Question

Помогите найти производную, дам 35 баллов, пожалуйста

Answer 1

Ответ:

1.

$f'(x) = 4 {( \cos(x) \times {5}^{x} + {x}^{3} )}^{3} \times ( - \sin(x) \times {5}^{x} + ln(5) \times {5}^{x} \times \cos(x) + 3 {x}^{2} ) = \\ = 4 {( {5}^{x} \cos(x) + {x}^{3} ) }^{2} \times ({5}^{x} ln(5) \cos(x) - {5}^{x} \sin(x) + 3 {x}^{2} )$

2.

$f'(x) = \frac{ \frac{1}{1 + {x}^{2} } \sqrt{4 - {x}^{2} } - \frac{1}{2 \sqrt{4 - {x}^{2} } } \times ( - 2x) \times arctgx}{4 - {x}^{2} } = \\ = \frac{1}{4 - {x}^{2} } \times (\frac{ \sqrt{4 - {x}^{2} } }{1 + {x}^{2} } + \frac{xarctgx}{ \sqrt{4 - {x}^{2} } } ) = \\ = \frac{1}{(1 + {x}^{2} ) \sqrt{4 - {x}^{2} } } + \frac{xarctgx}{ \sqrt{ {(4 - {x}^{2} )}^{3} } }$

3.

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