Home
О нас
Products
Services
Регистрация
Войти
Поиск
незнайка2901
@незнайка2901
November 2021
1
10
Report
СРОЧНО ПОМОГИТЕ!!!!!ПОЖАЛУЙСТА
Please enter comments
Please enter your name.
Please enter the correct email address.
Agree to
terms of service
You must agree before submitting.
Send
Answers & Comments
yarovoe
A).y'=(2x·(x²+1)-(x²-1)·2x)/(x²+1)²=(2x³+2x-2x³+2x)/(x²+1)²=4x/(x²+1)²
б) y'=(-x²-(3-x)·2x)/x⁴=(-x²-6x+2x²)/x⁴=(x²-6x)/x⁴=x(x-6)/x⁴=(x-6)/x³
в) y'=(2x·3x-(1+x²)·3)/9x²=(6x²-3-3x²)/9x²=(3x²-3)/9x²=3(x²-1)/9x²=(x²-1)/3x²
г)y'=(2x-1)x²-(x²-x+2)·2x)/x⁴=(2x³-x²-2x³+2x²-4x))/x⁴=(x²-4x)/x⁴=x(x-4)/x⁴=(x-4)/x³
д)/ y'=(-5x⁴(1+x⁵)-(1-x⁵)·5x⁴)/(1+x⁵)²=(-5x⁴-5x⁹-5x⁴+5x⁹)/(1+x⁵)²=-10x⁴/(1+x⁵)²
е). y'=(2x·x²-(x²-1)·2x)/x⁴=(2x³-2x³+2x²)/x⁴=2x²/x⁴=2/x²
ж). y'=(3x²·2x-(2+x³)·2)/4x²=(6x³-4-2x³)/4x²=(4x³-4)/4x²=4(x³-1)/4x²=(x³-1)/x²
з). y'=(-1/(1-x²)²)·(-2x)=2x/(1-x²)²
0 votes
Thanks 1
рекомендуемые вопросы
rarrrrrrrr
August 2022 | 0 Ответы
o chem dolzhny pozabotitsya v pervuyu ochered vzroslye pri organizacionnom vyvoze n
danilarsentev
August 2022 | 0 Ответы
est dva stanka na kotoryh vypuskayut odinakovye zapchasti odin proizvodit a zapcha
myachina8
August 2022 | 0 Ответы
najti po grafiku otnoshenie v3v1 v otvetah napisano 9 no nuzhno reshenie
ydpmn7cn6w
August 2022 | 0 Ответы
Choose the correct preposition: 1.I am fond (out,of,from) literature. 2.where ar...
millermilena658
August 2022 | 0 Ответы
opredelite kak sozdavalas i kto sozdaval arabskoe gosudarstvo v kracii
MrZooM222
August 2022 | 0 Ответы
ch ajtmanov v rasskaze krasnoe yabloko ispolzuet metod rasskaz v rasskaze opi
timobila47
August 2022 | 0 Ответы
kakovo bylo naznachenie kazhdoj iz chastej vizantijskogo hrama pomogite pozhalujsta
ivanyyaremkiv
August 2022 | 0 Ответы
moment. 6....
pozhalujsta8b98a56c0152a07b8f4cbcd89aa2f01e 97513
sarvinozwakirjanova
August 2022 | 0 Ответы
pomogite pozhalusto pzha519d7eb8246a08ab0df06cc59e9dedb 6631
×
Report "СРОЧНО ПОМОГИТЕ!!!!!ПОЖАЛУЙСТА..."
Your name
Email
Reason
-Select Reason-
Pornographic
Defamatory
Illegal/Unlawful
Spam
Other Terms Of Service Violation
File a copyright complaint
Description
Helpful Links
О нас
Политика конфиденциальности
Правила и условия
Copyright
Контакты
Helpful Social
Get monthly updates
Submit
Copyright © 2024 SCHOLAR.TIPS - All rights reserved.
Answers & Comments
б) y'=(-x²-(3-x)·2x)/x⁴=(-x²-6x+2x²)/x⁴=(x²-6x)/x⁴=x(x-6)/x⁴=(x-6)/x³
в) y'=(2x·3x-(1+x²)·3)/9x²=(6x²-3-3x²)/9x²=(3x²-3)/9x²=3(x²-1)/9x²=(x²-1)/3x²
г)y'=(2x-1)x²-(x²-x+2)·2x)/x⁴=(2x³-x²-2x³+2x²-4x))/x⁴=(x²-4x)/x⁴=x(x-4)/x⁴=(x-4)/x³
д)/ y'=(-5x⁴(1+x⁵)-(1-x⁵)·5x⁴)/(1+x⁵)²=(-5x⁴-5x⁹-5x⁴+5x⁹)/(1+x⁵)²=-10x⁴/(1+x⁵)²
е). y'=(2x·x²-(x²-1)·2x)/x⁴=(2x³-2x³+2x²)/x⁴=2x²/x⁴=2/x²
ж). y'=(3x²·2x-(2+x³)·2)/4x²=(6x³-4-2x³)/4x²=(4x³-4)/4x²=4(x³-1)/4x²=(x³-1)/x²
з). y'=(-1/(1-x²)²)·(-2x)=2x/(1-x²)²