Question

Сделайте пожалуйста СРОЧНО НУЖНО

Answer 1

1.

а)

$y' = 3 \times 7 {x}^{6} - 6 \times 6 {x}^{5} - 4 \times 3 {x}^{2} + 5 \times 2x + 0 = \\ = 21 {x}^{6} - 36 {x}^{5} - 12 {x}^{2} + 10x$

б)

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

в)

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

г)

$y '= (x)' \times ln(x) + ( ln(x) ) '\times x = \\ = ln(x) + \frac{1}{x} \times x = ln(x) + 1$

д)

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

е)

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

2.

$y '= - \sin(6x) \times (6x) '+ \cos(6x) \times (6x)' = \\ = - 6 \sin(6x) + 6 \cos(6x)$

$y'( \frac{\pi}{8} ) = - 6 \sin( \frac{3\pi}{4} ) + 6 \cos( \frac{3\pi}{4} ) = \\ = - 6 \times \frac{ \sqrt{2} }{2} + 6 \times \frac{ \sqrt{2} }{2} = 0$

3.

$v(t) = s'(t) = 3 {t}^{2} + 6 t \\ v(1) = 3 + 6 = 9$

Ответ: 9 м/с