Question

знайти похідні функцій.
зробіть хоч 8
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,
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за ахінею - бан )0)​

Answer 1

Ответ:

2

$y' = 0 -3 \times ( - 9) {x}^{ - 10} + 5 \times 6 {x}^{5} - 8x = \\ = \frac{27}{ {x}^{10} } + 30 {x}^{5} - 8x$

3

$y' = 2 \times ( - 5) {x}^{ - 6} + \frac{2}{8} \times ( - 4) {x}^{ - 5} - 0 = \\ = - \frac{10}{ {x}^{6} } - \frac{2}{ {x}^{5} }$

4

$y '= \frac{1}{9} \times ( - 9) {x}^{ - 10} - \sqrt{2} \sin(x) + 0 = \\ = - \frac{1}{ {x}^{10} } - \sqrt{2} \sin(x)$

5

$y' = - 4 \cos(x) - \frac{3}{ \sin {}^{2} (x) } - ( - \frac{1}{3} ) {x}^{ - \frac{4}{3} } = \\ = - 4 \cos(x) - \frac{3}{ \sin {}^{2} (x) } + \frac{1}{3x \sqrt[3]{x} }$

6

$y' = \frac{1}{4} \times ( - 8) {x}^{ - 9} + 6 \times \frac{1}{2} {x}^{ - \frac{1}{2} } - 7 \times \frac{2}{3} {x}^{ - \frac{1}{3} } = \\ = - \frac{2}{ {x}^{9} } + \frac{3}{ \sqrt{x} } - \frac{14}{3 \sqrt[3]{x} }$

7

$y' = (3 {x}^{4} - 6 {x}^{ - 2} + 2 {x}^{ \frac{1}{3} } ) = \\ = 12 {x}^{3} + 12 {x}^{ - 3} + \frac{2}{3} {x}^{ - \frac{2}{3} } = \\ = 12 {x}^{3} + \frac{12}{ {x}^{3} } + \frac{2}{3 \sqrt[3]{ {x}^{2} } }$

8

$y '= - 9 {x}^{ - 4} - 2 \times ( - \frac{3}{5} ) {x}^{ - \frac{8}{5} } = \\ = - \frac{9}{ {x}^{4} } + \frac{6}{5x \sqrt[5]{ {x}^{3} } }$

9

$y '= (2x - 3)'(1 - {x}^{3} ) + (1 - {x}^{3} )'(2x - 3) = \\ = 2(1 - {x}^{3}) - ( - 3 {x}^{2} )(2x - 3) = \\ = 2 - 2 {x}^{3} + 6 {x}^{3} - 9 {x}^{2} = \\ = 4 {x}^{3} - 9 {x}^{2} + 2$

10

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

11

$y' = (2 \sqrt{x} (3 {x}^{2} + 2))' = (6 {x}^{2} \sqrt{x} + 4 \sqrt{x} )' = \\ = (6 {x}^{ \frac{5}{2} } + 4 {x}^{ \frac{1}{2} } ) '= 15 {x}^{ \frac{3}{2} } + 2 {x}^{ - \frac{1}{2} } = 15x \sqrt{x} + \frac{2}{ \sqrt{x} }$