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Mathematics 19 Online
OpenStudy (anonymous):

use the chain rule to find dy/dx given y=2u/u^2-1 ,u=x^2

zepdrix (zepdrix):

\[\Large\rm y=\frac{2u}{u^2-1}\]So I guess uhh, start with the quotient rule.\[\Large\rm y'=\frac{(2u)'(u^2-1)-(2u)(u^2-1)'}{(u^2-1)^2}\]

OpenStudy (yanasidlinskiy):

\(\Huge\bf \color{yellow}{Welcome~to~OpenStudy!!}\hspace{-310pt}\color{cyan}{Welcome~to~OpenStudy!!}\hspace{-307.1pt}\color{midnightblue}{Welcome~to~\color{purple}{Open}}\color{blue}{Study!!!!}\)

zepdrix (zepdrix):

oo that's pretty =o

zepdrix (zepdrix):

\[\Large\rm y'=\frac{(2)u'(u^2-1)-(2u)(u^2-1)'}{(u^2-1)^2}\]For this first term, derivative of 2u gives us 2, chain rule tells us to multiply by the derivative of u.

zepdrix (zepdrix):

So if \(\Large\rm u=x^2\) Our derivative with respect to u is,\[\Large\rm u'=2x\]Yes?

zepdrix (zepdrix):

with respect to x*

OpenStudy (yanasidlinskiy):

Yep. It's really pretty:D

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