Question

найдите производное заданных функций​

Answer 1

Ответ:

1.

$f'(x) = \cos(x) + 63 {x}^{6} - 1$

2.

$f(x) = 4 \sqrt{x} {e}^{x} = 4 {x}^{ \frac{1}{2} } \times {e}^{x}$

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

3.

$f'(x) = \frac{(6 - {x}^{4}) '\times {4}^{x} - ( {4}^{x} )'(6 - {x}^{4}) }{ {4}^{2x} } = \\ = \frac{ - 4 {x}^{3} \times {4}^{x} - ln(4) \times {4}^{x} (6 - {x}^{4}) }{ {4}^{2x} } = \\ = - \frac{4 {x}^{3} + ln(4) \times (6 - {x}^{4} ) }{ {4}^{x} }$

4.

$f'(x) = ( {x}^{3} - 4x + 5)'tg( {x}^{2} + 1) + (tg( {x}^{2} + 1) )'\times ( {x}^{2} + 1) '\times ( {x}^{3} - 4x + 5) = \\ =( 3 {x}^{2} - 4)tg( {x}^{2} + 1) + \frac{1}{ \cos {}^{2} ( {x}^{2} + 1) } \times 2x \times ( {x}^{3} - 4x + 5) = \\ = (3 {x}^{2} - 4)tg( {x}^{2} + 1) + \frac{2x( {x}^{3} - 4x + 5) }{ \cos {}^{2} ( {x}^{2} + 1) }$

5.

$f'(x) = - \sin( ln(6x) ) \times ( ln(6x)) ' \times (6x) '= \\ = - \sin( ln(6x) ) \times \frac{1}{6x} \times 6 = - \frac{ \sin( ln(6x) ) }{x}$