ПОМОГИТЕ ПОЖАЛУЙСТА С МАТЕМАТИКОЙ 11 КЛАСС Задание 8 А. Created by step31121981. matematika-ru

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ПОМОГИТЕ ПОЖАЛУЙСТА С МАТЕМАТИКОЙ 11 КЛАСС Задание 8 А...

step31121981 @step31121981

August 2022 1 17 Report

ПОМОГИТЕ ПОЖАЛУЙСТА С МАТЕМАТИКОЙ 11 КЛАСС
Задание 8 А

Answers & Comments

Miroslava227

Ответ:

$y '= {e}^{x}$

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

$y' = {e}^{ - x} \times ( - x)' = - e { }^{ - x}$

$y' = {e}^{ {x}^{4} } \times ( {x}^{4} )' = {e}^{ {x}^{4} } \times 4 {x}^{3}$

$y' = {e}^{ {x}^{2} - 1 } \times ( {x}^{2} - 1)' = e {}^{ {x}^{2} - 1} \times 2x \\$

$y' = {e}^{2 - 3x} \times (2 - 3x) '= - 3 e {}^{2 - 3x}$

$y' = {5}^{x} ln(5)$

$y '= ln(10) \times {10}^{1 - x} \times (1 - x) '= \\ = - ln( 10) \times {10}^{1 - x}$

$y '= ln(3) \times {3}^{2x + 1} \times (2x + 1)' = \\ = 2 ln(3) \times {3}^{2x + 1}$

$y' = 3 ln(7) \times {7}^{x}$

$y '= ln(2.1) \times {(2.1)}^{3 + 4x} \times (3 + 4x) '= \\ = 4 ln(2.1) \times {(2.1)}^{3 + 4x}$

$y' = ln(2) \times {2}^{5 {x}^{2} + x - 1 } \times (5 {x}^{2} + x - 1) '= \\ = ln(2) \times {2}^{5 {x}^{2} + x - 1} \times (10x + 1)$

$y '= \frac{1}{2x} \times (2x) '= \frac{1}{2x} \times 2 = \frac{1}{x} \\$

$y' = \frac{1}{2x - 3} \times (2x - 3) '= \frac{2}{2x - 3} \\$

$y' = ( {x}^{3} )' \times ln(x) + ( ln(x)) ' \times {x}^{3} = \\ = 3 {x}^{2} ln(x) + \frac{1}{x} \times {x}^{3} = \\ = 3 {x}^{2} ln(x) + {x}^{2} = {x}^{2} (3 ln(x) + 1)$

$y' = \frac{1}{x \times ln(7) } \\$

$y '= \frac{1}{ ln(5) \times (7x)} \times (7x) '= \frac{7}{7x \times ln(5) } = \frac{1}{x ln(5) } \\$

$y' = \frac{1}{ ln(0.3) \times x} = \frac{1}{ ln( {( \frac{10}{3} )}^{ - 1} ) \times x} = - \frac{1}{x ln( \frac{10}{3} ) } \\$

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