} ) }{ {x}^{2} } = \\ = \frac{ \cos( {x}^{2} ) \times 2x \times x - \sin( {x}^{2} ) }{ {x}^{2} } = \\ = \frac{2 {x}^{2} \cos( { x }^{2} ) }{ {x}^{2} } - \frac{ \sin( {x}^{2} ) }{ {x}^{2} } = \\ = 2 \cos( {x}^{2} ) - \frac{ \sin( {x}^{2} ) }{ {x}^{2} }$

9.

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

10.

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