Question

Помогите решить производную.Пожалуйста!!!!!!!!!!!!!!!!!!!​

Answer 1

Ответ:

1.

$y' = (x + 7)'( {x}^{2} + 3x) + ( {x}^{2} + 3x)'(x + 7) = \\ = 1 \times ( {x}^{2} + 3x) + (2x + 3)(x + 7) = \\ = {x}^{2} + 3x + 2 {x}^{2} + 14x + 3x + 21 = \\ = 3 {x}^{2} + 20x + 21$

2.

$y' = ( {x}^{2} + 8x) '\times \sqrt{x} + ( {x}^{ \frac{1}{2} } )'( {x}^{2} + 8) = \\ = (2x + 8) \sqrt{x} + \frac{1}{2 \sqrt{x} } ( {x}^{2} + 8x) = \\ = 2x \sqrt{x} + 8 \sqrt{x} + \frac{1}{2} x \sqrt{x} + 4 \sqrt{x} = \\ = 12 \sqrt{x} + 2.5x \sqrt{x}$

3.

$y = \frac{1}{x} ( {x}^{6} - \frac{3}{ {x}^{2} } ) = {x}^{5} - \frac{3}{ {x}^{3} } = \\ = {x}^{5} - 3 {x}^{ - 3}$

$y' = 5 {x}^{4} + 9 {x}^{ - 4} = 5 {x}^{4} + \frac{9}{ {x}^{4} }$

4.

$y '= ( {x}^{ \frac{1}{2} } + 6)'( \frac{1}{x} - 2 {x}^{2} ) + ( {x}^{ - 1} - 2 {x}^{2} )'( \sqrt{x} + 6) = \\ = \frac{1}{2 \sqrt{x} } ( \frac{1}{x} - 2 {x}^{2} ) + ( - \frac{1}{ {x}^{2} } - 4x)( \sqrt{x} + 6) = \\ = \frac{1}{2x \sqrt{x} } - x \sqrt{x} - \frac{1}{ x \sqrt{x} } - \frac{6}{ {x}^{2} } - 4x \sqrt{x} - 24x = \\ = - \frac{6}{ {x}^{2} } - 5x \sqrt{x} + \frac{1}{x \sqrt{x} } - 24x$

5.

$y' = ( {x}^{2} - 3x)'(10 {x}^{ - 3} + \sqrt{x} ) + (10 {x}^{ - 3} + {x}^{ \frac{1}{2} } )'( {x}^{2} - 3x) = \\ = (2x - 3)( \frac{10}{ {x}^{3} } + \sqrt{x} ) + ( - \frac{30}{ {x}^{4} } + \frac{1}{2 \sqrt{x} } )( {x}^{2} - 3x) = \\ = \frac{20}{ {x}^{2} } + 2x \sqrt{x} - \frac{30}{ {x}^{3} } - 3 \sqrt{x} - \frac{30}{ {x}^{2} } + \frac{90}{ {x}^{3} } + \frac{1}{2} x \sqrt{x} - \frac{3}{2} \sqrt{x} = \\ = \frac{60}{ {x}^{3} } - \frac{10}{ {x}^{2} } + 2.5x \sqrt{x} - 4.5 \sqrt{x}$

6.

$y' = ({x}^{6} (3 {x}^{2} + 13x - 8)) '= \\ = (3 {x}^{8} + 13 {x}^{7} - 8 {x}^{6} ) '= \\ = 24 {x}^{7} + 91 {x}^{6} - 48 {x}^{5}$