Question

Помогите срочно надо

Answer 1

Ответ:

1.

$f'(x) = 7 \times 7 {x}^{6} + 13 \times 3 {x}^{2} + 0 = \\ = 49 {x}^{6} + 39 {x}^{2}$

2.

$f'(x) = \frac{1}{5} + (4 {x}^{ - 3} )' + ( {x}^{ \frac{1}{2} } )' = \\ = \frac{1}{5} + 4 \times ( - 3) {x}^{ - 4} + \frac{1}{2} {x}^{ - \frac{1}{2} } = \\ = 0.2 - \frac{12}{ {x}^{4} } + \frac{1}{2 \sqrt{x} }$

3.

$f'(x) = (x(x + 2)) '= ( {x}^{2} + 2x)' = \\ = 2x + 2$

4.

$f'(x) = (x + 5)'( {x}^{2} + 7) + ( {x}^{2} + 7)'(x + 5) = \\ = {x}^{2} + 7 + 2x(x + 5) = \\ = {x}^{2} + 7 + 2 {x}^{2} + 10x = \\ = 3 {x}^{2} + 10x + 7$

5.

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

6.

$f'(x) = \frac{(x + 1)'(x - 2) - (x - 2)'(x + 1)}{ {(x - 2)}^{2} } = \\ = \frac{x - 2 - (x + 1)}{ {(x - 2)}^{2} } = \frac{x - 2 - x - 1}{ {(x - 2)}^{2} } = \\ = - \frac{3}{ {(x - 2)}^{2} }$