Question

Вычислить производную dy/dx функций

Answer 1

Ответ:

а)

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

б)

$y' = \frac{ \cos( ln(2x) ) \times \frac{1}{2x} \times 2 \sqrt[4]{3 - 2x} - \frac{1}{4 \sqrt[4]{ {(3 - 2x)}^{3} } } \times ( - 2) \times \sin( ln(2x) ) }{ \sqrt{3 - 2x} } = \\ = \frac{ \frac{ \cos( ln(2x) ) \sqrt[4]{3 - 2x} }{x} + \frac{ \sin( ln(2x) ) }{2 \sqrt[4]{ {(3 - 2x)}^{3} } } }{ \sqrt{3 - 2x} } = \\ = \frac{ \cos( ln(2x) ) }{x \sqrt[4]{3 - 2x} } + \frac{ \sin( ln(2x) ) }{2 \sqrt[8]{ {(3 - 2x)}^{5} } }$

в)

$y' = ln(2) \times {2}^{ {arccos}^{4}(8x) } \times 4 {arccos}^{3} (8x) \times ( - \frac{1}{ \sqrt{1 - 64 {x}^{2} } } ) \times 8 = \\ = - \frac{32 ln(2) \times {2}^{ {arccos}^{4}(8x) } {arccos}^{3}(8x) }{ \sqrt{1 - 64 {x}^{2} } }$

г)

$y = {(x - 5)}^{ \sin(x) }$

$y' = ( ln(y)) ' \times y$

$( ln(y)) '= ( ln( {(x - 5)}^{ \sin(x) } ) ) ' = ( \sin(x) \times ln(x - 5)) ' = \\ = \cos( x ) ln(x - 5) + \frac{ \sin(x) }{x - 5}$

$y' = {(x - 5)}^{ \sin(x) } \times ( \cos(x) ln(x - 5) + \frac{ \sin(x) }{x - 5} )$

д)

$y'x = \frac{y't}{x't}$

$x't = 6 \cos(t) \times ( - \sin(t) )$

$y't = 6 { \sin }^{2} (t) \times \cos(t)$

$yx = \frac{6 { \sin }^{2}(t) \times \cos(t) }{ - 6 \sin(t) \cos(t) } = \\ = - \sin(t)$