Question

Вычислить производную.⁰

Answer 1

Ответ:

1.

$y '= 7 \times 6 {x}^{5} - 2 \times 3 {x}^{2} + 0 = 42 {x}^{5} - 6 {x}^{2}$

2.

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

3.

$y' = \frac{(x + 1) '{x}^{6} - ( {x}^{6} )'(x + 1) }{ {( {x}^{6} )}^{2} } = \\ = \frac{ {x}^{6} - 6 {x}^{5} (x + 1)}{ {x}^{12} } = \frac{ {x}^{5}(x - 6(x + 1)) }{ {x}^{12} } = \\ = \frac{x - 6x - 6}{ {x}^{7} } = - \frac{5x + 6}{ {x}^{7} }$

4.

$y' = 5 - 2x + 0 = 5 - 2x$

5.

$y' = \frac{( {x}^{2} + 6x)'(x - 2) - (x - 2)'( {x}^{2} + 6x) }{ {(x - 2)}^{2} } = \\ = \frac{(2x + 6)(x - 2) - 1 \times ( {x}^{2} + 6x) }{ {(x - 2)}^{2} } = \\ = \frac{2 {x}^{2} - 4x + 6x - 12 - {x}^{2} - 6 {x}^{} }{ {(x - 2)}^{2} } = \\ = \frac{ {x}^{2} - 4x - 12 }{ {(x - 2)}^{2} }$

6.

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