Question

найти производные!!! помогите пожалуйста, оченььььь прошу))(​

Answer 1

Ответ:

1.

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

в точке х = 4

$\frac{1}{2 \sqrt{4} } = \frac{1}{4} = 0.25$

2.

$(7 {x}^{5} - {x}^{3} + 2x - 4)' = 35 {x}^{4} - 3 {x}^{2} + 2$

3.

$( {x}^{4} + 2 {x}^{2} )'( {x}^{3} + 5) + ( {x}^{3} + 5)'( {x}^{4} + 2 {x}^{2}) = \\ = (4 {x}^{3} + 4x)( {x}^{3} + 5) + 3 {x}^{2} ( {x}^{4} + 2 {x}^{2} ) = \\ = 4 {x}^{6} + 20 {x}^{3} + 4 {x}^{4} + 20x + 3 {x}^{6} + 6 {x}^{4} = \\ = 7 {x}^{6} + 10 {x}^{4} + 20 {x}^{3} + 20x$

4.

$\frac{(2 {x}^{3} - 21)'(x - {x}^{2} ) - (x - {x}^{2} )'(2 {x}^{3} - 21) }{ {(x - {x}^{2} )}^{2} } = \\ = \frac{6 {x}^{2}(x - {x}^{2} ) - (1 - 2x)(2 {x}^{3} - 21)}{ {(x - {x}^{2}) }^{2} } = \\ = \frac{6 {x}^{3} - 6 {x}^{4} - 2 {x}^{3} + 21 + 4 {x}^{4} - 42x }{ {(x - {x}^{2} )}^{2} } = \\ = \frac{4 {x}^{3} - 2 {x}^{4} - 42x + 21 }{ {(x - {x}^{2} )}^{2} }$

5.

$( {(5x - 3)}^{ \frac{1}{3} } )' = \frac{1}{3} {(5x - 3)}^{ - \frac{2}{3} } \times (5x - 3)' \\ = \frac{5}{3 \sqrt[3]{ {(5x - 3)}^{2} } }$