срочно нужна помощь. Created by katrin04sd. matematika-ru

срочно нужна помощь...

katrin04sd @katrin04sd

July 2022 1 2 Report

срочно нужна помощь

Answers & Comments

QDominus

1. a)

$(x {}^{21} )' = 21 {x}^{20}$

б)

$(5 \cos(x) )' = 5( \cos(x) )' = - 5 \sin(x)$

в)

$( \sqrt{x} + \ctg(x))' = ( \sqrt{x} )' + ( \ctg(x))' = \frac{1}{2 \sqrt{x} } + \frac{1}{ \sin^{2} (x) }$

г)

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

1. а)

$f(x) = \frac{2}{3} {x}^{3} + 6 {x}^{2} - 4x + 8.4 \\ f'(x) = 2 {x}^{2} + 12x - 4$

б)

$f(x) = (5x - 4) \sqrt{x} = 5x \sqrt{x} - 4 \sqrt{x} \\ f(x) = 5 {x}^{ \frac{3}{2} } - 4 \sqrt{x} \\ f'(x) = \frac{15}{2} \sqrt{x} - \frac{2 \sqrt{x} }{x }$

в)

$f(x) = \frac{6x + 7}{2 - x} \\ f'(x) = \frac{(6x + 7)'(2 - x) - (2 - x)'(6x + 7)}{ {(2 - x)}^{2} } = \\ = \frac{6(2 - x) + 6x + 7}{ {(2 - x)}^{2} } = \frac{12 - 6 + 6x + 7}{ {(2 - x)}^{2} } = \frac{19}{ {(2 - x)}^{2} }$

0 votes Thanks 0

184b804e9b88dff8818b31389566858c

Answer

katrin04sd July 2022 | 0 Ответы

5 i 6 srochno pozhalujstaa3177df1fbeb92a9522dfc8fd9f2fe60 88061

Answer

Answers & Comments

184b804e9b88dff8818b31389566858c

5 i 6 srochno pozhalujstaa3177df1fbeb92a9522dfc8fd9f2fe60 88061

рекомендуемые вопросы

Helpful Links

Helpful Social