Ответ:
Производная дроби равна [tex]\bf \Big(\dfrac{u}{v}\Big)'=\dfrac{u'v-uv'}{v^2}[/tex] , [tex]\bf (uv)'=u'v+uv'[/tex] ,
[tex]\bf (\sqrt[9]{u})'=(u^{1/9})'=\dfrac{1}{9}\cdot u^{\frac{1}{9}-1}\cdot u'\ \ ,\ \ \ (log_{a}u)'=\dfrac{1}{u\cdot lna}\cdot u'[/tex]
[tex]y=\sqrt[9]{\dfrac{x+3}{x-3}}\cdot log_{a}(2x-3)\ \ \ ,\ \ \ a=const\\\\y'=\dfrac{1}{9}\cdot \Big(\dfrac{x+3}{x-3}\Big)^{-\frac{8}{9}}\cdot \dfrac{1\cdot (x-3)-(x+3)\cdot 1}{(x-3)^2}\cdot log_{a}(2x-3)+\\\\\\+\sqrt[9]{\dfrac{x+3}{x-3}}\cdot \dfrac{1}{(2x-3)\, lna}\cdot (2x-3)'=\\\\\\=\dfrac{1}{9}\cdot \sqrt[9]{\Big(\dfrac{x-3}{x+3}\Big)^8}\cdot \dfrac{-6}{(x-3)^2}\cdot log_{a}(2x-3)+\sqrt[9]{\dfrac{x+3}{x-3}}\cdot \dfrac{1}{(2x-3)\, lna}\cdot 2=[/tex]
[tex]\bf =-\dfrac{2}{3}\cdot \sqrt[9]{\Big(\dfrac{x-3}{x+3}\Big)^8}\cdot \dfrac{1}{(x-3)^2}\cdot log_{a}(2x-3)+\sqrt[9]{\dfrac{x+3}{x-3}}\cdot \dfrac{2}{(2x-3)\, lna}[/tex]
Copyright © 2024 SCHOLAR.TIPS - All rights reserved.
Answers & Comments
Ответ:
Производная дроби равна [tex]\bf \Big(\dfrac{u}{v}\Big)'=\dfrac{u'v-uv'}{v^2}[/tex] , [tex]\bf (uv)'=u'v+uv'[/tex] ,
[tex]\bf (\sqrt[9]{u})'=(u^{1/9})'=\dfrac{1}{9}\cdot u^{\frac{1}{9}-1}\cdot u'\ \ ,\ \ \ (log_{a}u)'=\dfrac{1}{u\cdot lna}\cdot u'[/tex]
[tex]y=\sqrt[9]{\dfrac{x+3}{x-3}}\cdot log_{a}(2x-3)\ \ \ ,\ \ \ a=const\\\\y'=\dfrac{1}{9}\cdot \Big(\dfrac{x+3}{x-3}\Big)^{-\frac{8}{9}}\cdot \dfrac{1\cdot (x-3)-(x+3)\cdot 1}{(x-3)^2}\cdot log_{a}(2x-3)+\\\\\\+\sqrt[9]{\dfrac{x+3}{x-3}}\cdot \dfrac{1}{(2x-3)\, lna}\cdot (2x-3)'=\\\\\\=\dfrac{1}{9}\cdot \sqrt[9]{\Big(\dfrac{x-3}{x+3}\Big)^8}\cdot \dfrac{-6}{(x-3)^2}\cdot log_{a}(2x-3)+\sqrt[9]{\dfrac{x+3}{x-3}}\cdot \dfrac{1}{(2x-3)\, lna}\cdot 2=[/tex]
[tex]\bf =-\dfrac{2}{3}\cdot \sqrt[9]{\Big(\dfrac{x-3}{x+3}\Big)^8}\cdot \dfrac{1}{(x-3)^2}\cdot log_{a}(2x-3)+\sqrt[9]{\dfrac{x+3}{x-3}}\cdot \dfrac{2}{(2x-3)\, lna}[/tex]