Bonjour je cherche quelqu’un d’aimable pouvant m’aider pour faire cet exercice de maths, merci d’ava.... Pergunta de ideia deverniereapo

verniereapo @verniereapo

October 2020 1 84 Report

Bonjour je cherche quelqu’un d’aimable pouvant m’aider pour faire cet exercice de maths, merci d’avance :)

Lista de comentários

aymanemaysae

Bonjour ;

$g^{'}(x)=(\dfrac{\sqrt x}{x+1})^{'}=\dfrac{(\sqrt x)^{'}(x+1)-\sqrt x(x+1)^{'}}{(x+1)^2}\\\\\\=\dfrac{\dfrac{1}{2\sqrt x}(x+1)-\sqrt x\times (1)}{(x+1)^2}=\dfrac{\dfrac{1}{2\sqrt x}(x+1)-\dfrac{2x}{2\sqrt x}}{(x+1)^2} \\\\\\=\dfrac{\dfrac{x+1-2x}{2\sqrt x}}{(x+1)^2} =\dfrac{1-x}{2\sqrt x(x+1)^2} =\dfrac{(1-x)\sqrt x}{2x(x+1)^2} \ .$

2 votes Thanks 1

verniereapo merci !

More Questions From This User See All

verniereapo April 2021 | 0 Respostas

verniereapo April 2021 | 0 Respostas

verniereapo December 2020 | 0 Respostas

verniereapo October 2020 | 0 Respostas

verniereapo October 2020 | 0 Respostas

verniereapo October 2020 | 0 Respostas

verniereapo October 2020 | 0 Respostas

verniereapo October 2020 | 0 Respostas

verniereapo October 2020 | 0 Respostas

verniereapo October 2020 | 0 Respostas

Bonjour ! Je chercher quelqu'un d'aimable qui pourrai m'aider pour mon DM de maths svp ? Merci beaucoup :)

Helpful Links

Política de Privacidade

Termos e Condições

direito autoral

Helpful Social

Copyright © 2024 ELIBRARY.TIPS - All rights reserved.