Question

Найти похидну функции ,вместо n подставить число - 12
Буду рада помощи!!!
Дам 50 б!!!!!
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Answer 1

Ответ:

7.

$y = \sqrt{ {e}^{12x} } = {e}^{6x}$

$y '= 6 {e}^{6x}$

8.

$y = {31}^{12 {x}^{2} + 3x}$

$y' = ln(31) \times {31}^{12 {x}^{2} + 3x } \times (12 {x}^{2} + 3x) '= \\ = ln(31) \times {31}^{12 {x}^{2} + 3x } \times (24x + 3)$

9.

$f(x) = \frac{x}{ \sqrt{144 + {x}^{2} } }$

$f'(x) = \frac{x' \sqrt{ {x}^{2} + 144 } -( ( {x}^{2} + 144) {}^{ \frac{1}{2} } ) '\times ( {x}^{2} + 144)' \times x }{ {x}^{2} + 144} = \\ = \frac{ \sqrt{ {x}^{2} + 144 } - \frac{1}{2 \sqrt{ {x}^{2} + 144} } \times 2x \times x }{ {x}^{2} + 144 } = \\ = \frac{1}{ \sqrt{ {x}^{2} + 144} } - \frac{ {x}^{2} }{ \sqrt{ {( {x}^{2} + 144)}^{3} } }$

10.

$y = {(36 + 8 \sqrt{x}) }^{5}$

$y '= 5 {(36 + 8 \sqrt{x}) }^{4} \times (36 + 8 \sqrt{x} )' = \\ = 5 {(36 + 8 \sqrt{x} )}^{2} + \frac{8}{2 \sqrt{x} } = \frac{20}{ \sqrt{x} } (36 + 8 \sqrt{x} ) {}^{2}$

11.

$y = \sqrt{x} \times x \times \sqrt[15]{x} = {x}^{ \frac{1}{2} + 1 + \frac{1}{15} } = {x}^{ \frac{47}{30} }$

$y '= \frac{47}{30} {x}^{ \frac{17}{30} } = \frac{47}{30} \sqrt[30]{ {x}^{17} }$

12.

$y = \frac{24}{3} {x}^{2} \sqrt[12]{x} = 8 {x}^{2 + \frac{1}{12} } = 8 {x}^{ \frac{25}{12} }$

$y '= 8 \times \frac{25}{12} {x}^{ \frac{13}{12} } = \frac{2 \times 25}{3} x \sqrt[12]{x} = \frac{50}{3} x \sqrt[12]{x}$