Question

Выполните дефференцирование функций , пожалуйста

Answer 1

Ответ:

1

$( \frac{14}{19} {x}^{19} ) '= \frac{14}{19} \times 19 {x}^{18} = 14 {x}^{18}$

2

$( {9}^{x} + {x}^{9} ) '= ln(9) \times {9}^{x} + 9 {x}^{8}$

3

$( {x}^{ - 1} + {x}^{ - 10} ) '= - {x}^{ - 2} - 10 {x}^{ - 11} = - \frac{1}{ {x}^{2} } - \frac{10}{ {x}^{11} }$

4

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

5

$( {x}^{7} )' \times tgx + (tgx)' \times {x}^{7} = 7 {x}^{6} tgx + \frac{1}{ \cos {}^{2} (x) } \times {x}^{7} = \\ = {x}^{6} (7tgx + \frac{x}{ \cos {}^{2} (x) } )$

6

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

7

$( {e}^{4} + \sqrt{3} - \pi)' = 0$

8

$( ln(x) \times \cos(x)) ' = (ln(x)) '\times \cos(x) + (\cos(x))' \times ln(x) = \\ = \frac{ \cos(x) }{x} - \sin(x) ln(x)$

9

$( \frac{ \sin(x) }{9} ) '= \frac{1}{9} (\sin(x)) ' = \frac{1}{9} \cos(x)$

10

$( \frac{9}{ \sin(x) })' = 9( \sin(x)) {}^{ - 1})' = \\ = - 9 { \sin }^{ - 2} (x) \times \cos(x) = - \frac{9 \cos(x) }{ \sin {}^{2} (x) }$

11

$( \frac{x + 4}{4} ) '= \frac{1}{4} (x + 4)' = \frac{1}{4}$

12

$\frac{(ctgx )' \times ln(x) + (ln(x)) ' \times ctgx}{ {ln}^{2} (x)} = \\ = \frac{ - \frac{1}{ \sin {}^{2} (x) } ln(x) + \frac{1}{x} \times ctgx}{ln {}^{2} x} = \\ = - \frac{1}{ {ln}^{2}(x) } = ( \frac{ ln(x) }{ \sin {}^{2} (x) } - \frac{ctgx}{x} )$