Question

Найдите производные следущих функций

Answer 1

Ответ:

а)

$y = 3 \times {e}^{ \frac{x}{9} } + arctg( {x}^{2} - 1)$

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

б)

$y = ln(3x + 2) \times \sqrt{ \cos(x) }$

$y '= U'V + V'U = \\ = \frac{1}{3x + 2} \times 3 \times \sqrt{ \cos(x) } + \frac{1}{2} {( \cos(x)) }^{ - \frac{1}{2} } \times ( - \sin(x)) \times ln(3x + 2) = \\ = \frac{3 \sqrt{ \cos(x) } }{3x + 2} - \frac{ \sin(x) ln(3x + 2) }{2 \sqrt{ \cos(x) } }$

в)

$y = \frac{ \sin(5x) }{ {ctg}^{2} (x)}$

$y' = \frac{U'V - V'U}{ {v}^{2} } = \\ = \frac{5 \cos(5x) \times {ctg}^{2} (x) - 2ctg(x) \times ( - \frac{1}{ { \sin }^{2}(x) } ) \times \sin(5x) }{ {ctg}^{4}(x) } = \\ = \frac{5 \cos(5x) } { {ctg}^{2} (x) } + \frac{2 \sin(5x) }{ { \sin}^{2} (x) {ctg}^{3}(x) } = \\ = \frac{5 \cos(5x) }{ {ctg}^{2}(x) } + \frac{2 \sin(5x) }{ { \cos}^{2}(x) {ctg}^{2}(x) }$

г)

$y = {2}^{arccos( {x}^{3}) }$

$y' = ln(2) \times {2}^{arccos( {x}^{3} )} \times ( - \frac{1}{ \sqrt{1 - {x}^{6} } } ) \times 3 {x}^{2} = \\ = - \frac{3 {x}^{2} ln(2) \times {2}^{arccos( {x}^{3)} } }{ \sqrt{1 - {x}^{6} } }$