С МАТЕМАТИКОЙ 11 КЛАСС
Задание 8 А


С МАТЕМАТИКОЙ 11 КЛАСС Задание 8 А

saneok67p086je saneok67p086je    2   06.04.2021 09:48    0

Ответы
natchiy03 natchiy03  06.05.2021 09:52

1

y '= {e}^{x}

2

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

3

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

4

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

5

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

6

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

7

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

8

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

9

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

10

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

11

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

12

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)

13

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

14

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

15

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)

16

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

17

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

18

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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