Задание на фото, хелп). Created by artem12431. matematika-ru

Ответ:1.2.3.4.5.6.7.8.9.10.11.12.13.14.15.16.17.18.19.20.непонятна функция в знаменателе

Задание на фото, хелп)...

artem12431 @artem12431

September 2021 1 15 Report

Задание на фото, хелп)

Answers & Comments

Miroslava227

Ответ:

$y' = 18 {x}^{2} + 3$

$y' = 42 {x}^{6} - 8x + \cos(x)$

$y' = \cos(x) + {e}^{x}$

$y' = \frac{1}{ \cos {}^{2} (x) } + 14 {x}^{6} - e \\$

$y' = ( {x}^{ - 1} + 2 {x}^{ -5 } + 3 {x}^{2} ) '= - {x}^{ - 2} - 10 {x}^{ - 6} + 6x = \\ = - \frac{1}{ {x}^{2} } - \frac{10}{ {x}^{6} } + 6x$

$y' = \frac{1}{x} + (3 {x}^{ - 4} ) '+ 0 = \frac{1}{x} - 12 {x}^{ - 5} = \frac{1}{x} - \frac{12}{ {x}^{5} } \\$

$y' = ( {x}^{7} )' \times \sin(x) + ( \sin(x) ) '\times {x}^{7} + ( {x}^{ \frac{1}{2} } ) = \\ = 7 {x}^{6} \sin(x) + \cos(x) \times {x}^{7} + \frac{1}{2} {x}^{ - \frac{1}{2} } = \\ = {x}^{6} (7 \sin(x) + x\cos(x) ) + \frac{1}{2 \sqrt{x} }$

$y' = ( {x}^{8} ) '\times {e}^{x} + ( {e}^{x} ) '\times {x}^{ 8} = \\ = 8 {x}^{7} {e}^{x} + {e}^{x} \times {x}^{8} = {x}^{7} {e}^{x} (8 + x)$

$y '= 3 \sin(x) + 3x \cos(x)$

10.

$y '= 6xtgx + \frac{3 {x}^{2} }{ \cos {}^{2} (x) } \\$

11.

$y' = 16 {x}^{3} \sin(x) + {x}^{4} \cos(x) = \\ = {x}^{3} (16 \sin(x) + x\cos(x))$

12.

$y' = 6 \cos(x) \times tgx + 6 \sin(x) \times \frac{1}{ \cos {}^{2} (x) } = \\ = 6 \sin(x) + \frac{6tgx}{ \cos(x) }$

13.

$y' = (2x + 3)'(3x - 1) + (3x - 1)'(2x + 3) = \\ = 2(3x - 1) + 3(2x + 3) = 6x - 2 +6x + 9 = \\ = 12x + 7$

14.

$y' = (4 {x}^{2} (2x - 1)) '= (8 {x}^{3} - 4 {x}^{2} )' = \\ = 24 {x}^{2} - 8x$

15.

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

16.

$y' = \frac{( {x}^{6})'( {x}^{2} - 2 x- 2) - ( {x}^{2} - 2x - 2) '\times {x}^{6} }{ {( {x}^{2} - 2x - 2)}^{2} } = \\ = \frac{6 {x}^{5}( {x}^{2} - 2x - 2) - (2x - 2) {x}^{6} }{ {( {x}^{2} - 2x - 2)}^{2} } = \\ = \frac{6 {x}^{7} - 12 {x}^{6} - 12 {x}^{5} - 2 {x}^{7} + 2 {x}^{6} }{ {( {x}^{2} - 2x - 2)}^{2} } = \\ = \frac{4 {x}^{7} - 10 {x}^{6} - 12 {x}^{5} }{ {( {x}^{2} - 2x - 2) }^{2} }$

17.

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